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Elements of Group 1 and 2 Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 8 Elements of Group 1 and 2 Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 8 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 8 Exercise Solutions

1. Explain the following

Question A.
Hydrogen shows similarity with alkali metals as well as halogens.
Answer:

• The electronic configuration of hydrogen is 1s¹ which is similar to the outer electronic configuration of alkali metals of group 1 i.e., ns¹.
• However, 1s¹ also resembles the outer electronic configuration of group 17 elements i.e., ns² np⁵.
• By adding one electron to H, it will attain the electronic configuration of the inert gas He which is 1s^2, and by adding one electron to ns² np⁵ we get ns² np⁶ which is the outer electronic configuration of the remaining inert gases.
• Therefore, some chemical properties of hydrogen are similar to those of alkali metals while some resemble halogens.

Hence, hydrogen shows similarity with alkali metals as well as halogens.

Question B.
Standard reduction potential of alkali metals have high negative values.
Answer:

• The general outer electronic configuration of alkali metals is ns¹.
• They readily lose one valence shell electron to achieve stable noble gas configuration and hence, they are highly electropositive and are good reducing agents.

Hence, standard reduction potentials of alkali metals have high negative values.

Question C.
Alkaline earth metals have low values of electronegativity; which decrease down the group.
Answer:

• Electronegativity represents attractive force exerted by the nucleus on shared electrons.
• The general outer electronic configuration of alkaline earth metals is ns². They readily lose their two valence shell electrons to achieve stable noble gas configuration. They are electropositive and hence, they have low values of electronegativity.

Question D.
Sodium dissolves in liquid ammonia to form a solution which shows electrical conductivity.
Answer:
i. Sodium dissolves in liquid ammonia giving deep blue coloured solutions which is electrically conducting in nature.
Na + (x + y) NH₃ → [Na(NH₃)_x]+ + [e(NH₃)_y]^–
ii. Due to formation of ions, the solution shows electrical conductivity.

Question E.
BeCl₂ is covalent while MgCl₂ is ionic.
Answer:

• Be²⁺ ion has very small ionic size and therefore, it has very high charge density.
• Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl^–) which is larger in size.
• This results in partial covalent character of the bond in BeCl₂.
• Mg²⁺ ion has very less tendency to distort the electron cloud of Cl^– due to the bigger size of Mg²⁺ as compared to Be²⁺.

Hence, BeCl₂ is covalent while MgCl₂ is ionic.

Question F.
Lithium floats an water while sodium floats and catches fire when put in water.
Answer:

• When lithium and sodium react with water, hydrogen gas is released. Due to these hydrogen gas bubbles, lithium and sodium floats on water.
  eg. 2Na + 2H₂O → 2Na⁺ + 2OH^– + H₂↑
• The reactivity of group 1 metals increases with increasing atomic radius and lowering of ionization enthalpy down the group.
• Thus, sodium having lower ionization enthalpy, is more reactive than lithium.
• Hence, lithium reacts slowly while sodium reacts vigorously with water.
• Since the reaction of sodium with water is highly exothermic, it catches fire when put in water.

2. Write balanced chemical equations for the following.

Question A.
CO₂ is passed into concentrated solution of NaCl, which is saturated with NH₃.
Answer:

Question B.
A 50% solution of sulphuric acid is subjected to electrolyte oxidation and the product is hydrolysed.
Answer:

Question C.
Magnesium is heated in air.
Answer:

Question D.
Beryllium oxide is treated separately with aqueous HCl and aqueous NaOH solutions.
Answer:
Beryllium oxide (BeO) is an amphoteric oxide and thus, it reacts with both acid (HCl) as well as base (NaOH) to give the corresponding products.
i. \(\mathrm{BeO}+\underset{(\text { Acid })}{2 \mathrm{HCl}} \longrightarrow \mathrm{BeCl}_{2}+\mathrm{H}_{2} \mathrm{O}\)
ii. \(\mathrm{BeO}+\underset{(\text { Base })}{2 \mathrm{NaOH}} \longrightarrow \mathrm{Na}_{2} \mathrm{BeO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

3. Answer the following questions

Question A.
Describe the diagonal relationship between Li and Mg with the help of two illustrative properties.
Answer:
a. The relative placement of these elements with similar properties in the periodic table is across a diagonal and is called diagonal relationship.
b. Lithium is placed in the group 1 and period 2 of the modem periodic table. It resembles with magnesium which is placed in the group 2 and period 3.

ii. Li and Mg show similarities in many of their properties.
e. g.
a. Reaction with oxygen:
1. Group 1 elements except lithium, react with oxygen present in the air to form oxides (M₂O) as well as peroxides (M₂O₂) and superoxides (MO₂) on further reaction with excess of oxygen.
2. This anomalous behaviour of lithium is due to its resemblance with magnesium as a result of diagonal relationship.
3. As group 2 elements form monoxides i.e., oxides, lithium also form monoxides.

b. Reaction with nitrogen:
1. All the group 1 elements react only with oxygen present in the air to form oxides while group 2 elements react with both nitrogen and oxygen present in the air forming corresponding oxides and nitrides.
2. However, lithium reacts with oxygen as well as nitrogen present in the air due to its resemblance with magnesium.

Question B.
Describe the industrial production of dihydrogen from steam. Also write the chemical reaction involved.
Answer:
Three stages are involved in the industrial production of dihydrogen from steam.
i. Stage 1:
a. Reaction of steam on hydrocarbon or coke (C) at 1270 K temperature in presence of nickel catalyst gives water-gas which is a mixture of carbon monoxide and hydrogen.
1. Reaction of steam with hydrocarbon:

2. Reaction of steam with coke or carbon (C):

b. Sawdust, scrapwood, etc. can also be used in place of carbon.

ii. Stage 2:
Water-gas shift reaction: When carbon monoxide in the water-gas reacts with steam in the presence of iron chromate (FeCrO₄) as catalyst, it gets transformed into carbon dioxide. This is called water-gas shift reaction.

iii. Stage 3: In the last stage, carbon dioxide is removed by scrubbing with sodium arsenite solution.

Question C.
A water sample, which did not give lather with soap, was found to contain Ca(HCO₃)₂ and Mg(HCO₃)₂. Which chemical will make this water give lather with soap? Explain with the help of chemical reactions.
Answer:

• Soap does not lather in hard water due to presence of the soluble salts of calcium and magnesium in it. So, the given water sample is hard water.
• Hardness of hard water can be removed by removal of these calcium and magnesium salts.
• Sodium carbonate is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates. Thus, it will make water give lather with soap.
  e.g. Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaHCO_3(aq)

Question D.
Name the isotopes of hydrogen. Write their atomic composition schematically and explain which of these is radioactive ?
Answer:
i. Hydrogen has three isotopes i.e., hydrogen \(\left({ }_{1}^{1} \mathrm{H}\right)\), deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) and tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) with mass numbers 1, 2 and 3 respectively.
ii. They all contain one proton and one electron but different number of neutrons in the nucleus.
iii. Atomic composition of isotopes of hydrogen:

iv. Tritium is a radioactive nuclide with half-life period 12.4 years and emits low energy β^– particles.
v. Schematic representation of isotopes of hydrogen is as follows:

4. Name the following

Question A.
Alkali metal with smallest atom.
Answer:
Lithium (Li)

Question B.
The most abundant element in the universe.
Answer:
Hydrogen (H)

Question C.
Radioactive alkali metal.
Answer:
Francium (Fr)

Question D.
Ions having high concentration in cell sap.
Answer:
Potassium ions (K⁺)

Question E.
A compound having hydrogen, aluminium and lithium as its constituent elements.
Answer:
Lithium aluminium hydride (LiAlH₄)

5. Choose the correct option.

Question A.
The unstable isotope of hydrogen is …..
a. H-1
b. H-2
c. H-3
d. H-4
Answer:
c. H-3

Question B.
Identify the odd one.
a. Rb
b. Ra
c. Sr
d. Be
Answer:
a. Rb

Question C.
Which of the following is Lewis acid ?
a. BaCl₂
b. KCl
c. BeCl₂
d. LiCl
Answer:
c. BeCl₂

Question D.
What happens when crystalline Na₂CO₃ is heated ?
a. releases CO₂
b. loses H₂O
c. decomposes into NaHCO₃
d. colour changes.
Answer:
b. loses H₂O

Activity :

1. Collect the information of preparation of dihydrogen and make a chart.
2. Find out the s block elements compounds importance/uses.
Answer:
1.

2. Uses of s-block elements:
Group 1 elements (alkali metals):
a. Lithium: Lithium is widely used in batteries.
b. Sodium:

• Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
• Sodium is also used as an important reagent in the Wurtz reaction.
• It is used in the manufacture of sodium vapour lamp.

c. Potassium:

• Potassium has a vital role in biological system.
• Potassium chloride (KCl) is used as a fertilizer.
• Potassium hydroxide (KOH) is used in the manufacture of soft soaps and also as an excellent absorbent of carbon dioxide.
• Potassium superoxide (KO₂) is used as a source of oxygen.

d. Caesium: Caesium is used in devising photoelectric cells.

Group 2 elements (alkaline earth metals):
a. Magnesium: Magnesium hydroxide [Mg(OH)₂] in its suspension form is used as an antacid.
b. Calcium: Compounds of calcium such as limestone and gypsum are used as constituents of cement and mortar.
c. Barium: BaSO₄ being insoluble in H₂O and opaque to X-rays is used as ‘barium meal’ to scan the X-ray of human digestive system.
[Note: Students are expected to collect additional information about preparation of dihydrogen and uses of s-block elements on their own.]

11th Chemistry Digest Chapter 8 Elements of Group 1 and 2 Intext Questions and Answers

Can you recall? (Textbook Page No. 110)

Question 1.
Which is the first element in the periodic table?
Answer:
Hydrogen is the first element in the periodic table.

Question 2.
What are isotopes?
Answer:
Many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element. Isotopes of an element have the same atomic number (number of protons) but different atomic mass numbers due to different number of neutrons in their nuclei.

Question 3.
Write the formulae of the compounds of hydrogen formed with sodium and chlorine.
Answer:
Hydrogen combines with sodium to form sodium hydride (NaH) while it reacts with chlorine to form hydrogen chloride (HCl).

Can you tell? (Textbook Page No. 110)

Question 1.
In which group should hydrogen be placed? In group 1 or group 17? Why?
Answer:

• Hydrogen contains one valence electron in its valence shell and thus, its valency is one. Therefore, hydrogen resembles alkali metals (group 1 elements) as they also contain one electron in their valence shell (alkali metals tend to lose their valence electron).
• However, hydrogen also shows similarity with halogens (group 17 elements) as their valency is also one because halogens tend to accept one electron in their valence shell.
• Due to this unique behaviour, it is difficult to assign any definite position to hydrogen in the modem periodic table.

Just think! (Textbook Page No. 112)

Question 1.
\(2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaH}_{(\mathrm{s})}\)
In the above chemical reaction which element does undergo oxidation and which does undergo reduction?
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + H_2(g) → 2Na⁺ + 2H^–
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and each hydrogen atom gains one electron to form H. This can be represented as follows:

iii. Na is oxidised to NaH by loss of electrons while the elemental hydrogen is reduced to NaH by a gain of electrons.

Can you recall? (Textbook Page No. 113)

Question i.
What is the name of the family of reactive metals having valency one?
Answer:
The family of reactive metals having valency one is known as alkali metals (group 1).

Question ii.
What is the name of the family of reactive metals having valency two?
Answer:
The family of reactive metals having valency two is known as alkaline earth metals (group 2).

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions
• States of Matter Class 11 Chemistry Textbook Solutions
• Adsorption and Colloids Class 11 Chemistry Textbook Solutions
• Chemical Equilibrium Class 11 Chemistry Textbook Solutions
• Nuclear Chemistry and Radioactivity Class 11 Chemistry Textbook Solutions
• Basic Principles of Organic Chemistry Class 11 Chemistry Textbook Solutions
• Hydrocarbons Class 11 Chemistry Textbook Solutions
• Chemistry in Everyday Life Class 11 Chemistry Textbook Solutions

Modern Periodic Table Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 7 Modern Periodic Table Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 7 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 7 Exercise Solutions

1. Explain the following

Question A.
The elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.
Answer:

• Li, B, Be and N belong to the same period.
• As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily and therefore, electronegativity increases.

Hence, the elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.

Question B.
The atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.
Answer:

• Cl, I and Br belong to group 17 (halogen group) in the periodic table.
• As we move down the group from top to bottom in the periodic table, a new shell gets added in the atom of the elements.
• As a result, the effective nuclear charge decreases due to increase in the atomic size as well as increased shielding effect.
• Therefore, the valence electrons experience less attractive force from the nucleus and are held less tightly resulting in the increased atomic radius.
• Thus, their atomic radii increases in the following order down the group.
  Cl (99 pm) < Br (114 pm) < I (133 pm)

Hence, the atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.

Question C.
The ionic radii of F^– and Na⁺ are 133 and 98 pm, respectively.
Answer:

• F^– and Na⁺ are isoelectronic ions as they both have 10 electrons.
• However, the nuclear charge on F^– is +9 while that of Na⁺ is +11.
• In isoelectronic species, larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.

Thus, F^– has larger ionic radii (133 pm) than Na⁺ (98 pm).

Question D.
₁₃Al is a metal, ₁₄Si is a metalloid and ₁₅P is a nonmetal.
Answer:

• Electronic configuration of Al is [Ne] 3s² 3p¹, ₁₄Si is [Ne] 3s² 3p² and that of ₁₅P is [Ne] 3s² 3p³.
• Metals are characterized by the ability to form compounds by loss of valence electrons.
• ‘Al’ has 3 valence electrons, thus shows tendency to lose 3 valence electrons to complete its octet. Hence, Al is a metal.
• Nonmetals are characterized by the ability to form compounds by gain of valence electrons in valence shell.
• ‘P’ has 5 valence electrons thus, shows tendency to gain 3 electrons to complete its octet. Hence, ‘P’ is a nonmetal.
• Si has four valence electrons, thus it can either lose/gain electrons to complete its octet. Hence, behaves as a metalloid.

Question E.
Cu forms coloured salts while Zn forms colourless salts.
Answer:

• Electronic configuration of ₂₉CU is [Ar] 3d¹⁰4s¹ while that of Zn is [Ar] 3d¹⁰4s².
• Electronic configuration of Cu in its +1 oxidation state is [Ar] 3d¹⁰ while that in +2 oxidation state is [Ar] 3d⁹.
• Therefore, Cu contains partially filled d orbitals in +2 oxidation state and thus, Cu²⁺ salts are coloured.
• However, Zn has completely filled d orbital which is highly stable and hence, it does not form coloured ions.

Hence, Cu forms coloured salts while Zn forms colourless salts.

2. Write the outer electronic configuration of the following using orbital notation method. Justify.
A. Ge (belongs to period 4 and group 14)
B. Po (belongs to period 6 and group 16)
C. Cu (belongs to period 4 and group 11)
Answer:
A. a. Ge belongs to period 4. Therefore, n = 4.
b. Group 14 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 14 elements is ns² np².
d. Thus, the outer electronic configuration of Ge is 4s² 4p².

B. a. Po belongs to period 6. Therefore, n = 6.
b. Group 16 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 16 elements is ns² np⁴.
d. Thus, the outer electronic configuration of Po is 6s² 6p⁴.

C. a. Cu belongs to period 4. Therefore, n = 4.
b. Group 11 indicates that the element belongs to the d-block of the modem periodic table.
c. The general outer electronic configuration of the d-block elements is ns^0-2(n-1)d^1-10.
d. The expected configuration of Cu is 4s²3d⁹. However, the observed configuration of Cu is 4s¹3d¹⁰. This is due to the extra stability associated with completely filled d-subshell. Thus, the outer electronic configuration of Cu is 4s¹3d¹⁰.

3. Answer the following

Question A.
La belongs to group 3 while Hg belongs to group 12 and both belong to period 6 of the periodic table. Write down the general outer electronic configuration of the ten elements from La to Hg together using orbital notation method.
Answer:
i. La and Hg both belongs to period 6. Therefore, n = 6.
ii. Elements of group 3 to group 12 belong to the d-block of the modem periodic table.
iii. The general outer electronic configuration of the d-block elements is ns^0-2 (n -1 )^1-10.
iv. Therefore, the outer electronic configuration of all ten elements from La to Hg is as given in the table below.

[Note: There are 14 elements between La and Hf which are called lanthanides. Therefore, after La, electrons are filled in 4f subshell of lanthanide elements. Once all the 14 elements of lanthanide series are filled, next electron enters 5d subshell of Hf. Hence, the outer electronic configurations of Hf to Hg often include completely filled 4f subshell. For example, the electronic configuration of Hf ‘5d²6s²’ can also be written as ‘4f¹⁴5d²6s²’.]

Question B.
Ionization enthalpy of Li is 520 kJ mol^-1 while that of F is 1681 kJ mol^-1. Explain.
Answer:

• Both Li and F belong to period 2.
• Across a period, the screening effect is the same while the effective nuclear charge increases.
• As a result, the outer electron is held more tightly and therefore, the ionization enthalpy increases across a period.
• Hence, F will have higher ionization enthalpy than Li.

Thus, ionization enthalpy of Li is 520 kJ mol^-1 while that of F is 1681 kJ mol^-1.

Question C.
Explain the screening effect with a suitable example.
Answer:
i. In a multi-electron atom, the electrons in the inner shells tend to prevent the attractive influence of the nucleus from reaching the outermost electron.
ii. Thus, they act as a screen or shield between the nuclear attraction and outermost or valence electrons. This effect of the inner electrons on the outer electrons is known as screening effect or shielding effect.
iii. Across a period, screening effect due to inner electrons remains the same as electrons are added to the same shell.
iv. Down the group, screening effect due to inner electrons increases as a new valence shell is added.
e.g. Potassium (19K) has electronic configuration 1s²2s²2p⁶3s²3p⁶4s¹.
K has 4 shells and thus, the valence shell electrons are effectively shielded by the electrons present in the inner three shells. As a result of this, valence shell electron (4s¹) in K experiences much less effective nuclear charge and can be easily removed.

Question D.
Why the second ionization enthalpy is greater than the first ionization enthalpy ?
Answer:
The second ionization enthalpy (Δ_iH₂) is greater than the first ionization enthalpy (Δ_iH₁) as it involves removal of electron from the positively charged species.

Question E.
Why the elements belonging to the same group do have similar chemical properties ?
Answer:

• Chemical properties of elements depend upon their valency.
• Elements belonging to the same group have the same valency.

Hence, the elements belonging to the same group show similar chemical properties.

Question F.
Explain : electronegativity and electron gain enthalpy. Which of the two can be measured experimentally?
Answer:
i. The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN). Electronegativity cannot be measured experimentally. However, various numerical scales to express electronegativity were developed by many scientists. Pauling scale of electronegativity is the one used most widely.

ii. Electron gain enthalpy is a quantitative measure of the ease with which an atom adds an electron forming the anion and is expressed in kJ mol^-1. Thus, it is an experimentally measurable quantity.

4. Choose the correct option

Question A.
Consider the elements B, Al, Mg and K predict the correct order of metallic character :
a. B > Al > Mg > K
b. Al > Mg > B > K
c. Mg > Al > K > B
d. K > Mg > Al > B
Answer:
d. K > Mg > Al > B

Question B.
In modern periodic table, the period number indicates the :
a. atomic number
b. atomic mass
c. principal quantum number
d. azimuthal quantum number
Answer:
c. principal quantum number

Question C.
The lanthanides are placed in the periodic table at
a. left hand side
b. right hand side
c. middle
d. bottom
Answer:
d. bottom

Question D.
If the valence shell electronic configuration is ns²np⁵, the element will belong to
a. alkali metals
b. halogens
c. alkaline earth metals
d. actinides
Answer:
b. halogens

Question E.
In which group of elements of the modern periodic table are halogen placed ?
a. 17
b. 6
c. 4
d. 2
Answer:
a. 17

Question F.
Which of the atomic number represent the s-block elements ?
a. 7, 15
b. 3, 12
c. 6, 14
d. 9, 17
Answer:
b. 3, 12

Question G.
Which of the following pairs is NOT isoelectronic ?
a. Na⁺ and Na
b. Mg²⁺ and Ne
c. Al³⁺ and B³⁺
d. P3^– and N^3-
Answer:
b. Mg²⁺ and Ne

Question H.
Which of the following pair of elements has similar properties ?
a. 13, 31
b. 11, 20
c. 12, 10
d. 21, 33
Answer:
a. 13, 31

5. Answer the following questions

Question A.
The electronic configuration of some elements are given below:
a. 1s²
b. 1s²2s²2p⁶
In which group and period of the periodic table they are placed ?
Answer:
a. 1s²
Here n = 1. Therefore, the element belongs to the 1st period.
The outer electronic configuration 1s² corresponds to the maximum capacity of 1s, the complete duplet. Therefore, the element is placed at the end of the 1st period in the group 18 of inert gases in the modem periodic table,

b. 1s²2s²2p⁶
Here n = 2. Therefore, the element belongs to the 2nd period.
The outer electronic configuration 2s²2p⁶ corresponds to complete octet. Therefore, the element is placed in the 2nd period of group 18 in the modem periodic table.

Question B.
For each of the following pairs, indicate which of the two species is of large size :
a. Fe²⁺ or Fe³⁺
b. Mg²⁺ or Ca²⁺
Answer:
a. Fe²⁺ has a larger size than Fe³⁺.
b. Ca²⁺ has a larger size than Mg²⁺.

Question C.
Select the smaller ion form each of the following pairs:
a. K⁺, Li⁺
b. N^3-, F^–
Answer:
i. Li⁺ has smaller ionic radius than K⁺
ii. F^– has smaller ionic radius than N^3-.

Question D.
With the help of diagram answer the questions given below:

a. Which atom should have smaller ionization enthalpy, oxygen or sulfur?
b. The lithium forms +1 ions while berylium forms +2 ions ?
Answer:
Sulfur should have smaller ionization energy than oxygen.
a. Lithium has electronic configuration 1s²2s¹ while that of beryllium is 1s²2s².
b. Li can achieve a noble gas configuration by losing one electron while Be can do so by losing two electrons. Hence, lithium forms +1 ions while beryllium forms +2 ions.

Question E.
Define : a. Ionic radius
b. Electronegativity
Answer:
a. Ionic radius: Ionic radius is defined as the distance of valence shell of electrons from the centre of the nucleus in an ion.

b. Electronegativity: The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN).

Question F.
Compare chemical properties of metals and non-metals.
Answer:
i. Metals (like alkali metals) react vigorously with oxygen to form oxides which reacts with water to form strong bases.
e. g. Sodium (Na) reacts with oxygen to form Na₂O which produces NaOH on reaction with water.

ii. Nonmetals (like halogens) react with oxygen to form oxides which on reaction with water form strong acids.
e.g. Chlorine reacts with oxygen to form Cl₂O₇ which produces HClO₄ on reaction with water.

Question G.
What are the valence electrons ? For s-block and p-block elements show that number of valence electrons is equal to its group number.
Answer:

• Electrons present in the outermost shell of the atom of an element are called valence electrons.
• 3Li is an s-block element and its electronic configuration is 1s²2s¹. Since it has one valence electron, it is placed in group 1.
• Therefore, for s-block elements, group number = number of valence electrons.
• However, for p-block elements, group number = 18 – number of electrons required to attain complete octet.
• 7N is a p-block element and its electronic configuration is 1s²2s²2p³. Since it has five electrons in its valence shell, it is short of three electrons to complete its octet.
• Therefore, its group number = 18 – 3 = 15.

Question H.
Define ionization enthalpy. Name the factors on which ionisation enthalpy depends? How does it vary down the group and across a period?
Answer:
i. The energy required to remove an electron from the isolated gaseous atom in its ground state is called ionization enthalpy (Δ_iH).
Ionization enthalpy is the quantitative measure of tendency of an element to lose electron and expressed in kJ mol^-1.

ii. Ionization energy depends on the following factors

• Size (radius) of an atom
• Nuclear charge
• The shielding or screening effect of inner electrons
• Nature of electronic configuration

iii. Variation of ionization energy down the group: On moving down the group, the ionization enthalpy decreases. This is because electron is to be removed from the larger valence shell. Screening due to core electrons goes on increasing and the effective nuclear charge decreases down the group. As a result, the removal of the outer electron becomes easier down the group.

iv. Variation of ionization energy across a period: The screening effect is the same while the effective nuclear charge increases across a period. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period. Therefore, the alkali metal shows the lowest first ionization enthalpy while the inert gas shows the highest first ionization enthalpy across a period.

Note: First ionization enthalpy values of elements of group 1.

Note: First ionization enthalpy values of elements of period 2.

Question I.
How the atomic size vary in a group and across a period? Explain with suitable example.
Answer:
i. Variation in atomic size down the group:
a. As we move down the group from top to bottom in the periodic table, the atomic size increases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases but simultaneously the number of shells in the atoms also increases.
c. Asa result, the effective nuclear charge decreases due to increase in the size of the atom and shielding effect increases down the group. Thus, the valence electrons experience less attractive force from nucleus and are held less tightly.
d. Hence, the atomic size increases in a group from top to bottom.

e. g.

• In group 1, as we move from top to bottom i.e., from Li to Cs, a new shell gets added in the atom of the elements and the electrons are added in this new shell.
• As a result of this, the effective nuclear charge goes on decreasing and screening effect goes on increasing down a group.
• Therefore, the atomic size is the largest for Cs and is the smallest for Li in group 1.

[Note: Atomic radii of Li and Cs are 152 pm and 262 pm respectively.]

ii. Variation in atomic size across a period:
a. As we move across a period from left to right in the periodic table, the atomic size of an element decreases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.
c. Therefore, as we move across a period, the effective nuclear charge increases but screening effect caused by the core electrons remains the same.
d. As a result of this, attraction between the nucleus and the valence electrons increases. Therefore, valence electrons are more tightly bound and hence, the atomic radius goes on decreasing along a period resulting in decrease in atomic size.

e. g.

• In the second period, as we move from left towards right i.e., from Li to F, the electrons are added in the second shell of all the elements in second period (except noble gas Ne).
• As a result of this, the effective nuclear charge goes on increasing from Li to F, however, screening effect remains the same.
• Therefore, the atomic size is the largest for Li (alkali metal) and is the smallest for F (halogen).

[Note: Atomic radii of Li and F are 152 pm and 64 pm respectively.]

Question J.
Give reasons.
a. Alkali metals have low ionization energies.
b. Inert gases have exceptionally high ionization energies.
c. Fluorine has less electron affinity than chlorine.
d. Noble gases possess relatively large atomic size.
Answer:
a. i. Across a period, the screening effect is the same while the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Since the alkali metals are present in the group 1 of the modem periodic table, they have low ionization energies.

b. i. Across a period, the screening effect is the same and the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Inert gases are present on the extreme right of the periodic table i.e., in group 18. Also, inert gases have stable electronic configurations i.e., complete octet or duplet. Due to this, they are extremely stable and it is very difficult to remove electrons from their valence shell.
Hence, inert gases have exceptionally high ionization potential.

c. The less electron affinity of fluorine is due to its smaller size. Adding an electron to the 2p orbital in fluorine leads to a greater repulsion than adding an electron to the larger 3p orbital of chlorine.
Hence, fluorine has less electron affinity than chlorine.

d. i. Noble gases have completely filled valence shell i.e., complete octet (except He with complete duplet).
ii. Since their valence shell contains eight electrons, they experience greater electronic repulsion and this results in increased atomic size (atomic radii) of the noble gas elements.
Hence, noble gases possess

Question K.
Consider the oxides Li₂O, CO₂, B₂O₃.
a. Which oxide would you expect to be the most basic?
b. Which oxide would be the most acidic?
c. Give the formula of an amphoteric oxide.
Answer:
a. Li₂O is the most basic oxide.
b. CO₂ is the most acidic oxide.
c. Formula of an amphoteric oxide: Al₂O₃.
[Note: Both B₂O₃ and CO₂ are acidic oxides. But CO₂ is more acidic oxide as compared to B₂O₃. Hence, CO₂ is most acidic oxide amongst the given.]

Activity :

Question 1.
Prepare a wall mounting chart of the modern periodic table.
Answer:
Students can scan the adjacent Q.R. Code to visualise the modern periodic table and are expected to prepare the chart on their own.

11th Chemistry Digest Chapter 7 Modern Periodic Table Intext Questions and Answers

Can you recall? (Textbook Page No. 93)

Question 1.
What was the basis of classification of elements before the knowledge of electronic structure of atom?
Answer:
Elements were classified on the basis of their physical properties before the knowledge of electronic structure of atom.

Question 2.
Name the scientists who made the classification of elements in the nineteenth century.
Answer:
Dmitri Mendeleev, John Newlands and Johann Doberiener were the scientists who made the classification of elements based on their atomic mass in the nineteenth century.

Question 3.
What is Mendeleev’s periodic law?
Answer:
Mendeleev’s periodic law: “The physical and chemical properties of elements are the periodic function of their atomic masses

Question 4.
How many elements are discovered until now?
Answer:
Including manmade elements, total 118 elements are discovered until now.

Question 5.
How many horizontal rows and vertical columns are present in the modern periodic table?
Answer:
The modem periodic table consists of seven horizontal rows called periods numbered from 1 to 7 and eighteen vertical columns called groups numbered from 1 to 18.

Just think. (Textbook Page No. 93)

Question 1.
How many days pass between two successive full moon nights?
Answer:
29.5 days i.e., approximately 30 days pass between two successive full moon nights.

Question 2.
What type of motion does a pendulum exhibit?
Answer:
A pendulum exhibits periodic motion since it traces the same path after regular interval of time.

Question 3.
Give some other examples of periodic events.
Answer:
Following are some other examples of periodic events:

• Motion of earth around the sun.
• Rotation of earth around its own axis.
• Day and night.

Can you recall? (Textbook Page No. 95)

Question i.
What does the principal quantum number ‘n’ and azimuthal quantum number ‘l’ of an electron belonging to an atom represent?
Answer:
The principal quantum number ‘n’ represents the outermost or valence shell of an element (which corresponds to period number) while azimuthal quantum number ‘l’ constitutes a subshell belonging to the shell for the given ‘n’.

Question ii.
Which principle is followed in the distribution of electrons in an atom?
Answer:
The distribution of electrons in an atom is according to the following three principles:

1. Aufbau principle
2. Pauli’s exclusion principle
3. Hund’s rule of maximum multiplicity

[Note: According to aufbau principle, electrons are filled in the subshells in the increasing order of their energies which follows the following order: s < p < d < f.]

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Redox Reactions Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 6 Redox Reactions Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 6 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 6 Exercise Solutions

1. Choose the most correct option

Question A.
Oxidation numbers of Cl atoms marked as Cl^a and Cl^b in CaOCl₂ (bleaching powder) are

a. zero in each
b. -1 in Cl^a and +1 in Cl^b
c. +1 in Cl^a and -1 in Cl^b
d. 1 in each
Answer:
b. -1 in Cl^a and +1 in Cl^b

Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H₂ → Cu + H₂O
b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂
c. 2K + F₂ → 2KF
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Answer:
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A₂(BC₃)₂
b. A₃(BC₄)₂
c. A₃(B₄C)₂
d. ABC₂
Answer:
b. A₃(BC₄)₂

Question D.
The coefficients p, q, r, s in the reaction
\(\mathrm{pCr}_{2} \mathrm{O}_{7}^{2-}\) + q Fe^2⊕ → r Cr^3⊕ + s Fe^3⊕ + H₂O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6

Question E.
For the following redox reactions, find the correct statement.
Sn^2⊕ + 2Fe^3⊕ → Sn^4⊕ + 2Fe^2⊕
a. Sn^2⊕ is undergoing oxidation
b. Fe^3⊕ is undergoing oxidation
c. It is not a redox reaction
d. Both Sn^2⊕ and Fe^3⊕ are oxidised
Answer:
a. Sn^2⊕ is undergoing oxidation

Question F.
Oxidation number of carbon in H₂CO₃ is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4

Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O₂
b. Mn(II)O₂
c. Mn(III)O₂
d. Mn(IV)O₂
Answer:
d. Mn(IV)O₂

Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –\(\frac {1}{2}\)
d. 0
Answer:
c. –\(\frac {1}{2}\)

Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F

Question K.
The process SO₂ → S₂Cl₂ is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction

2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl₂
ii. Tl₂SO₄
iii. SnO₂
iv. Cr₂O₃

3. Answer the following questions

Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
In CH₄, the oxidation state of carbon is -4 while in CO₂, the oxidation state of carbon is +4.

Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:

C. Calculate the oxidation number of underlined atoms.
a. H₂SO₄
b. HNO₃
c. H₃PO₃
d. K₂C₂O₄
e. H₂S₄O₆
f. Cr₂O₇^2-
g. NaH₂PO₄
Answer:
i. H₂SO₄
Oxidation number of H = +1
Oxidation number of O = -2
H₂SO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H₂SO₄ = +6

ii. HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
HNO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO₃ = +5

iii. H₃PO₃
Oxidation number of O = -2
Oxidation number of H = +1
H₃PO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H₃PO₃ = +3

iv. K₂C₂O₄
Oxidation number of K = +1
Oxidation number of O = -2
K₂C₂O₄ is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +\(\frac {6}{2}\)
∴ Oxidation number of C in K₂C₂O₄ = +3

v. H₂S₄O₆
Oxidation number of H = +1
Oxidation number of O = -2
H₂S₄O₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +\(\frac {10}{4}\)
∴ Oxidation number of S in H₂S₄O₆ = +2.5

vi. Cr₂O₇^2-
Oxidation of O = -2
Cr₂O₇^2- is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +\(\frac {12}{2}\)
∴ Oxidation number of Cr in Cr₂O₇^2- = +6

vii. NaH₂PO₄
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH₂PO₄ is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH₂PO₄ = +5

Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
b. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
c. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
Answer:
i. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agents (Reduced species): Cu₂O/ Cu₂S
3. Reductant/reducing agent (Oxidised species): Cu₂S

[Note: Cu in both Cu₂O and Cu₂S undergoes reduction. Hence, both Cu₂O and Cu₂S can be termed as oxidising agents in the given reaction.]

ii. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.

iii. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent (Reduced species): I₂
3. Reductant/reducing agent (Oxidised species): S₂O₃^2-

Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–
Answer:
a. Mg / Mg²⁺
Here, Mg loses two electrons to form Mg²⁺ ion.
\(\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Mg / Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺
Here, Cu loses two electrons to form Cu²⁺ ion.
\(\mathrm{Cu}_{(\mathrm{s})} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O^2-
Here, each O gains two electrons to form O^2- ion.
\(\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}_{(\mathrm{aq})}^{2-}\)
Hence, O₂ / O^2- is in a reduced state.

d. Cl₂ / Cl^–
Here, each Cl gains one electron to form Cl^– ion.
\(\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence, Cl₂ / Cl^– is in a reduced state.

Question F.
Justify the following reaction as redox reaction.
2 Na_2(s) + S_(s) → Na₂S_(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + S_(s) → 2Na⁺ + S^2-
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S^2-. This can be represented as follows:

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electrons to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S^2- in Na₂S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

Question G.
Provide the stock notation for the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, MnO and CuO.
Answer:

Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)₄^–
b. Na₂S₂O₃
c. H₃BO₃
Answer:
i. Cr(OH)₄^–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)₄^– is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)₄^– = +3

ii. Na₂S₂O₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂S₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +\(\frac {4}{2}\)
∴ Oxidation number of S in Na₂S₂O₃ = +2

iii. H₃BO₃
Oxidation number of H = +1
Oxidation number of O = -2
H₃BO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H₃BO₃ = +3

Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)
b. \(\mathrm{MnO}_{4}^{\Theta}\) (E0 = 1.51 V) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\) (E⁰ = 1.33 V)
Answer:
a. Cl₂ has a larger positive value of E⁰ than Br₂. Thus, Cl₂ is a stronger oxidizing agent than Br₂.
b. \(\mathrm{MnO}_{4}^{\Theta}\) has larger positive value of E⁰ than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\). Thus, \(\mathrm{MnO}_{4}^{\Theta}\) is stronger oxidizing agent than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\)

Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)
b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)
Answer:
a. Li has a larger negative value of E⁰ than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E⁰ than Fe. Thus, Zn is a stronger reducing agent than Fe.

4. Balance the reactions/equations :

Question A.
Balance the following reactions by oxidation number method

Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \quad(\text { acidic })\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(a q)}^{2-}+\mathrm{SO}_{3(a)}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{(a q)} \quad \text { (basic) }\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{ }_{(\mathrm{aq})}\)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Mn.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}\)
Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
\(2 \mathrm{MnO}_{4(a q)}^{-}+\mathrm{Br}_{(2 q)}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3 \text { (aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

iii. H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.


To make the net increase and decrease equal, we must take 2 atoms of S.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l) + H₂O_(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H₂SO_4(aq) + C_(s) → CO₂ + 2SO_2(g) + H₂O_(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)

iv. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.

Question B.
Balance the following redox equation by half reaction method

Answer:
i. H₂C₂O_4(aq) + \(\mathrm{MnO}_{4(a q)}^{-}\) → CO_2(g) + \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

ii. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{SnO}_{2(\mathrm{aq})}^{2-} \longrightarrow \mathrm{SnO}_{3(\mathrm{aq})}^{2-}+\mathrm{Bi}_{(\mathrm{s})}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1H₂O to left side of oxidation half equation and 3H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 3H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH^– ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:

5. Complete the following table :

Assign oxidation number to the underlined species and write Stock notation of compound

Answer:

11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers

Can you tell? (Textbook Page No. 81)

Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.

Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.

Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.

Can you recall? (Textbook Page No. 81)

Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.

Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.

Try this. (Textbook Page No. 82)

Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

Answer:

Try this (Textbook Page No. 88)

Question 1.
Classify the following unbalanced half equations as oxidation and reduction.

Answer:

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Chemical Bonding Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 5 Exercise Solutions

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO₃
b. CO₂
c. H₂S
d. Cl₂O
Answer:
b. CO₂

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. energy levels in an atom
b. the shapes of molecules and ions.
c. the electron negetivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO₂?

Answer:

Question E.
Which O₂ molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH₄
b. C₂H₂
c. NH₃
d. H₂O
Answer:
d. H₂O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H₂)
b. Water (H₂O)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H₂ and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp² Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C₂H₆
c. C₂H₄
d. CF₃Cl
e. SO₂
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
b. Boron: The electronic configuration of boron is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\). The valency is expected to be 1 but it is 3 as in BF₃.
c. Carbon: The electronic configuration of carbon is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) . The valency is expected to be 2, but observed valency is 4 as in CH₄.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH₃ (107°18′) and H₂O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH₄) molecule on the basis of sp³ hybridization:
i. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\);
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH₄ molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

• One lone pair and three bond pairs are present in ammonia molecule.
• The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
• Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

• Two lone pairs and two bond pairs are present in water molecule.
• The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
• Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s² 2s² 2p_x¹ 2p_y¹
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH₃) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H₂
b. F₂
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\). During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF₂ molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H₂O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF₃ molecule is planar but NH₃ pyramidal. Explain.
Answer:
i. In the BF₃ molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH₃ molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH₃ molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH₃ is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF₄
b. NF₃
c. HCN
d. H₂S
Answer:
a. CF₄: There are four bond pairs on the central atom. Hence, shape of CF₄ is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF₃: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF₃ is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H₂S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C₂H₂).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF₂, AlCl₃, LiCl, CaCl₂, NaCl.
Answer:
AlCl₃ > BeF₂ > CaCl₂ > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF₆ compound.
Answer:
The total number of electrons around sulphur (S) in SF₆ is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

• formation of the excited state and
• mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
where, N_b is the number of electrons present in bonding MOs and N_a is the number of electrons present in antibonding MOs.

Question I.
Why is O₂ molecule paramagnetic?
Answer:
The electronic configuration of O₂ molecule is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

1. by loss and gain of electrons
2. by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl₂, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K⁺ + e^–
Cl + e^– → Cl^–
K⁺ + Cl^– → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in \(\mathrm{NH}_{4}^{+}\) will have formal charge +1?
Answer:
In \(\mathrm{NH}_{4}^{+}\), nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF₇?
Answer:
Lewis structure of IF₇ is:

In IF₇, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H₂ stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s¹. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H₂ molecule attains stability due to duplet formation. Hence, H₂ is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

• Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
• Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases, and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br₂, N₂, I₂, NH₃
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br₂: Nonpolar
v. N₂: Nonpolar
vi. I₂: Nonpolar
vii. NH₃: Polar

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Structure of Atom Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 4 Structure of Atom Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 4 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 4 Exercise Solutions

1. Choose the correct option.

Question A.
The energy difference between the shells goes on ……….. when moved away from the nucleus.
a. Increasing
b. decreasing
c. equalizing
d. static
Answer:
b. decreasing

Question B.
The value of Plank’s constant is
a. 6.626× 10^-34 Js
b. 6.023× 10^-24 Js
c. 1.667 × 10^-28 Js
d. 6.626× 10^-28 Js
Answer:
a. 6.626× 10^-34 Js

Question C.
p-orbitals are ……. in shape.
a. spherical
b. dumbbell
c. double dumbbell
d. diagonal
Answer:
b. dumbbell

Question D.
“No two electrons in the same atoms can have an identical set of four quantum numbers”. This statement is known as
a. Pauli’s exclusion principle
b. Hund’s rule
c. Aufbau rule
d. Heisenberg uncertainty principle
Answer:
a. Pauli’s exclusion principle

Question E.
Principal Quantum number describes
a. shape of orbital
b. size of the orbital
c. spin of electron
d. orientation of in the orbital electron cloud
Answer:
b. size of the orbital

2. Make the pairs:

Answer:
a – iv,
b – i,
c – ii,
d – iii

3. Complete the following information about the isotopes in the chart given below :

(Hint: Refer to Periodic Table if required)
Answer:

4. Match the following :

Answer:
a – iv,
b – iii,
c – ii,
d – i

5. Answer in one sentence :

Question A.
If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.
Answer:
The representation of the given element is \({ }_{5}^{11} \mathrm{X}\).

Question B.
Name the element that shows simplest emission spectrum.
Answer:
The element that shows simplest emission spectrum is hydrogen.

Question C.
State Heisenberg uncertainty principle.
Answer:
Heisenberg uncertainty principle states that “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron”.

Question D.
Give the names of quantum numbers.
Answer:
The four quantum numbers are: principal quantum number (n), azimuthal or subsidiary quantum number (l), magnetic quantum number (m_l) and electron spin quantum number (m_s).

Question E.
Identify from the following the isoelectronic species:
Ne, O^2-, Na⁺ OR Ar, Cl^2-, K⁺
Answer:
Atoms and ions having the same number of electrons are isoelectronic.

Hence, Ne, O^2-, Na⁺ are isoelectronic species.

6. Answer the following questions.

Question A.
Differentiate between Isotopes and Isobars.
Answer:

Question B.
Define the terms:
i. Isotones
ii. Isoelectronic species
iii. Electronic configuration
Answer:
i. Isotones: Isotones are defined as the atoms of different elements having same number of neutrons in their nuclei. e.g. \({ }_{5}^{11} \mathrm{B}\) and \({ }_{6}^{12} \mathrm{C}\) having 6 neutrons each are isotones.

ii. Isoelectronic species:
soelectronic species are defined as atoms and ions having the same number of electrons.
e. g. Ar, Ca²⁺ and K⁺ containing 18 electrons each.

iii. Electronic configuration:
Electronic configuration of an atom is defined as the distribution of its electrons in orbitals.

Question C.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle:
i. Statement: “No two electrons in an atom can have the same set of four quantum numbers”. OR “Only two electrons can occupy the same orbital and they must have opposite spins. ”
ii. The capacity of an orbital to accommodate electrons is decided by Pauli’s exclusion principle.
iii. According to this principle, for an electron belonging to the same orbital, the spin quantum number must be different since the other three quantum numbers are the same.
iv. The spin quantum number can have two values: +\(\frac {1}{2}\) and –\(\frac {1}{2}\).
v. Example, consider helium (He) atom with electronic configuration 1 s².
For the two electrons in Is orbital, the four quantum numbers are as follows:
Electron number Quantum number Set of values of quantum numbers

Thus, in an atom, any two electrons can have the same three quantum numbers, but the fourth quantum number must be different.
vi. This leads to the conclusion that an orbital can accommodate maximum of two electrons and if it has two electrons, they must have opposite spin.

Question D.
State Hund’s rule of maximum multiplicity with suitable example.
Answer:
Hund’s rule of maximum multiplicity:
i. Statement: “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
ii. Example, according to Hund’s rule, each of the three-degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, the configuration of four electrons occupying p-orbitals is represented as

iii. As a result of Hund’s rule, the atom with fully filled and half-filled set of degenerate orbitals has extra stability.

Question E.
Write the drawbacks of Rutherford’s model of an atom.
Answer:
Drawbacks of Rutherford’s model of an atom:
i. Rutherford’s model of an atom resembles the solar system with the nucleus playing the role of the massive sun and the electrons are lighter planets. Thus, according to this model, electrons having negative charge revolve in various orbits around the nucleus. However, the electrons revolving about the nucleus in fixed orbits pose a problem. Such orbital motion is an accelerated motion accompanied by a continuous change in the velocity of electron as noticed from the continuously changing direction. According to Maxwell’s theory of electromagnetic radiation, accelerated charged particles would emit electromagnetic radiation. Hence, an electron revolving around the nucleus should continuously emit radiation and lose equivalent energy. As a result, the orbit would shrink continuously and the electron would come closer to the nucleus by following a spiral path. It would ultimately fall into the nucleus. Thus, Rutherford’s model has an intrinsic instability of atom. However, real atoms are stable.

ii. Rutherford’s model of an atom does not describe the distribution of electrons around the nucleus and their energies.

Question F.
Write postulates of Bohr’s Theory of hydrogen atom.
Answer:
Postulates of Bohr’s theory of hydrogen atom:
i. The electron in the hydrogen atom can move around the nucleus in one of the many possible circular paths of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus in an increasing order of energy.

ii. The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state if and when the required amount of energy is absorbed by the electron. Energy is emitted when electron moves from a higher stationary state to a lower stationary state. The energy change does not take place in a continuous manner.

iii. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ΔE is given by the following expression:
ν = \(\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\) ………….(1)
Where E₁ and E₂ are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.

iv. The angular momeñtum of an electron in a given stationary state can be expressed as mvr = n × h/2π
where, n 1,2, 3
Thus, an electron can move only in those orbits for which its angular momentum is integral multiple of h/2π.
Thus, only certain fixed orbits are allowed.

Question G.
Mention demerits of Bohr’s Atomic model.
Answer:
Demerits of Bohr’s atomic model:

• Bohr’s atomic model (theory) failed to account for finer details of the atomic spectrum of hydrogen as observed in sophisticated spectroscopic experiments.
• Bohr’s atomic model (theory) was unable to explain the spectrum of atoms other than hydrogen.
• Bohr’s atomic model (theory) could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or electric field (Stark effect).
• Bohr’s atomic model (theory) failed to explain the ability of atoms to form molecules by chemical bonds.

Question H.
State the order of filling atomic orbitals following Aufbau principle.
Answer:
Aufbau principle:
i. Aufbau principle gives the sequence in which various orbitals are filled with electrons.
ii. In the ground state of an atom, the orbitals are filled with electrons based on increasing order of energies of orbitals, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
iii. Increasing order of energies of orbitals:

• Orbitals are filled in order of increasing value of (n + l)
• In cases where the two orbitals have same value of (n + l), the orbital with lower value of n is filled first.

iv. The increasing order of energy of different orbitals in a multi-electron atom is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s and so on.

Question I.
Explain the anomalous behavior of copper and chromium.
Answer:
i. Copper:

• Copper (Cu) has atomic number 29.
• Its expected electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹.
• The 3d orbital is neither half-filled nor fully filled. Hence, it has less stability.
• Due to interelectronic repulsion forces, one 4s electron enters into 3d orbital. This makes 3d orbital completely filled and 4s orbital half-filled which gives extra stability and the electronic configuration of Cu becomes, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.

ii. Chromium:

• Chromium (Cr) has atomic number 24.
• Its expected electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²3d⁴.
• The 3d orbital is less stable as it is not half-filled.
• Due to inter electronic repulsion forces, one 4s electron enters into 3d orbital. This makes 4s and 3d orbitals half-filled which gives extra stability and the electronic configuration of Cr becomes, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵.

Question J.
Write orbital notations for electrons in orbitals with the following quantum numbers.
a. n = 2, l =1
b. n = 4, l = 2
c. n = 3, l = 2
Answer:
i. 2p
ii. 4d
iii. 3d

Question K.
Write electronic configurations of Fe, Fe²⁺, Fe³⁺
Answer:

Question L.
Write condensed orbital notation of electonic configuration of the following elements:
a. Lithium (Z = 3)
b. Carbon (Z=6)
c. Oxygen (Z = 8)
d. Silicon (Z = 14)
e. Chlorine (Z = 17)
f. Calcium (Z = 20)
Answer:

Question M.
Draw shapes of 2s and 2p orbitals.
Answer:
2s orbital:

2p orbital:

Question N.
Explain in brief, the significance of azimuthal quantum number.
Answer:
Azimuthal quantum number (l):

• Azimuthal quantum number is also known as subsidiary quantum number and is represented by letter l.
• It represents the subshell to which the electron belongs. It also defines the shape of the orbital that is occupied by the electron.
• Its value depends upon the value of principal quantum number ‘n’. It can have only positive values between 0 and (n – 1).
• Atomic orbitals with the same value of ‘n’ but different values of ‘l’ constitute a subshell belonging to the shell for the given ‘n’ The azimuthal quantum number gives the number of subshells in a principal shell. The subshells have l to be 0, 1, 2,3 … which are represented by symbols s, p, d, f, … respectively.

Question O.
If n = 3, what are the quantum number l and m_l?
Answer:
: For a given n, l = 0 to (n – 1) and for given l, m_l = -l……, 0…….. + l
Therefore, the possible values of l and m_l for n = 3 are:

Question P.
The electronic configuration of oxygen is written as 1s² 2s² 2p_x² 2p_y¹ 2p_z¹ and not as 1s² 2s² 2p_x² 2p_y² 2p_z⁰. Explain.
Answer:

• According to Hund’s rule of maximum multiplicity “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
• Oxygen has 8 electrons. The first two electrons will pair up in the Is orbital, the next two electrons will pair up in the 2s orbital and this leaves 4 electrons, which must be placed in the 2p orbitals.
• Each of the three degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, two p orbitals have one electron each and one p-orbital will have two electrons.

Thus, the electronic configuration of oxygen is written as 1s² 2s² 2p_x² 2p_y¹ 2p_z¹ and not as 1s² 2s² 2p_x² 2p_y² 2p_z⁰.

Question Q.
Write note on ‘Principal Quantum number.
Answer:
Principal quantum number (n):
i. Principal quantum number indicates the principal shell or main energy level to which the electron belongs.
ii. It is denoted by ‘n’ and is a positive integer with values 1, 2, 3, 4, 5, 6, ….
iii. A set of atomic orbitals with given value of ‘n’ constitutes a single shell. These shells are also represented by the letters K, L, M, N, etc.
iv. With increase of ‘n’, the number of allowed orbitals in that shell increases and is given by n².
v. The allowed orbitals in the first four shells are given below:

vi. As the value of ‘n’ increases, the distance of the shell from the nucleus increases and the size of the shell increases. Its energy also goes on increasing.

Question R.
Using concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells and orbitals.
Answer:
i. Each main shell contains a maximum of 2n² electrons.
For ‘M’ shell, n = 3.
Therefore, the maximum numbers of electrons present in the ‘M’ shell = 2 × (3)² = 18.

ii. The distribution of these electrons in shells, subshells and orbitals can be given as follows:

Note: Orbital distribution in the first four shells:

Question S.
Indicate the number of unpaired electrons in :
a. Si (Z = 14)
b. Cr (Z = 24)
Answer:
i. . Si (Z = 14): 1s² 2s² 2p⁶ 3s² 3p²
Orbital diagram:

Number of unpaired electrons = 2

ii. Cr (Z = 24): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
Orbital diagram:

Number of unpaired electrons = 6

Question T.
An atom of an element contains 29 electrons and 35 neutrons. Deduce-
a. the number of protons
b. the electronic configuration of that element.
Answer:
a. In an atom, number of protons is equal to number of electrons.
The given atom contains 29 electrons.
∴ Number of protons = 29

b. The electronic configuration of an atom of an element containing 29 electrons is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
[Note: Given element is copper (Cu) with Z = 29]

11th Chemistry Digest Chapter 4 Structure of Atom Intext Questions and Answers

Can you recall? (Textbook Page No. 35)

Question i.
What is the smallest unit of matter?
Answer:
The smallest unit of matter is atom.

Question ii.
What is the difference between molecules of an element and those of a compound?
Answer:
The molecules of an element are made of atoms of same element while the molecules of a compound are made of atoms of different elements.

Question iii.
Does an atom have any internal structure or is it indivisible?
Answer:
Yes, an atom has internal structure. Different subatomic particles such as protons, electrons and neutrons constitute an atom. So, it is divisible.

Question iv.
Which particle was identified by J. J. Thomson in the cathode ray tube experiment?
Answer:
Electron was identified by J.J. Thomson in the cathode ray tube experiment.

Question v.
Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil?
Answer:
Nucleus of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil.

Just Think (Textbook Page No. 41)

Question 1.
What does the negative sign of electron energy convey?
Answer:
Negative sign for the energy of an electron in any orbit in a hydrogen atom indicates that the energy of the electron in the atom is lower than the energy of a free electron at rest. A free-electron at rest is an electron that is infinitely far away from the nucleus and is assigned the energy value of zero.

As the electron gets close to the nucleus, value of ‘n’ decreases and E_n becomes large in absolute value and more negative. The negative sign corresponds to attractive forces between electron and nucleus.

Internet my friend (Textbook Page No. 44)

Question 1.
Collect information about the structure of atom.
Answer:
Students can use links given below as references and collect information about structure of atom on their own.
https://www.livescience.com/65427-fundamental-elementary-particles.html http://www.chemistryexplained.com/Ar-Bo/Atomic-Structure.html
https://www.thoughtco.com/basic-model-of-the-atom-603799

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Basic Analytical Techniques Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 3 Basic Analytical Techniques Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 3 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 3 Exercise Solutions

1. Choose the correct option

Question A.
Which of the following methods can be used to separate two compounds with different solubilities in the same solvent?
a. Fractional crystallization
b. Crystallization
c. Distillation
d. Solvent extraction
Answer:
a. Fractional crystallization

Question B.
Which of the following techniques is used for the separation of glycerol from soap in the soap industry?
a. Distillation under reduced pressure
b. Fractional distillation
c. Filtration
d. Crystallization
Answer:
a. Distillation under reduced pressure

Question C.
Which technique is widely used in industry to separate components of the mixture and also to purify them?
a. Steam distillation
b. Chromatography
c. Solvent extraction
d. Filtration
Answer:
b. Chromatography

Question D.
A mixture of acetone and benzene can be separated by the following method :
a. Simple distillation
b. Fractional distillation
c. Distillation under reduced pressure
d. Sublimation
Answer:
b. Fractional distillation

Question E.
Colourless components on chromatogram can not be observed by the following :
a. Using UV light
b. Using iodine chamber
c. Using the spraying reagent
d. Using infrared light
Answer:
d. Using infrared light

2. Answer the following

Question A.
Which of the following techniques is used for purification of solid organic compounds?
a. Crystallisation
b. Distillation
Answer:
Solid (crude/impure) organic compounds can be purified by crystallization.

Question B.
What do you understand by the terms
a. residue
b. filtrate.
Answer:
a. Residue: In the process of filtration, the insoluble (undissolved) impurities which remain on the filter paper are called residue.

b. Filtrate: In the process of filtration, the liquid which pass through the filter paper and collected in the beaker is called filtrate.

Question C.
Why is a condenser used in distillation process?
Answer:
In the process of distillation, a liquid is converted into its vapour and the vapour is then condensed back to liquid on cooling. The condenser has a jacket with two outlets through which water is circulated. Hence, to provide efficient cooling, a condenser is used.

Question D.
Why is paper moistened before filtration?
Answer:
Before filtration, filter paper is moistened with appropriate solvent to ensure that it sticks to the funnel and does not let the air to pass through the leaks.

Question E.
What is the stationary phase in Paper Chromatography?
Answer:
Paper chromatography is a type of partition chromatography in which a special quality paper, namely Whatman paper 1 is used. The water trapped in the fibres of the paper acts as stationary phase.

Question F.
What will happen if the upper outlet of the condenser is connected to the tap instead of the lower outlet?
Answer:

• If water enters through upper outlet of condenser, the water will quickly flow down under the influence of gravity. This allows only a small section of the condenser to be cooled enough.
• If water enters through lower outlet of condenser, the entire condenser will be filled with water before it leaves out providing maximum cooling to the condenser. This results in maximum recovery of purified liquid.

Hence, water must be allowed to enter through lower outlet of condenser during distillation process.

Question G.
Give names of two materials used as stationary phase in chromatography.
Answer:

1. Alumina
2. Silica gel

Question H.
Which properties of solvents are useful for solvent extraction?
Answer:

• Organic compound must be more soluble in the organic solvent, than in water.
• Solvent should be immiscible with water and be able to form two distinct layers.

Question I.
Why should spotting of mixture be done above the level of mobile phase ?
Answer:

• If spotting of a mixture is done at the level of mobile phase, then solvent will come in contact with the sample spot.
• Sample spot will dissolve in the mobile phase and its components will move all over the plate resulting in no distinct separation.

Hence, spotting of mixture should be done above the level of mobile phase.

Question J.
Define : a. Stationary phase b. Saturated solution
Answer:
a. Stationary phase:
Stationary phase is a solid or a liquid supported on a solid which remains fixed in a place and on which different solutes are adsorbed to a different extent.

b. Saturated solution:
A saturated solution is a solution which cannot dissolve additional quantity of a solute.

Question K.
What is the difference between simple distillation and fractional distillation?
Answer:

Question L.
Define a. Solvent extraction
b. Distillation.
Answer:
a. Solvent extraction:
Solvent extraction is a method used to separate an organic compound present in an aqueous solution, by shaking it with a suitable organic solvent in which the compound is more soluble than water.

b. Distillation:
The process in which liquid is converted into its vapour phase at its boiling point and the vapour is then condensed back to liquid on cooling is known as distillation.

Question M.
List the properties of solvents which make them suitable for crystallization.
Answer:
The solvent to be used for crystallization should have following properties:

• The compound to be crystallized should be least or sparingly soluble in the solvent at room temperature but highly soluble at high temperature.
• Solvent should not react chemically with the compound to be purified.
• Solvent should be volatile so that it can be removed easily.

Question N.
Name the different types of Chromatography and explain the principles underlying them.
Answer:
Depending on the nature of the stationary phase i.e., whether it is a solid or a liquid, chromatography is classified into adsorption chromatography and partition chromatography.
i. Adsorption chromatography: This technique is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.

Adsorption chromatography is further classified into two types:

1. Column chromatography
2. Thin-layer chromatography

ii. Partition chromatography: This technique is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. For example, paper chromatography

Question O.
Why do we see bands separating in column chromatography?
Answer:

• In column chromatography, the solutes get adsorbed on the stationary phase and depending on the degree to which they are adsorbed, they get separated from each other.
• The component which is readily adsorbed are retained on the column and others move down the column to various distances forming distinct bands.

Hence, we see bands separating in column chromatography.

Question P.
How do you visualize colourless compounds after separation in TLC and Paper Chromatography?
Answer:
i. Thin-layer chromatography (TLC): If components are colourless but have the property of fluorescence then they can be visualized under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The iodine vapours are adsorbed by the components and the spots appear brown. Also, spraying agent like ninhydrin can also be used (for amino acids).

ii. Paper Chromatography: The spots of the separated colourless components may be observed either under ultra-violet light or by the use of an appropriate spraying agent.

Question Q.
Compare TLC and Paper Chromatography techniques.
Answer:

3. Label the diagram and explain the process in your words.

Answer:
When filtration is carried out using a vacuum pump it is called filtration under suction. It is a faster and more efficient technique than simple filtration. The diagram is as follows:

ii. Procedure:

• The assembly for filtration under suction consists of a thick wall conical flask with a sidearm (Buchner flask).
• The flask is connected to a safety bottle by rubber tube through the side arm.
• Buchner funnel (a special porcelain funnel with a porous circular bottom) is fitted on the conical flask with the help of a rubber cork.
• A circular filter paper of correct size is placed on the circular porous bottom of the Buchner funnel and the funnel is placed on the flask.
• Filter paper is moistened with a few drops of water or solvent.
• Suction is created by starting the pump and filtration is carried out.

iii. Crystals are collected on the filter paper and filtrate in the flask.

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Introduction to Analytical Chemistry Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 2 Introduction to Analytical Chemistry Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 2 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 2 Exercise Solutions

1. Choose the correct option

Question A.
The branch of chemistry which deals with the study of separation, identification and quantitative determination of the composition of different substances is called………………..
a. Physical chemistry
b. Inorganic chemistry
c. Organic chemistry
d. Analytical chemistry
Answer:
d. Analytical chemistry

Question B.
Which one of the following properties of matter is Not quantitative in nature?
a. Mass
b. Length
c. Colour
d. Volume
Answer:
c. Colour

Question C.
SI unit of mass is ……..
a. kg
b. mol
c. pound
d. m³
Answer:
a. kg

Question D.
The number of significant figures in 1.50 × 10⁴ g is ………..
a. 2
b. 3
c. 4
d. 6
Answer:
b. 3

Question E.
In Avogadro’s constant 6.022 × 10²³ mol^-1, the number of significant figures is ……….
a. 3
b. 4
c. 5
d. 6
Answer:
b. 4

Question F.
By decomposition of 25 g of CaCO₃, the amount of CaO produced will be ……………….
a. 2.8 g
b. 8.4 g
c. 14.0 g
d. 28.0 g
Answer:
c. 14.0 g

Question G.
How many grams of water will be produced by complete combustion of 12g of methane gas
a. 16
b. 27
c. 36
d. 56
Answer:
b. 27

Question H.
Two elements A (At. mass 75) and B (At. mass 16) combine to give a compound having 75.8 % of A. The formula of the compound is
a. AB
b. A₂B
c. AB₂
d. A₂B₃
Answer:
d. A₂B₃

Question I.
The hydrocarbon contains 79.87 % carbon and 20.13 % of hydrogen. What is its empirical formula ?
a. CH
b. CH₂
c. CH₃
d. C₂H₅
Answer:
c. CH₃

Question J.
How many grams of oxygen will be required to react completely with 27 g of Al? (Atomic mass : Al = 27, O = 16)
a. 8
b. 16
c. 24
d. 32
Answer:
c. 24

Question K.
In CuSO₄.5H₂O the percentage of water is ……
(Cu = 63.5, S = 32, O = 16, H = 1)
a. 10 %
b. 36 %
c. 60 %
d. 72 %
Answer:
b. 36 %

Question L.
When two properties of a system are mathematically related to each other, the relation can be deduced by
a. Working out mean deviation
b. Plotting a graph
c. Calculating relative error
d. all the above three
Answer:
b. Plotting a graph

2. Answer the following questions

Question A.
Define : Least count
Answer:
The smallest quantity that can be measured by the measuring equipment is called least count.

Question B.
What do you mean by significant figures? State the rules for deciding significant figures.
Answer:
i. The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
ii. Rules for deciding significant figures:
a. All non-zero digits are significant.
e.g. 127.34 g contains five significant figures which are 1, 2, 7, 3 and 4.
b. All zeros between two non-zero digits are significant, e.g. 120.007 m contains six significant figures.
c. Zeros on the left of the first non-zero digit are not significant. Such a zero indicates the position of the decimal point.
e.g. 0.025 has two significant figures, 0.005 has one significant figure.
d. Zeros at the end of a number are significant if they are on the right side of the decimal point,
e. g. 0.400 g has three significant figures and 400 g has one significant figure.
e. In numbers written is scientific notation, all digits are significant.
e.g. 2.035 × 10² has four significant figures and 3.25 × 10^-5 has three significant figures.

Question C.
Distinguish between accuracy and precision.
Answer:
Accuracy:

1. Accuracy refers to nearness of the measured value to the true value.
2. Accuracy represents the correctness of the measurement.
3. Accuracy is expressed in terms of absolute error and relative error.
4. Accuracy takes into account the true or accepted value.
5. Accuracy can be determined by a single measurement.
6. High accuracy implies smaller error.

Precision:

1. Precision refers to closeness of multiple readings of the same quantity.
2. Precision represents the agreement between two or more measured values.
3. Precision is expressed in terms of absolute deviation and relative deviation.
4. Precision does not take into account the true or accepted value.
5. Several measurements are required to determine precision.
6. High precision implies reproducibility of the readings.

Question D.
Explain the terms percentage composition, empirical formula and molecular formula.
Answer:
Percentage Composition:

• The percentage composition of a compound is the percentage by weight of each element present in the compound.
• Quantitative determination of the constituent elements by suitable methods provides the percent elemental composition of a compound.
• If the percent total is not 100, the difference is considered as percent oxygen.
• From the percentage composition, the ratio of the atoms of the constituent elements in the molecule is calculated.

Empirical Formula:
The simplest ratio of atoms of the constituent elements in a molecule is called the empirical formula of that compound.
e.g. The empirical formula of benzene is CH.

Molecular Formula:
1. Molecular formula of a compound is the formula which indicates the actual number of atoms of the constituent elements in a molecule.
e.g. The molecular formula of benzene is C6H6.
2. It can be obtained from the experimentally determined values of percent elemental composition and molar mass of that compound.
3. Molecular formula can be obtained from the empirical formula if the molar mass is known.
Molecular formula = r × Empirical formula

Question E.
What is a limiting reagent ? Explain.
Answer:
Limiting reagent:

• The reactant which gets consumed and limits the amount of product formed is called the limiting reagent.
• When a chemist carries out a reaction, the reactants are not usually present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation.
• This is because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials. Frequently, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product.
• The reactant which is present in lesser amount gets consumed after some time and subsequently, no further reaction takes place, whatever be the amount left of the other reactant present.

Hence, limiting reagent is the reactant that gets consumed entirely and limits the reaction.

Question F.
What do you mean by SI units ? What is the SI unit of mass ?
Answer:
i. In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI units, abbreviated from its French name.
ii. The SI unit of mass is kilogram (kg).

Question G.
Explain the following terms
(a) Mole fraction
(b) Molarity
(c) Molality
Answer:
(a) Mole fraction: Mole fraction is the ratio of number of moles of a particular component of a solution to the total number of moles of the solution.

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are n_A and n_B, respectively, then the mole fraction of A and B are given as:

(b) Molarity: Molarity is defined as the number of moles of the solute present in 1 litre of the solution. It is the most widely used unit and is denoted by M.
Molarity is expressed as follows:
Molarity (M) = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}\)

Molality: Molality is the number of moles of solute present in 1 kg of solvent. It is denoted by m. Molality is expressed as follows:
Molality (m) = \(\frac{\text { Number of moles of solute }}{\text { Mass of solvent in kilograms }}\)

Question H.
Define : Stoichiometry
Answer:
The study of quantitative relations between the amount of reactants and/or products is called stoichiometry.

Question I.
Why there is a need of rounding off figures during calculation ?
Answer:

• When performing calculations with measured quantities, the rule is that the accuracy of the final result is limited to the accuracy of the least accurate measurement.
• In other words, the final result cannot be more accurate than the least accurate number involved in the calculation.
• Sometimes, the final result of a calculation often contains figures that are not significant.
• When this occurs, the final result is rounded off.

Question J.
Why does molarity of a solution depend upon temperature ?
Answer:

• Molarity is the number of moles of the solute present in 1 litre of the solution. Therefore, molarity depends on the volume of the solution.
• Volume of the solution varies with the change in temperature.

Hence, molarity of a solution depends upon temperature.

Question M.
Define Analytical chemistry. Why is accurate measurement crucial in science?
Answer:
The branch of chemistry which deals with the study of separation, identification, qualitative and quantitative determination of the compositions of different substances, is called analytical chemistry.

1. The accuracy of measurement is of great concern in analytical chemistry. This is because faulty equipment, poor data processing, or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in analytical measurement.
2. When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.
3. Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
4. Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation, and properly express the quantitative error in the result.

3. Solve the following questions

Question A.
How many significant figures are in each of the following quantities ?
a. 45.26 ft
b. 0.109 in
c. 0.00025 kg
d. 2.3659 × 10^-8 cm
e. 52.0 cm³
f. 0.00020 kg
g. 8.50 × 10⁴ mm
h. 300.0 cg
Answer:
a. 4
b. 3
c. 2
d. 5
e. 3
f. 2
g. 3
h. 4

Question B.
Round off each of the following quantities to two significant figures :
a. 25.55 mL
b. 0.00254 m
c. 1.491 × 10⁵ mg
d. 199 g
Answer:
a. 26 mL
b. 0.0025 m
c. 1.5 × 10⁵ mg
d. 2.0 × 10² g

Question C.
Round off each of the following quantities to three significant figures :
a. 1.43 cm³
b. 458 × 10² cm
c. 643 cm²
d. 0.039 m
e. 6.398 × 10^-3 km
f. 0.0179 g
g. 79,000 m
h. 42,150
i. 649.85
j. 23,642,000 mm
k. 0.0041962 kg
Answer:
a. 43 cm³
b. 4.58 × 10⁴ cm
c. 643 cm² (or 6.43 × 10² cm²)
d. 0.0390 m (or 3.90 × 10^-2 m)
e. 6.40 × 10^-3 km
f. 0.0179 g (or 1.79 × 10^-2 m)
g. 7.90 × 10⁴ m
h. 4.22 × 10⁴ (or 42,200)
i. 6.50 × 10²
j. 2.36 × 10⁷ mm
k. 0.00420 kg (or 4.20 × 10^-3 kg)

Question D.
Express the following sum to appropriate number of significant figures :
a. 2.3 × 10³ mL + 4.22 × 10⁴ mL + 9.04 × 10³ mL + 8.71 × 10⁵ mL;
b. 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g
Answer:
To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.
a. (0.23 × 10⁴ mL) + (4.22 × 10⁴ mL) +(0.904 × 10⁴ mL) + (87.1 × 10⁴ mL)
= (0.23 + 4.22 + 0.904 + 87.1) × 10⁴ mL
= 92.454 × 10⁴ mL
= 9.2454 × 10⁵
= 9.2 × 10⁵ mL
b. 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g
= – 20190.3789 g
= – 20190 g
Ans: Sum to appropriate number of significant figures = 9.2 × 10⁵ mL
ii. Sum to appropriate number of significant figures = – 20190 g
[Note: In addition and subtraction, the final answer is rounded to the minimum number of decimal point of the number taking part in calculation. If there is no decimal point, then the final answer will have no decimal point.]

4. Solve the following problems

Question A.
Express the following quantities in exponential terms.
a. 0.0003498
b. 235.4678
c. 70000.0
d. 1569.00
Answer:
a. 0.0003498 = 3.498 × 10^-4
b. 235.4678 = 2.354678 × 10²
c. 70000.0 = 7.00000 × 10⁴
d. 1569.00 = 1.56900 × 10³

Question B.
Give the number of significant figures in each of the following
a. 1.230 × 10⁴
b. 0.002030
c. 1.23 × 10⁴
d. 1.89 × 10^-4
Answer:
a. 4
b. 4
c. 3
d. 3

Question C.
Express the quantities in above (B) with or without exponents as the case may be.
Answer:
a. 12300
b. 2.030 × 10^-3
c. 12300
d. 0.000189

Question D.
Find out the molar masses of the following compounds :
a. Copper sulphate crystal (CuSO₄.5H₂O)
b. Sodium carbonate, decahydrate (Na₂CO₃.10H₂O)
c. Mohr’s salt [FeSO₄(NH₄)₂SO₄.6H₂O]
(At. mass : Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)
Answer:
a. Molar mass of CuSO₄.5H₂O
= (1 × At. mass Cu) + (1 × At. mass S) + (9 × At. mass O) + (10 × At. mass H)
= (1 × 63.5) + (1 × 32) + (9 × 16) + (10 × 1)
= 63.5 + 32 + 144 + 10
= 249.5 g mol^-1
Molar mass of CuSO₄.5H₂O = 249.5 g mol^-1

b. Molar mass of Na₂CO₃.10H₂O
= (2 × At. mass Na) + (1 × At. mass C) + (13 × At. mass O) + (20 × At. mass H)
= (2 × 23) + (1 × 12) + (13 × 16) + (20 × 1)
= 46 + 12 + 208 + 20
= 286 g mol^-1
Molar mass of Na₂CO₃.10H₂O = 286 g mol^-1

c. Molar mass of [FeSO₄(NH₄)₂SO₄.6H₂O]
= (1 × At. mass Fe) + (2 × At. mass S) + (2 × At. mass N) + (14 × At. mass O) + (20 × At. mass H)
= (1 × 56) + (2 × 32) + (2 × 14) + (14 × 16) + (20 × 1)
= 56 + 64 + 28 + 224 + 20
= 392 g mol^-1
Molar mass of [FeSO₄(NH₄)₂SO₄.6H₂O] = 392 g mol^-1

Question E.
Work out the percentage composition of constituents elements in the following compounds :
a. Lead phosphate [Pb₃(PO₄)₂],
b. Potassium dichromate (K₂Cr₂O₇),
c. Macrocosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄.4H₂O)
(At. mass : Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14)
Answer:
Given: Atomic mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14
To find: The percentage composition of constituent elements
Formula:

Calculation:
i. Lead phosphate [Pb₃(PO₄)₂]
Molar mass of Pb₃(PO₄)₂ = 3 × (207) + 2 × (31) + 8 × (16) = 621 + 62 + 128 = 811 g mol^-1
Percentage of Pb = \(\frac {621}{811}\) × 100 = 76.57%
Percentage of P = \(\frac {621}{811}\) × 100 = 7.64%
Percentage of O = \(\frac {128}{811}\) × 100 = 15.78%

ii. Potassium dichromate (K₂Cr₂O₇)
Molar mass of K₂Cr₂O₇ = 2 × (39) + 2 × (52) + 7 × (16) = 78 + 104 + 112 = 294 g mol^-1
Percentage of K = \(\frac {78}{294}\) × 100 = 26.53%
Percentage of Cr = \(\frac {104}{294}\) × 100 = 35.37%
Percentage of O = \(\frac {112}{294}\) × 100 = 38.10%

iii. Microcosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄.4H₂O)
Molar mass of NaNH₄HPO₄.4H₂O = 1 × (23) + 1 × (14) + 1 × (31) + 13 × (1) + 8 × (16)
= 23 + 14 + 31 + 13 + 128 = 209 g mol^-1
Percentage of Na = \(\frac {23}{209}\) × 100 = 11.00%
Percentage of N = \(\frac {14}{209}\) × 100 = 6.70%
Percentage of P = \(\frac {31}{209}\) × 100 = 14.83%
Percentage of H = \(\frac {13}{209}\) × 100 = 6.22%
Percentage of O = \(\frac {128}{209}\) × 100 = 61.24%
Ans: i. Mass percentage of Pb, P and O in lead phosphate [Pb₃(PO₄)₂] are 76.57%, 7.64% and 15.78% respectively.
ii. Mass percentage of K, Cr and O in potassium dichromate (K₂Cr₂O₇) are 26.53%, 35.37% and 38.10% respectively.
iii. Mass percentage of Na, N, P, H and O in NaNH₄HPO₄.4H₂O are 11.00%, 6.70%, 14.83%, 6.22% and 61.24% respectively.

Question F.
Find the percentage composition of constituent green vitriol crystals (FeSO₄.7H₂O). Also find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals. (At. mass : Fe = 56; S = 32; O = 16)
Answer:
Given: i. Atomic mass: Fe = 56; S = 32; O = 16
ii. Mass of crystal = 4.54 kg
To find: i. Mass percentage of Fe, S, H and O
ii. Mass of iron and water of crystallisation in 4.54 kg of crystal
Formula:

i. Molar mass of FeSO₄.7H₂O = 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16)
= 56 + 32 + 14+ 176
= 278 g mol^-1
Percentage of Fe = \(\frac {56}{278}\) × 100 = 20.14%
Percentage of S = \(\frac {32}{278}\) × 100 = 11.51%
Percentage of H = \(\frac {14}{278}\) × 100 = 5.04%
Percentage of O = \(\frac {176}{278}\) × 100 = 63.31%

ii. 278 kg green vitriol = 56 kg iron
∴ 4.54 kg green vitriol = x
∴ x = \(\frac{56 \times 4.54}{278}\)
Mass of 7H₂O in 278 kg green vitriol = 7 × 18 = 126 kg
∴ 4.54 kg green vitriol = y
∴ y = \(\frac{126 \times 4.54}{278}\)
Ans: i. Mass percentage of Fe, S, H and O in FeSO₄.7H₂O are 20.14%, 11.51%, 5.04% and 63.31% respectively.
ii. Mass of iron in 4.54 kg green vitriol = 0.915 kg
Mass of water of crystallisation in 4.54 kg green vitriol = 2.058 kg

Question G.
The red colour of blood is due to a compound called “haemoglobin”. It contains 0.335 % of iron. Four atoms of iron are present in one molecule of haemoglobin. What is its molecular weight ? (At. mass : Fe = 55.84)
Answer:
Given: Iron percentage in haemoglobin = 0.335%
To find: Molecular weight of haemoglobin
Calculation: There are four atoms of iron in a molecule of haemoglobin. Four atoms of iron contribute 0.335% mass to a molecule of haemoglobin.
Mass of one Fe atom = 55.84 u
∴ Mass of 4 Fe atoms = 55.84 × 4 = 223.36 u = 0.335%
Let molecular weight of haemoglobin be x.
Hence,
\(\frac{223.36}{x}\) × 100 = 0.335%
∴ x = \(\frac{223.36}{0.335}\) × 100 = 66674.6 g mol^-1
Ans: Molecular weight of haemoglobin = 66674.6 g mol^-1

Question H.
A substance, on analysis, gave the following percent composition:
Na = 43.4 %, C = 11.3 % and O = 45.3 %. Calculate the empirical formula. (At. mass Na = 23 u, C = 12 u, O = 16 u).
Answer:
Given: Atomic mass of Na = 23 u, C = 12 u, and O = 16 u
Percentage of Na, C and O = 43.4%, 11.3% and 45.3% respectively.
To find: The empirical formula of the compound
Calculation:

Hence, empirical formula is Na₂CO₃.
Ans: Empirical formula of the compound = Na₂CO₃

Question I.
Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.
Answer:
Given: Atomic mass of M = 56
Percentage of M = 70.0%
To find: The empirical formula of the compound
Calculation: % M = 70.0%
Hence, % O = 30.0%, Atomic mass of O = 16 u

Convert the ratio into whole number by multiplying by the suitable coefficient, i.e., 2.
Therefore, the ratio of number of moles of M : O is 2 : 3.
Hence, the empirical formula is M₂O₃.
Ans: Empirical formula of the compound = M₂O₃

Question J.
1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt. : Fe = 56; S = 32; O = 16)
Answer:
Given: Atomic mass of Fe = 56, S = 32, and O = 16
Mass of iron, sulphur, oxygen and water = 0.2014 g, 0.1153 g, 0.2301 g and 0.4532 respectively.
To find: The empirical formula of the compound
Calculation: Since the mass of crystal is 1 g, the % iron, sulphur, oxygen and water = 20.14%, 11.53%, 23.01% and 4.32 % respectively.

Hence, empirical formula is FeSO₄.7H₂O.
Ans: Empirical formula of the compound = FeSO₄.7H₂O.

Question K.
An organic compound containing oxygen, carbon, hydrogen and nitrogen contains 20 % carbon, 6.7 % hydrogen and 46.67 % nitrogen. Its molecular mass was found to be 60. Find the molecular formula of the compound.
Answer:
Given: Percentage of carbon, hydrogen, nitrogen = 20%, 6.7%, 46.67% respectively.
Molar mass of the compound = 60 g mol^-1
To find: The molecular formula of the compound
Calculation: % carbon + % hydrogen + % nitrogen = 20 + 6.7 + 46.67 = 73.37%
This is less than 100%. Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, % of oxygen = 100 – 73.37 = 26.63%

Hence, empirical formula is CH₄N₂O.
Empirical formula mass = 12 + 4 + 28 + 16 = 60 g mol^-1
Hence,
Molar mass = Empirical formula mass
∴ Molecular formula = Empirical formula = CH₄N₂O
Ans: Molecular formula of the compound = CH₄N₂O

Question L.
A compound on analysis gave the following percentage composition by mass : H = 9.09; O = 36.36; C = 54.55. Mol mass of compound is 88. Find its molecular formula.
Answer:
Given: Percentage of H, O, C = 9.09%, 36.36%, 54.55% respectively.
Molar mass of the compound = 88 g mol^-1
To find: The molecular formula of the compound
Calculation:

Hence, empirical formula is C₂H₄O.
Empirical formula mass = 24 + 4 + 16 = 44 g mol^-1
Hence,

Molecular formula = r × empirical formula
Molecular formula = 2 × C₂H₂O = C₄H₈O₂
Ans: Molecular formula of the compound = C₄H₈O₂

Question M.
Carbohydrates are compounds containing only carbon, hydrogen and oxygen. When heated in the absence of air, these compounds decompose to form carbon and water. If 310 g of a carbohydrate leave a residue of 124 g of carbon on heating in absence of air, what is the empirical formula of the carbohydrate ?
Answer:
Given: Mass of carbon residue = 124 g, mass of carbohydrate = 310 g
To find: Empirical formula of the carbohydrate
Calculation: Since the 310 g of compound decomposes to carbon and water and the mass of carbon produced is 124 g, the remaining mass would be of water.
∴ Molar mass of water = 310 – 124 = 186 g

The ratio of number of moles of C : water = C : H₂O = 1 : 1
Hence, empirical formula = CH₂O
Ans: Empirical formula of the carbohydrate = CH₂O

Question N.
Write each of the following in exponential notation :
a. 3,672,199
b. 0.000098
c. 0.00461
d. 198.75
Answer:
a. 3,672,199 = 3.672199 × 10⁶
b. 0.000098 = 9.8 × 10^-5
c. 0.00461 = 4.61 × 10^-3
d. 198.75 = 1.9875 × 10²

Question O.
Write each of the following numbers in ordinary decimal form :
a. 3.49 × 10^-11
b. 3.75 × 10^-1
c. 5.16 × 10⁴
d. 43.71 × 10^-4
e. 0.011 × 10^-3
f. 14.3 × 10^-2
g. 0.00477 × 10⁵
h. 5.00858585
Answer:
a. 3.49 × 10^-11 = 0.0000000000349
b. 3.75 × 10^-1 = 0.375
c. 5.16 × 10⁴ = 51,600
d. 43.71 × 10^-4 = 0.004371
e. 0.011 × 10^-3 = 0.000011
f. 14.3 × 10^-2 = 0.143
g. 0.00477 × 10⁵ = 477
h. 5.00858585 = 5.00858585

Question P.
Perform each of the following calculations. Round off your answers to two digits.

Answer:

Question Q.
Perform each of the following calculations. Round off your answers to three digits.
a. (3.26 × 10⁴) (1.54 × 10⁶)
b. (8.39 × 10⁷) (4.53 × 10⁹)
c. \(\frac{8.94 \times 10^{6}}{4.35 \times 10^{4}}\)
d. \(\frac{\left(9.28 \times 10^{9}\right) \times\left(9.9 \times 10^{-7}\right)}{(511) \times\left(2.98 \times 10^{-6}\right)}\)
Answer:
i. (3.26 × 10⁴) (1.54 × 10⁶) = 5.0204 × 10⁴⁺⁶ = 5.02 × 10¹⁰
ii. (8.39 × 10⁷) (4.53 × 10⁹) = 38.0067 × 10⁷⁺⁹ = 38.0067 × 10¹⁶ = 3.80 x 10¹⁷

Question R.
Perform the following operations :
a. 3.971 × 10⁷ + 1.98 × 10⁴;
b. 1.05 × 10^-4 – 9.7 × 10^-5;
c. 4.11 × 10^-3 + 8.1 × 10^-4;
d. 2.12 × 10⁶ – 3.5 × 10⁵.
Answer:
Solution:
To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.
a. 3.971 × 10⁷ + 1.98 × 10⁴ = 3.971 × 10⁷ + 0.00198 × 10⁷ = (3.971 + 0.00198) × 10⁷
= 3.97298 × 10⁷
b. 1.05 × 10^-4 – 9.7 × 10^-5 = 10.5 × 10^-5 – 9.7 × 10^-5 = (10.5 – 9.7) × 10^-5 = 0.80 × 10^-5
= 8.0× 10^-6
c. 4.11 × 10^-3 + 8.1 × 10^-4 = 41.1 × 10^-4 + 8.1 × 10^-4 = (41.1 + 8.1) × 10^-4 = 49.2 × 10^-4
= 4.92 × 10^-3
d. 2.12 × 10⁶ – 3.5 × 10⁵ = 21.2 × 10⁵ – 3.5 × 10⁵ = (21.2 – 3.5) × 10⁵ = 17.7 × 10⁵
= 1.77 × 10⁶

Question S.
A 1.000 mL sample of acetone, a common solvent used as a paint remover, was placed in a small bottle whose mass was known to be 38.0015 g. The following values were obtained when the acetone – filled bottle was weighed : 38.7798 g, 38.7795 g and 38.7801 g. How would you characterise the precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g ?
Answer:
Precision:

Mean = \(\frac{0.7783+0.7780+0.7786}{3}\) = 0.7783 g

Mean absolute deviation = \(\frac{0+0.0003+0.0003}{3}\) = 0.0002
∴ Mean absolute deviation = ±0.0002 g

ii. Accuracy:
Actual mass of acetone = 0.7791 g
Observed value (average) = 0.7783 g
a. Absolute error = Observed value – True value
= 0.7783 – 0.7791
= – 0.0008 g

Ans: These observed values are close to each other and are also close to the actual mass. Therefore, the results are precise and as well accurate.
i. Relative deviation = 0.0257%
ii. Relative error = 0.1027%
[Note: i. As per the method given in textbook, the calculated value of relative deviation is 0.0257%.
ii. The negative sign in -0.1027% indicates that the experimental result is lower than the true value.]

Question T.
Your laboratory partner was given the task of measuring the length of a box (approx 5 in) as accurately as possible, using a metre stick graduated in milimeters. He supplied you with the following measurements: 12.65 cm, 12.6 cm, 12.65 cm, 12.655 cm, 126.55 mm, 12 cm.
a. State which of the measurements you would accept, giving the reason.
b. Give your reason for rejecting each of the others.
Answer:
a. The metre stick is graduated in millimetres i.e. 1 mm to 1000 mm, and 1 mm = 0.1 cm. Therefore, if the length is measured in centimetres, the least count of metre stick is 0.1 cm. The results 12.6 cm has the least count of 0.1 cm and is an acceptable result.

b. Since, the least count of metre stick is 0.1 cm or 1mm, the results such as 12.65 cm, 12.655 cm, 126.55 mm cannot be measured using this stick, and hence, these results are rejected. The result, 12 cm doesn’t include the least count and is rejected.

Question U.
What weight of calcium oxide will be formed on heating 19.3 g of calcium carbonate?
(At. wt. : Ca = 40; C = 12; O = 16)
Answer:
Given: Mass of CaCO₃ consumed in reaction = 19.3 g
To find: Mass of CaO formed
Calculation: Calcium carbonate decomposes according to the balanced equation,

So, 100 g of CaCO3 produce 56 g of CaO.

Ans: Mass of CaO formed = 10.81 g

[Calculation using log table:
56 × 0.193
= Antilog₁₀ [log₁₀ (56) + log₁₀ (0.193)]
= Antilog₁₀ [1.7482 + \(\overline{1} .2856\)]
= Antilog₁₀ [1.0338] = 10.81]

Question V.
The hourly energy requirements of an astronaut can be satisfied by the energy released when 34 grams of sucrose are “burnt” in his body. How many grams of oxygen would be needed to be carried in a space capsule to meet his requirement for one day?
Answer:
34 g of sucrose provides energy for an hour.
Hence, for a day, the mass of sucrose needed = 34 × 24 = 816g
The balanced equation is,

Thus, 342 g of sucrose requires 384 g of oxygen.
∴ 816 g of sucrose will require = \(\frac{816}{342}\) × 384 = 916 g of O₂
Ans: Astronaut needs to carry 916 g of O₂.

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Some Basic Concepts of Chemistry Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 1 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 1 Exercise Solutions

1. Choose the most correct option.

Question A.
A sample of pure water, whatever the source always contains …………. by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
Answer:
a. 88.9

Question B.
Which of the following compounds can NOT demonstrate the law of multiple proportions?
a. NO, NO₂
b. CO, CO₂
c. H₂O, H₂O₂
d. Na₂S, NaF
Answer:
d. Na₂S, NaF

Question C.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
a. – 40°
b. + 40°
c. – 80°
d. – 20°
Answer:
a. – 40°

Question D.
SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
Answer:
b. Ampere

Question E.
In the reaction N₂ + 3H₂ → 2NH₃, the ratio by volume of N₂, H₂ and NH₃ is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
Answer:
d. gaseous volumes

Question F.
Which of the following has maximum number of molecules ?
a. 7 g N₂
b. 2 g H₂
c. 8 g O₂
d. 20 g NO₂
Answer:
b. 2 g H₂

Question G.
How many g of H₂O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
Answer:
a. 4.5

Question H.
The number of molecules in 22.4 cm³ of nitrogen gas at STP is
a. 6.022 × 10²⁰
b. 6.022 × 10²³
c. 22.4 × 10²⁰
d. 22.4 × 10²³
Answer:
a. 6.022 × 10²⁰

Question I.
Which of the following has the largest number of atoms ?
a. 1g Au_(s)
b. 1g Na_(s)
c. 1g Li_(s)
d. 1g Cl_2(g)
Answer:
c. 1g Li_(s)

2. Answer the following questions.

Question A.
State and explain Avogadro’s law.
Answer:
i. In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.

ii. Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:

According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:

Question B.
Point out the difference between 12 g of carbon and 12 u of carbon.
Answer:
12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom.

Question C.
How many grams does an atom of hydrogen weigh ?
Answer:
The mass of a hydrogen atom is 1.6736 × 10^-24 g.

Question D.
Calculate the molecular mass of the following in u.
a. NH₃
b. CH₃COOH
c. C₂H₅OH
Answer:
i. Molecular mass of NH₃ = (1 × Average atomic mass of N) + (3 × Average atomic mass of H)
= (1 × 14.0 u) +(3 × 1.0 u)
= 17 u

ii. Molecular mass of CH₃COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u)
= 60 u

iii. Molecular mass of C₂H₅OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u)
= 46 u
Ans: i. The molecular mass of NH₃ = 17 u
ii. The molecular mass of CH₃COOH = 60 u
iii. The molecular mass of C₂H₅OH = 46 u

Question E.
How many particles are present in 1 mole of a substance ?
Answer:
The number of particles in one mole is 6.0221367 × 10²³.

Question F.
What is the SI unit of amount of a substance ?
Answer:
The SI unit for the amount of a substance is mole (mol).

Question G.
What is meant by molar volume of a gas ?
Answer:
The volume occupied by one mole of a gas at standard temperature (0 °C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm³.

Question H.
State and explain the law of conservation of mass.
Answer:
Law of conservation of mass:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• Antoine Lavoisier who is often referred to as the father of modem chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
• Both his experiments resulted in increased weight of products.
• After several experiments, in burning of phosphorus, he found that the weight gained by the phosphoms was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
• When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.

Question I.
State the law of multiple proportions.
Answer:
The law states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.

3. Give one example of each

Question A.
Homogeneous mixture
Answer:
Homogeneous mixture: Solution (An aqueous solution of sugar)

Question B.
Heterogeneous mixture
Answer:
Heterogeneous mixture: Suspension (of sand in water)

Question C.
Element
Answer:
Element: Gold

Question D.
Compound
Answer:
Compound: Distilled water.

4. Solve problems :

Question A.
What is the ratio of molecules in 1 mole of NH₃ and 1 mole of HNO₃.
Answer:
One mole of any substance contains particles equal to 6.022 × 10²³.
1 mole of NH₃ = 6.022 × 10²³ molecules of NH₃
I mole of HNO₃ = 6.022 × 10²³ molecules of HNO₃
∴ Ratio = \(\frac{6.022 \times 10^{23}}{6.022 \times 10^{23}}\) = 1 : 1
Ans: The ratio of molecules is = 1 : 1.

Question B.
Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen

Molar volume of a gas = 22.4 dm³ mol^-1 = 22.4 L at STP

Ans: Number of moles of hydrogen = 0.02 mol

Question C.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
∴ Mass of 18 atoms of hydrogen = 18 × 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u

Question D.
Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ x = \(\frac{254 \mathrm{u}}{127 \mathrm{u}}\) = 2 atoms

b. 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol^-1
Now,

Now,
Number of atoms = Number of moles × Avogadro’s constant
= 2 mol × 6.022 × 10²³ atoms/mol
= 12.044 × 10²³ atoms
= 1.2044 × 10²⁴ atoms
Ans. i.Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = 1.2044 × 10²⁴ atoms

Question E.
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. 5 mg carbon = 5 × 10^-3 g carbon
Atomic mass of carbon = 12 u
∴ Molar mass of carbon 12 g mol^-1

b. 12 mg carbon = 12 × 10^-3 g carbon

Number of atoms = Number of moles × Avogadro’s constant
Number of atoms of carbon = 1 × 10^-3 mol × 6.022 × 10²³ atoms/mol
= 6.022 × 10²⁰ atoms
Ans: Number of moles of carbon in his homework writing = 4.167 × 10^-4 mol
Number of atoms of carbon in 12 mg homework writing = 6.022 × 10²⁰ atoms

Question F.
Arjun purchased 250 g of glucose (C₆H₁₂O₆) for Rs 40. Find the cost of glucose per mole.
Answer:
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C₆H₁₂O₆
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C₆H₁₂O₆).
Molecular mass of glucose
= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)
=180 u
∴ Molar mass of glucose = 180 g mol^-1

Now,
\(\frac {250}{180}\) mol of glucose cost = Rs 40
1 mol glucose cost = x
∴ x = \(\frac{40 \times 180}{250}\) = Rs 28.8/mol of glucose
Ans. The cost of glucose per mole is Rs 28.8.

[ Calculation using log table:
\(\frac{40 \times 180}{250}\)
= Antilog₁₀ [log₁₀(40) + log₁₀(180) + log₁₀(250)]
= Antilog₁₀ [1.6021 + 2.2553 – 2.3979]
= Antilog₁₀ [1.4595] = 28.80 ]

Question G.
The natural isotopic abundance of ¹⁰B is 19.60% and ¹¹B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron(B)

Ans. Average atomic mass of boron = 10.84 u

Question H.
Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 °C
b. 30 °C
Answer:
a. Given: Temperature in degree Celsius =40°C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 40 °C in the formula,
°F = \(\frac {9}{5}\) (°C)+32
= \(\frac {9}{5}\) (40) + 32
= 72 + 32
= 104 °F

b. Given: Temperature in degree Celsius = 30 °C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 30 °C in the formula,
°F = \(\frac {9}{5}\)(°C) + 32
= \(\frac {9}{5}\)(30) + 32
= 54 + 32
= 86 °F
Ans: i. The temperature 40 °C corresponds to 104 °F.
ii. The temperature 30 °C corresponds to 86 °F.

Question I.
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Calculator: Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH₃COOH
= (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12 u) + (4 × 1 u) + (2 × 16 u) = 60 u
∴ Molar mass of acetic acid = 60 g mol^-1

Now,
Number of molecules of acetic acid = Number of moles × Avogadro’s constant
= 0.367 mol × 6.022 × 10²³ molecules/mol
= 2.210 × 10²³ molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 × 10²³ molecules

Question J.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 32 = 64 g) of oxygen to give (2 × 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g

Question K.
Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles × Avogadro’s constant
= 0.4 mol × 6.022 × 10²³ atoms/mol
= 2.4088 × 10²³ atoms of N

b. 1.6 g of Sulphur (S)
Molar mass of sulphur = 32 g mol^-1

Number of atoms of S = Number of moles × Avogadro’s constant
= 0.05 mol × 6.022 × 10²³ atoms/mol
= 0.3011 × 10²³ atoms
= 3.011 × 10²² atoms of S
Ans: a. Number of nitrogen atoms in 0.4 mole = 2.4088 × 10²³ atoms of N
b. Number of sulphur atoms in 1.6 g = 3.011 × 10²² atoms of S

Question L.
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Answer:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
% of oxygen = \(\frac{1.2}{3.2}\) × 100 = 37.5%
% of metal = \(\frac{2.0}{3.2}\) × 100 = 62.5%
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g
% of oxygen = \(\frac{0.85}{2.27}\) × 100 = 37.44 ≈ 37.5%
% of metal = \(\frac{1.42}{2.27}\) × 100 = 62.56 ≈ 62.5%
Therefore, irrespective of the source, the given compound contains same elements in the same proportion. The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight”. Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.

Question M.
In two moles of acetaldehyde (CH₃CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer:
Molecular formula of acetaldehyde: C₂H₄O
Moles of acetaldehyde = 2 mol
a. Number of moles of carbon atoms = Moles of acetaldehyde × Number of carbon atoms
= 2 × 2
= 4 moles of carbon atoms

b. Number of moles of hydrogen atoms = Moles of acetaldehyde × Number of hydrogen atoms
= 2 × 4
= 8 moles of hydrogen atoms

c. Number of moles of oxygen atoms = Moles of acetaldehyde × Number of oxygen atoms
= 2 × 1
= 2 moles of oxygen atoms

d. Number of molecules of acetaldehyde = Moles of acetaldehyde × Avogadro number
= 2 mol × 6.022 × 10²³ molecules/mol
= 12.044 × 10²³ molecules of acetaldehyde
Ans: i. Number of moles of carbon, hydrogen and oxygen are 4, 8, 2 respectively,
ii. Number of molecules of acetaldehyde = 12.044 × 10²³

Question N.
Calculate the number of moles of magnesium oxide, MgO in
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of Mg = 24 and O = 16)
Answer:
Given: i. Mass of MgO = 80 g
ii. Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae: Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
Calculation: i. Molecular mass of MgO = (1 × Average atomic mass of Mg) + (1 × Average atomic mass of O)
= (1 × 24u) + (1 × 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol^-1
Mass of MgO = 80 g
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{80 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 2 mol

ii. Mass of MgO = 10 g, Molar mass of MgO = 40 g mol^-1
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{10 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.25 mol
Ans: i. The number of moles in 80 g of magnesium oxide, MgO = 2 mol
ii. The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol

Question O.
What is volume of carbon dioxide, CO₂ occupying by i. 5 moles and ii. 0.5 mole of CO₂ gas measured at STP.
Answer:
Given: i. Number of moles of CO₂ = 5 mol
ii. Number of moles of CO₂ = 0.5 mol
To find: Volume at STP
Formula: Number of moies of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Calculation: Molar volume of a gas 22.4 dm³ mol^-1 at STP.
Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
∴ i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5mol × 22.4 dm³ mol^-1 = 112 dm³
ii. Volume of the gas at STP Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm³ mol^-1 = 11.2 dm³
Ans: i. Volume of 5 mol of CO₂ = 112 dm³
ii. Volume of 0.5 mol of CO₂ = 11.2 dm³

Question P.
Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1.
Answer:
The molecular formula of potassium chlorate is KClO₃.
Required chemical equation:

2 moles of KClO₃ = 2 × 122.5 = 245 g
3 moles of O₂ at STP occupy = (3 × 22.4 dm³) = 67.2 dm³
Thus, 245 g of potassium chlorate will liberate 67.2 dm³ of oxygen gas.
Let ‘x’ gram of KClO₃ liberate 6.72 dm³ of oxygen gas at S.T.P.
∴ x = \(\frac{245 \times 6.72}{67.2}\) = 24.5 g
Ans: Mass of potassium chlorate required = 24.5 g

Question Q.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH₂)₂CO. Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH₂)₂CO
Molar mass of urea = 60 g mol^-1

∴ Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 2.247 × 10²³ atoms of hydrogen
∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 1.124 × 10²³ atoms of nitrogen
∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of carbon
∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 10²³ atoms of H, 1.124 × 10²³ atoms of N, 0.562 × 10²³ atoms of C and 0.562 × 10²³ atoms of O.

Question R.
Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Answer:
Given: Mass of sulphur (reactant) = 16 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 32 g oxygen to form 64 g of sulphur dioxide as follows:

Hence, (0.5 × 32 = 16 g) of sulphur will combine with (0.5 × 32 = 16 g) of oxygen to give (0.5 × 64 = 32 g) sulphur dioxide.
Ans: Mass of sulphur dioxide produced = 32 g

5. Explain

Question A.
The need of the term average atomic mass.
Answer:

• Several naturally occurring elements exist as a mixture of two or more isotopes.
• Isotopes have different atomic masses.
• The atomic mass of such an element is the average of atomic masses of its isotopes.
• For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.

Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.

Question B.
Molar mass.
Answer:
i. The mass of one mole of a substance (element/compound) in grams is called its molar mass.
ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
e.g.

iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
e.g.

Question C.
Mole concept.
Answer:

• Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
• Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon -12 isotope.
• One mole is the amount of substance which contains 6.0221367 × 10²³ particles/entities.

Question D.
Formula mass with an example.
Answer:

• The formula mass of a substance is the sum of atomic masses of the atoms present in the formula.
• In substances such as sodium chloride, positive (sodium), and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na⁺) ion is surrounded by six chlorides (Cl^–) ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units.
• Therefore, NaCl is just the formula that is used to represent sodium chloride though it is not a molecule.
• In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass.

e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

Question E.
Molar volume of gas.
Answer:
i. It is more convenient to measure the volume rather than mass of the gas.
ii. It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm³ at standard temperature (0 °C) and pressure (1 atm) (STP).
iii. The volume of 22.4 dm³ at STP is known as molar volume of a gas.
iv. The relationship between number of moles and molar volume can be expressed as follows:

[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol^-1]

Question F.
Types of matter (on the basis of chemical composition).
Answer:
Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.

They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
1. Metals:

• They have a lustre (a shiny appearance).
• They conduct heat and electricity.
• They can be drawn into wire (ductile).
• They can be hammered into thin sheets (malleable).
• e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.

2. Nonmetals:

• They have no lustre, (except diamond, iodine)
• They are poor conductors of heat and electricity, (except graphite)
• They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine

3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.

ii. Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.

Mixtures are of two types:

• Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
  e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
• Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
  e.g. Suspension, which contains insoluble solid in a liquid.

11th Chemistry Digest Chapter 1 Some Basic Concepts of Chemistry Intext Questions and Answers

Can you tell? (Textbook Page No. 1)

Question 1.
Which are mixtures and pure substances from the following?
i. Sea water
ii. Gasoline
iii. Skin
iv. A rusty nail
v. A page of textbook
vi. Diamond
Answer:

Can you tell? (Textbook Page No. 2)

Question 1.
Classify the following as element and compound.
i. Mercuric oxide
ii. Helium gas
iii. Water
iv. Table salt
v. Iodine
vi. Mercury
vii. Oxygen
viii. Nitrogen
Answer:

Can you tell? (Textbook Page No. 6)

Question 1.
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:

If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced.

Can you recall? (Textbook Page No. 6)

Question 1.
What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:

• The smallest indivisible particle of an element is called an atom.
• A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
• Every atom of an element has definite mass. The order of magnitude of mass of one atom is 10^-27 kg.
• Isotopes are the atoms of the same element having same atomic number but different mass number.

Try this (Textbook Page No. 8)

Question 1.
Find the formula mass of CaSO₄, if atomic mass of Ca = 40.1 u, S =32.1 u and O = 16.0 u.
Solution:
Formula mass of CaSO₄
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= (40.1) + 32.1 + (4 × 16.0) = 136.2 u
Ans: Formula mass of CaSO₄ = 136.2 u

Can you recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items.

Try this (Textbook Page No. 10)

Question 1.
Calculate the volume in dm³ occupied by 60.0 g of ethane at STP.
Solution:
Given: Mass of ethane at STP = 60.0 g
To find: Volume of ethane
Formulae:

Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP
Molecular mass of ethane = 30 g mol^-1

∴ Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 2 mol × 22.4 dm³ mol^-1 = 44.8 dm³
Ans: Volume of ethane = 44.8 dm³

Activity :

Activity 1.
Collect information from various scientists and prepare charts of their contributions to chemistry.
Answer:

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Semiconductor Devices Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 16 Semiconductor Devices Textbook Exercise Questions and Answers.

Class 12 Physics Chapter 16 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 16 Exercise Solutions

1. Choose the correct option

i.
In a BJT, the largest current flow occurs
(A) in the emitter
(B) in the collector
(C) in the base
(D) through CB junction.
Answer:
(A) in the emitter

ii.
A series resistance is connected in the Zener diode circuit to
(A) properly reverse bias the Zener
(B) protect the Zener
(C) properly forward bias the Zener
(D) protect the load resistance.
Answer:
(A) properly reverse bias the Zener

iii.
An LED emits visible light when its
(A) junction is reverse biased
(B) depletion region widens
(C) holes and electrons recombine
(D) junction becomes hot.
Answer:
(C) holes and electrons recombine

iv.
Solar cell operates on the principle of
(A) diffusion
(B) recombination
(C) photovoltaic action
(D) carrier flow.
Answer:
(C) photovoltaic action

v.
A logic gate is an electronic circuit which
(A) makes logical decisions
(B) allows electron flow only in one direction
(C) works using binary algebra
(D) alternates between 0 and 1 value.
Answer:
(A) makes logical decisions

2 Answer in brief.

i.
Why is the base of a transistor made thin and is lightly doped?
Answer:
The base of a transistor is lightly doped than the emitter and is made narrow so that virtually all the electrons injected from the emitter (in an npn tran-sistor) diffuse right across the base to the collector junction without recombining with holes. That is, the base width is kept less than the recombination distance. Also, the emitter is much heavily doped than the base to improve emitter efficiency and common-base current gain a.

ii.
How is a Zener diode different than an ordinary diode?
Answer:
A Zener diode is heavily doped-the doping con-centrations for both p- and n-regions is greater than 1018 cm-3 while those of an ordinary diode are voltage (PIV) of an ordinary diode is higher than a Zener diode and the breakdown occurs by impact ionization (avalanche process). Their I-V characteristics are otherwise similar.

iii.
On which factors does the wavelength of light emitted by a LED depend?
Answer:
The intensity of the emitted light is directly propor-tional to the recombination rate and hence to the diode forward current. The colour of the light emitted by an LED depends on the compound semiconductor material used and its composition (and doping levels) as given below :
Table: Typical semiconductor materials and emitted colours of LEDs

iv.
Why should a photodiode be operated in reverse biased mode?
Answer:
A photodiode is operated in a reverse biased mode because as photodetector or photosensor, it must conduct only when radiation is incident on it. In the reverse biased mode, the dark current for zero illumination is negligibly small—of the order of few picoamperes to nanoamperes. But when illuminated, the photocurrent is several orders of magnitude greater.

v.
State the principle and uses of a solar Cell.
Answer:
A solar cell is an unbiased pn-junction that converts the energy of sunlight directly into electricity with a high conversion efficiency.

Principle : A solar cell works on the photovoltaic effect in which an emf is produced between the two layers of a pn-junction as a result of irradiation.

Question 3.
Draw the circuit diagram of a half wave rectifier. Explain its working. What is the frequency of ripple in its output?
Answer:
A device or a circuit which rectifies only one-half of each. cycle of an alternating voltage is called a half-wave rectifier.
Electric circuit : The alternating voltage to be rectified is applied across the primary coil (P₁P₂) of a transformer. The secondary coil (S₁S₂) of the transformer is connected in series with the junction diode and a load resistance R_L, as shown in below figure. The alternating voltage across the secondary coil is the ac input voltage V_i. The dc voltage across the load resistance is called the output voltage V₀.

Working : Due to the alternating voltage V_i, the p-region of the diode becomes alternatively positive and negative with respect to the n-region.
During the half-cycle when the p-region is positive, the diode is forward biased and conducts. A current I_L passes through the load resistance RL in the direction shown.

During the next half cycle, when the p-region is negative, the diode is reverse biased and the forward current drops to zero.

Thus, the diode conducts only during one-half of the input cycle and thus acts as a half-wave rectifier. The intermittent output voltage V₀ has a fixed polarity but changes periodically with time between zero and a maximum value. I_L is unidirectional. Above figure shows the input and output voltage waveforms.

The pulsating dc output voltage of a half-wave rectifier has the same frequency as the input.

Question 4.
Why do we need filters in a power supply?
Answer:
A rectifier-half-wave or full-wave – outputs a pul-sating dc which is not directly usable in most electronic circuits. These circuits require something closer to pure dc as produced by batteries. Unlike pure dc waveform of a battery, a rectifier output has an ac ripple riding on a dc waveform.

The circuit used in a dc power supply to remove the ripple is called a filter. A filter circuit can produce a very smooth waveform that approximates the waveform produced by a battery. The most common technique used for filtering is a capacitor connected across the output of a rectifier.

Question 5.
Draw a neat diagram of a full wave rectifier and explain it’s working.
Answer:
A device or a circuit which rectifies both halves of each cycle of an alternating voltage is called a full-wave rectifier.
Electric circuit : The alternating voltage to be rectified is applied across the primary coil (P₁P₂) of a transformer with a centre-tapped secondary coil (S₁S₂). The terminals and S2 of the secondary are connected to the two p-regions of two junction diodes D₁ and D₂, respectively. The centre-tap T is connected to the ground. The load resistance RL is connected across the common n-regions and the

P₁P₂, S₁S₂ : Primary and secondary of transformer,
T : Centre-tap on secondary; D₁ D₂ : Junction diodes,
R_L : Load resistance, I_L : Load current,
V_i: AC input voltage, V₀ : DC output voltage
Above Figure : Full-wave rectifier circuit

Working : During one half cycle of the input, terminal S₁ of the secondary is positive while S₂ is negative with respect to the ground (the centre-tap T). During this half cycle, diode D₁ is forward biased and conducts, while diode D₂ is reverse biased and does not conduct. The direction of current Z_L through R_L is in the sense shown.

During the next half cycle of the input voltage, S2 becomes positive while S, is negative with respect to T. Diode D2 now conducts sending a current IL through RL in the same sense as before. Dt now does not conduct. Thus, the current through RL flows in the same direction, i.e., it is unidirectional, for both halves or the full-wave of the input. This is called full-wave rectification.

The output voltage has a fixed polarity but varies periodically with time between zero and a maximum value. Above figure shows the input and output voltage waveforms. The pulsating dc output voltage of a full-wave rectifier has twice frequency of the input.

Question 6.
Explain how a Zener diode maintains constant voltage across a load.
Answer:
Principle : In the breakdown region of a Zener diode, for widely changing Zener current, the voltage across the Zener diode remains almost constant.

Electric circuit : The circuit for regulating or stabilizing the voltage across a load resistance RL against change in load current and supply voltage is shown in above figure. The Zener diode is connected parallel to load Rh such that the current through the Zener diode is from the n to p region. The series resistance Rs limits the current through the diode below the maximum rated value.
From the circuit, I = I_Z + I_L and V = IR_s + V_Z
= (I_Z + I_L)R_s + V_Z
Working: When the input unregulated dc voltage V across the Zener diode is greater than the Zener voltage V_Z in magnitude, the diode works in the Zener breakdown region. The voltage across the diode and load Rh is then V_Z. The corresponding current in the diode is I_Z.

As the load current (I) or supply voltage (V) changes, the diode current (7Z) adjusts itself at constant V_Z. The excess voltage V-V_Zappears across the series resistance Rs.

For constant supply voltage, the supply current I and the voltage drop across R_s remain constant. If the diode is within its regulating range, an increase in load current is accompanied by a decrease in Iz at constant V_Z.
Since the voltage across R_L remains constant at V_Z, the Zener diode acts as a voltage stabilizer or voltage regulator.

Question 7.
Explain the forward and the reverse characteristic of a Zener diode.
Answer:
The forward bias region of a Zener diode is identical to that of a regular diode. There is forward current only after the barrier potential of the pn- junction is overcome. Beyond this threshold or cut in voltage, there is an exponential upward swing.

The typical forward voltage at room temperature with a current of around 1 mA is around 0.6 V.

In the reverse bias condition the Zener diode is an open circuit and only a small reverse saturation current flows as shown with change of scale. At the reverse breakdown voltage there is an abrupt rapid increase in the current-the knee is very sharp, followed by an almost vertical increase in current. The voltage across the Zener diode in the breakdown region is very nearly constant with only a small increase in voltage with increasing current. There is a minimum Zener current, I_Z (min), that places the operating point in the desired breakdown region. At some high current level, I_ZM, the power dissipation of the diode becomes excessive beyond which the diode can be damaged.

Zener diode characteristics

The I-V characteristics of a Zener diode is not totally vertical in the breakdown region. This means that for slight changes in current, there will be a small change in the voltage across the diode. The voltage change for a given change in current is the resistance R_Z of the Zener diode.

Question 8.
Explain the working of a LED.
Answer:
Working :
An LED is forward-biased with about 1.2 V to 3.6 V at 12 mA to 20 mA. Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p-layer, some of these excess minority carriers electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of diferent energies.

In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy hv.
[Note : The photons originate primarily in the p-side of the junction which has a bandgap E_Gp narrower than that of the n-side, E_Gn. Thus, with hv < E_Gn, the photons are emitted through the wide-bandgap n-region with essentially no absorption.]

Question 9.
Explain the construction and working of solar cell.
Answer:
Construction :
A simple pn-junction solar cell con-sists of a p-type semiconductor substrate backed with a metal electrode back contact. A thin n-layer (less than 2.5 pm, for silicon) is grown over the p-type substrate by doping with suitable donor impurity. Metal finger electrodes are prepared on top of the n-layer so that there is enough space between the fingers for sunlight to reach the n-layer and, subsequently, the underlying pn-junction.

Working : When exposed to sunlight, the absorption of incident radiation (in the range near-UV to infrared) creates electron-hole pairs in and near the depletion layer.

Consider light of frequency v incident on the pn-junction such that the incident photon energy hv is greater than the band gap energy EG of the semiconductor. The photons excite electrons from the valence band to the conduction band, leaving vacancies or holes in the valence band, thus generating electron-hole pairs.

The photogenerated electrons and holes move towards the n side and p side, respectively. If no external load is connected, these photogenerated charges get collected at the two sides of the junction and give rise to a forward photovoltage. In a closed- circuit, a current I passes through the external load as long as the solar cell is exposed to sunlight.

A solar cell module consists of several solar cells connected in series for a higher voltage output. For outdoor use with higher power output, these modules are connected in different series and parallel combinations to form a solar cell array.

[Note : Currently most of the crystalline solar cells are p-type as described above. This is because of a lower cost of production of p-type. But performance wise, n-type solar cells (a thin p-layer over an n-type substrate by doping with suitable acceptor impurity) can give much better efficiency compared to p-type solar cells.]

Question 10.
Explain the principle of operation of a photodiode.
Answer:
Construction:
A photodiode consists of an n-type silicon substrate with a metal electrode back contact. A thin p-type layer is grown over the n-type substrate by diffusing a suitable acceptor dopant.

The area of the p-layer defines the photodiode active area. An ohmic contact pad is deposited on the active area. The rest of the active area is left open with a protective antireflective coating of silicon nitride to minimize the loss of photons. The nonactive area is covered with an insulating opaque SiO₂ coating.

Depending on the required spectral sensitivity, i.e., the operating wavelength range, typical photodiode materials are silicon, germanium, indium gallium arsenide phosphide (InGaAsP) and indium gallium arsenide (InGaAs), of which silicon is the cheapest while the last two are expensive.

Working : The band gap energy of silicon is E_G = 1.12 eV at room temperature. Thus, photons or particles with energies greater than or equal to 1.12 eV, which corresponds to 110 nm, can transfer electrons from the valence band into the conduction band.

A photodiode is operated in the reverse bias mode which results in a wider depletion region. When operated in the dark (zero illumination), there is a reverse saturation current due solely to the thermally generated minority charge carriers. This is called the dark current. Depending on the minority carrier concentrations, the dark current in an Si photodiode may range from 5 pA to 10 nA.

When exposed to radiation of energy hv ≥ E_G (in the range near-UV to near-IR), electron-hole pairs are created in the depletion region. The electric field in the depletion layer accelerates these photogenerated electrons and holes towards the n side and p side, respectively, constituting a photocurrent l in the external circuit from the p side to the n side. Due to the photogeneration, more charge carriers are available for conduction and the reverse current is increased. The photocurrent is directly propor-tional to the intensity of the incident light. It is independent of the reverse bias voltage.
[Notes : Typical photodiode materials are :
(1) silicon (Si) : low dark current, high speed, good sensitivity between ~ 400 nm and 1000 run (best around 800 nm-900 nm)
(2) germanium (Ge) : high dark current, slow speed, good sensitivity between ~ 900 nm and 1600 nm (best around 1400 nm-1500 nm)
(3) indium gallium arsenide phosphide (InGaAsP) : expensive, low dark current, high speed, good sensitivity between ~ 1000 nm and 1350 nm (best around 1100 nm- 1300 nm)
(4) indium gallium arsenide (InGaAs): expensive, low dark current, high speed, good sensitivity between ~ 900 nm and 1700 nm (best around 1300 nm-1600 nm],

Question 11.
What do you mean by a logic gate, a truth table and a Boolean expression?
Answer:
A logic gate is a basic switching circuit used in digital circuits that determines when an input pulse can pass through to the output. It generates a single output from one or more inputs.

Explanation/Uses :
Any digital computation process consists in performing a sequence of arithmetical operations on the data of the problem. At each stage in the computation, the nature of the operation to be performed is decided partly by the pre-determined program and partly by the outcome of earlier stages in the process. We therefore need switches with multiple inputs to perform logical operations, i.e., the outputs of these switches are determined in specified ways by the condition (binary state) of their inputs. These arrangements are known as logic gates, and mostly they are extension of a simple transistor switch.

(1) Boolean expression : An equation expressing a logical compound statement in Boolean algebra is called a Boolean expression. A Boolean expression for a logic gate expresses the relation between input(s) and output of a logic gate.

(2) Truth table : The table which shows the truth values of a Boolean expression for a logic gate for all possible combinations of its inputs is called the truth table of logic gate.

The truth table contains one row for each input combination. Since a logical variable can assume only two possible values, 0 and 1, there are 2N combinations of N inputs so that the table has 2N rows.
[Note : Boolean algebra is a form of symbolic logic developed in 1847 by George Boole (1815-64) British mathematician.]

Question 12.
What is logic gate? Write down the truth table and Boolean expression for ‘AND’ gate.
Answer:
A logic gate is a basic switching circuit used in digital circuits that determines when an input pulse can pass through to the output. It generates a single output from one or more inputs.

The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.

The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.
Logic symbol:

Truth table:

Boolean expression:
Y = A ∙ B

Question 13.
What are the uses of logic gates? Why is a NOT gate known as an inverter?
Answer:
Explanation/Uses :
Any digital computation process consists in performing a sequence of arithmetical operations on the data of the problem. At each stage in the computation, the nature of the operation to be performed is decided partly by the pre-determined program and partly by the outcome of earlier stages in the process. We therefore need switches with multiple inputs to perform logical operations, i.e., the outputs of these switches are determined in specified ways by the condition (binary state) of their inputs. These arrangements are known as logic gates, and mostly they are extension of a simple transistor switch.

The NOT gate or INVERTER : It is a circuit with one input whose output is HIGH if the input is LOW and vice versa.

The NOT operation outputs an inverted version of the input. Hence, a NOT gate is also known as an INVERTER.

The small invert bubble on the output side of the inverter logic symbol, below figure and the over bar (^–) in the Boolean expression represent the invert function.

Question 14.
Write the Boolean expression for (i) OR gate, (ii) AND gate, and (iii) NAND Gate.
Answer:
(i) The OR gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if any one or more of the inputs is HIGH.
The OR operation represents a logical addition.
Below figure shows the 2-input OR gate logic sym-bol, and the Boolean expression and the truth table for the OR function.

(ii) The AND gate : It is a circuit with two or more inputs and one output in which the output signal is HIGH if and only if all the inputs are HIGH simultaneously.

The AND operation represents a logical multiplication.

Below figure shows the 2-input AND gate logic symbol and the Boolean expression and the truth table for the AND function.
Logic symbol:

Truth table:

Boolean expression:
Y = A ∙ B

(iii) The NAND gate : It is a circuit with two or more inputs and one output, whose output is HIGH if any one or more of the inputs is LOW; the output is LOW if all the inputs are HIGH.

The NAND gate is a combination of an AND gate followed by a NOT gate so that the truth table of the NAND function is obtained by inverting the outputs of the AND gate.

Question 15.
Why is the emitter, the base and the collector of a BJT doped differently?
Answer:
A BJT being a bipolar device, both electrons and holes participate in the conduction process. Under the forward-biased condition, the majority carriers injected from the emitter into the base constitute the largest current component in a BJT. For these carriers to diffuse across the base region with t negligible recombination and reach the collector junction, these must overwhelm the majority carriers of opposite charge in the base. The total emitter current has two components, that due to majority carriers in the emitter and that due to minority carriers diffused from base into emitter. The ratio of the current component due to the injected majority carriers from the emitter to the total emitter current is a measure of the emitter efficiency. To improve the emitter efficiency and the common-base current gain (a), it can be shown that’ the emitter should be much heavily doped than the base.

Also, the base width is a function of the base- collector voltage. A low doping level of the collector increases the size of the depletion region. This increases the maximum collector-base voltage and reduces the base width. Further, the large depletion region at the collector-base junction-extending mainly into the collector-corresponds to a smaller electric field and avoids avalanche breakdown of the reverse-biased collector-base junction.
[Note : Effective dopant concentrations of (a) npn transistor (b) pnp transistor are shown below.

The base doping is less than the emitter doping but greater than the collector doping. Contrary to the impression stressed in the Board’s and NCERT textbooks, collector doping is typically an order of magnitude lower than base doping. {Ref. : Semiconductor Devices Physics and Technology (3rd Edition), Simon M. Sze and M. K. Lee, p. 125}]

Question 16.
Which method of biasing is used for operating transistor as an amplifier?
Answer:
For use as an amplifier, the transistor should be in active mode. Therefore, the emitter-base junction is forward biased and the collector-base junction is reverse biased. Also, an amplifier uses an emitter bias rather than a base bias.

Question 17.
Define α and β. Derive the relation between then.
Answer:
The dc common-base current ratio or current gain (α_dc) is defined as the ratio of the collector current to emitter current.
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\)
The dc common-emitter current ratio or current gain (β_dc) is defined as the ratio of the collector current to base current.
β_dc = \(\frac{I_{C}}{I_{B}}\)
Since the emitter current I_E = I_B + I_C
\(\frac{I_{\mathrm{E}}}{I_{C}}=\frac{I_{\mathrm{B}}}{I_{\mathrm{C}}}+1\)
∴ \(\frac{1}{\alpha_{\mathrm{dc}}}=\frac{1}{\beta_{\mathrm{dc}}}+1\)
Therefore, the common-base current gain in terms of the common-emitter current gain is
α_dc = \(\frac{\beta_{\mathrm{dc}}}{1+\beta_{\mathrm{dc}}}\)
and the common-emitter current gain in terms of the common-base current gain is
β_dc = \(\frac{\alpha_{\mathrm{dc}}}{1-\alpha_{\mathrm{dc}}}\)
For a transistor, α_dc is close to but always less than 1 (about 0.92 to 0.98) and β_dc ranges from 20 to 200 for most general purpose transistors.

Question 18.
The common-base DC current gain of a transistor is 0.967. If the emitter current is 10mA. What is the value of base current?
Answer:
Data : α_dc = 0.967, I_E = 10 mA
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\) and I_E = I_B + I_C
The collector current,
I_C = α_dcI_E = 0.967 × 10 = 9.67 mA
Therefore, the base current,
I_B = I_E – I_C = 10 – 9.67 = 0.33 mA

Question 19.
In a comman-base connection, a certain transistor has an emitter current of 10mA and collector current of 9.8 mA. Calculate the value of the base current.
Answer:
DATA : I_E = 10 mA, I_C = 9.8 mA
I_E = I_B + I_C
Therefore, the base current,
I_B = I_E – I_C – 10 – 9.8 = 0.2 mA

Question 20.
In a common-base connection, the emitter current is 6.28mA and collector current is 6.20 mA. Determine the common base DC current gain.
Answer:
Data : I_E = 6.28 mA, I_C = 6.20 mA
α_dc = \(\frac{I_{C}}{I_{\mathrm{E}}}\) and β_dc = \(\frac{I_{\mathrm{C}}}{I_{\mathrm{B}}}=\frac{\alpha_{\mathrm{dc}}}{1-\alpha_{\mathrm{dc}}}\)
Common-emitter current gain, α_dc = \(\frac{6.20}{6.28}\) = 0.9873
Therefore, common-base current gain,
β_dc = \(\frac{0.9873}{1-0.9873}=\frac{0.9873}{0.0127}\) = 77.74
OR
I_E = I_B + I_C
∴ I_B = I_E – I_C = 6.28 – 6.20 = 0.08 mA
∴ β_dc = \(\frac{6.20}{0.08}\) = 77.5
[Note : The answer given in the textbook obviously refers to the common-emitter current gain.]

12th Physics Digest Chapter 16 Semiconductor Devices Intext Questions and Answers

Remember this (Textbook Page No. 346)

Question 1.
A full wave rectifier utilises both half cycles of AC input voltage to produce the DC output.
Answer:
A half-wave rectifier rectifies only one half of each cycle of the input ac wave while a full-wave rectifier rectifies both the halves. Hence the pulsating dc output voltage of a half-wave rectifier has the same frequency as the input but that of a full-wave rectifier has double the frequency of the ac input.

Do you know (Textbook Page No. 346)

Question 1.
The maximum efficiency of a full wave rectifier is 81.2% and the maximum efficiency of a half wave rectifier is 40.6%. It is observed that the maximum efficiency of a full wave rectifier is twice that of half wave rectifier.
Answer:
The ratio of dc power obtained at the output to the applied input ac power is known as rectifier efficiency. A half-wave rectifier can convert maximum 40.6% of ac power into dc power, and the remaining power of 59.4% is lost in the rectifier circuit. In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit. Hence, a half wave rectifier efficiency is 40.6%. The maximum efficiency of a full-wave rectifier is 81.2%, i.e., twice that of a half-wave rectifier.

Do you know (Textbook Page No. 349)

Question 1.
The voltage stabilization is effective when there is a minimum Zener current. The Zener diode must be always operated within its breakdown region when there is a load connected in the circuit. Similarly, the supply voltage V_s must be greater than V_z.
Answer:
A Zener diode is operated in the breakdown region. There is a minimum Zener current, I_z, that places the desired operating point in the breakdown region. There is a maximum Zener current, I_zM, at which the power dissipation drives the junction temperature to the maximum allowed. Beyond that current the diode can be damaged. Hence, the supply voltage must be greater than V_z and the current-limiting resistor must limit the diode current to less than the rated maxi mum, I_zM.

Remember this (Textbook Page No. 350)

Question 1.
Zener effect occurs only if the diode is heavily doped, because when the depletion layer is thin, breakdown occurs at low reverse voltage and the field strength will be approximately 3 × 10⁷ V/m. It causes an increase in the flow of free carriers and increase in the reverse current.
Answer:
Zener breakdown occurs only in heavily doped pn junctions (doping concentrations for both p- and n-regions greater than 1018 cm3) and can take place only if the electric field in the depletion region of the reverse-biased junction is very high. It is found that the critical field at which tunneling becomes probable, i.e., at which Zener breakdown commences, is approximately 10⁶ V/cm. [“internal Field Emissiot at Narrow Silicon and Germanium PN-Junctions,” Phys. Rev., 118, 425 (1960).]

Can you tell (Textbook Page No. 350)

Question 1.
How does a cell phone charger produce a voltage of 5.0 V form the line voltage of 230V?
Answer:
A phone charger is usually a 5 V power supply. A 4-diode bridge input rectifier rectifies the ac mains voltage a provide a high voltage dc. A transistor chopper switches this on and off at high frequency. This stage is required because this high frequency allows the transformer to be smaller, lighter and much lower in cost.

A small transformer steps this down to a low voltage high-frequency ac. An output rectifier and filter convert this to low-voltage (5 V) dc and smooths out the ripple. A chopper controller provides a feedback to the chopper through an optoisohitor and adjusts the chopping cycle to maintain the output voltage at 5 V.

Question 2.
Why is a resistance connected in series with a Zener diode when used in a circuit?
Answer:
The I-V characteristics in the breakdown region of a Zener diode is almost vertical. That is, the current I_Z can rapidly increase at constant V_Z. To prevent damage due to excessive heating, the Zener current should not exceed the rated maximum current, I_ZM. Hence, a current-limiting resistor R_s is connected in series with the diode.

I_Z and the power dissipated in the Zener diode will be large for I _L = 0 (no-load condition) or when I_L is less than the rated maximum (when R_s is small and R_L is large). The current-limiting resistor R_s is so chosen that the Zener current does not exceed the rated maximum reverse current, I_ZM when there is no load or when the load is very high.
The rated maximum power of a Zener diode is
P_ZM = I_ZM = V_Z

At n-load condition, the current through R is I = I_ZM and the voltage drop across it is V – V_Z, where V is the unregulated source voltage. The diode current will be maximum when V is maxi mum at V_max and I = I_ZM. Then, the minimum value of the series resistance should be
R_{s, min} = \(\frac{V_{\max }-V_{\mathrm{Z}}}{I_{\mathrm{ZM}}}\)

Question 3.
The voltage across a Zener diode does not remain strictly constant with the changes in the Zener current. This is due to R_Z, the Zener impedance, or the internal resistance of the Zener diode. R_Z acts like a small resistance in series with the Zener. Changes in I_Z cause small changes in V_Z .
Answer:
The I-V characteristics of a Zener diode in the breakdown region is not strictly vertical. Its slope is 1/R_Z, where R_Z is the Zener impedance.

Can you know (Textbook Page No. 354)

Question 1.
What is the difference between a photo diode and a solar cell?
Answer:
Both are semiconductor photovoltaic devices. A photodiode is a reverse-biased pn-junction diode while a solar cell is an unbiased pn-junction diode. Photod iodes, however, are optimized for light detection while solar cells are optimized for energy conversion efficiency.

Question 2.
When the intensity of light incident on a photo diode increases, how is the reverse current affected?
Answer:
The photocurrent increases linearly with increasing illuminance, limited by the power dissipation of the photodiode.

Do you know (Textbook Page No. 355)

Question 1.
LED junction does not actually emit that much light so the epoxy resin body is constructed in such a way that the photons emitted by the junction are reflected away from the surrounding substrate base to which the diode is attached and are focused upwards through the domed top of the LED, which itself acts like a lens concentrating the light. This is why the emitted light appears to be brightest at the top of the LED.
Answer:
The pn-junction of an LED is encased in a transparent, hard plastic (epoxy resin), not only for shock protection but also for enhancing the brightness in one direction. Light emitted by the pn-junction is not directional. The hemispherical epoxy lens focuses the light in the direction of the hemispherical part. This is why the emitted light appears to be brightest at the top of the LED.

Question 2.
The current rating of LED is of a few tens of milli-amps. Hence it is necessary to connect a high resistance in series with it. The forward voltage drop of an LED is much larger than an ordinary diode and is around 1.5 to 3.5 volts.
Answer:
Most common LEDs require a forward operating voltage of between approximately 1.2 V (for a standard red LED) to 3.6 V (for a blue LED) with a forward current rating of about 10 mA to 30 mA, with 12 mA to 20 mA being the most common range. Like any diode, the forward current is approximately an exponential function of voltage and the forward resistance is very small. A small voltage change may result in a large change in current. If the current exceeds the rated maximum, an LED may overheat and get destroyed. LEDs are current driven devices and a current-limiting series resistor is required to prevent burning up the LED.

Do you know (Textbook Page No. 356)

Question 1.
White Light LEDs or White LED Lamps:
Shuji Nakamura, a Japanese – born American electronic engineer invented the blue LED. He was awarded the Nobel prize for physics for 2014. He was also awarded the global energy prize in the year 2015. His invention of blue LED made the fabrication of white LED possible.
LED lamps, bulbs, street lighting are becoming very popular these days because of the very high efficiency of LEDs in terms of light output per unit input power(in milliWatts), as compared to the incandescent bulbs. So for general purpose lightings, white light is preferred.
Commercially available white LEDs are normally manufactured by using the technique of wavelength conversion. It is a process which partly or completely converts the radiation of a LED into white light. There are many ways of wavelength conversion. One of these methods uses blue LED and yellow phosphor. In this method of wavelength conversion, a LED which emits blue colour is used to excite a yellow colour phosphor. This results in the emission of yellow and blue light and this mixture of blue and yellow light gives the appearance of white light. This method is the least expensive method for producing white light.
Answer:

The all important blue LEDs
The development of LEDs has made more efficient light sources possible. Creating white light that can be used for lighting requires a combination of red, green and blue light. Blue LEDs proved to be much more difficult to create than red and green LEDs. During the 1980s and 1990s Isamu Akasaki, Hiroshi Amano, and Shuji Nakamura successfully used the difficult-to-handle semiconductor gallium nitride to create efficient blue LEDs. Isamu Akasaki is known for invent ing the bright gallium nitride (CaN) pn-junction blue LED in 1989 and subsequently the high-brightness CaN blue LED.

Using blue LEDs, highly efficient white light sources. became possible by converting part of the blue light emitted from an LED to yellow using a phosphor. To the human eye, the combination of blue and yellow light is perceived as white. A white LED can be created by embedding phosphors in the plastic cap which surrounds a blue LED. Higher quality white light can also be created by mixing blue light with other colors as well, including red and green

Isamu Akasaki, together with Shuji Nakamura and Hiroshi Amano, received the 2014 Nobel Prize in Physics for the invention of efficient blue light-emitting diodes which has enabled bright and energy saving white light sources.

Use your brain power (Textbook Page No. 357)

Question 1.
What would happen if both junctions of a BJT are forward biased or reverse biased?
Answer:
A BJT has four regimes of operation, depending on the four combinations of the applied biases (voltage polarities) to the emitter-base junction and the collector-base junction, as shown in the following table; ‘F’ and ‘R’ indicate forward bias and reverse bias, respectively.

Remember This (Textbook Page No. 358)

Question 1.
The lightly doped, thin base region sandwiched between the heavily doped emitter region and the intermediate doped collector region plays a crucial role in the transistor action.
Answer:
If the two junctions in a BJT are physically close compared with the minority carrier diffusion length (i.e., the distance within which recombination will take place), the careers injected from the emitter can diffuse through the base to reach the base-collector junction. The narrow width of the base is thus crucial for transistor action.

Use your brain power (Textbook Page No. 361)

Question 1.
If a transistor amplifies power, explain why it is not used to generate power.
The term ‘amplification’ is used as an abstraction of the transistor properties so that we have few equations which are useful for a large number of practical problems. Transistors use a small power to control a power supply which can output a huge power. The large output comes from the power supply, while the input signal valves the transistor on and off. The increased power comes from the power supply so that a transistor does not violate the law of conservation of energy.

Maharashtra State Board Class 12 Textbook Solutions

• Current Electricity Class 12 Physics Textbook Solutions
• Magnetic Fields due to Electric Current Class 12 Physics Textbook Solutions
• Magnetic Materials Class 12 Physics Textbook Solutions
• Electromagnetic Induction Class 12 Physics Textbook Solutions
• AC Circuits Class 12 Physics Textbook Solutions
• Dual Nature of Radiation and Matter Class 12 Physics Textbook Solutions
• Structure of Atoms and Nuclei Class 12 Physics Textbook Solutions
• Semiconductor Devices Class 12 Physics Textbook Solutions

Structure of Atoms and Nuclei Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 15 Structure of Atoms and Nuclei Textbook Exercise Questions and Answers.

Class 12 Physics Chapter 15 Exercise Solutions Maharashtra Board

Physics Class 12 Chapter 15 Exercise Solutions

In solving problems, use m_e = 0.00055 u = 0.5110 MeV/c², m_p = 1.00728 u, m_n = 1.00866u, m_H = 1.007825 u, u = 931.5 MeV, e = 1.602 × 10^-19 C, h = 6.626 × 10^-34 Js, ε₀ = 8.854 × 10^-12 SI units and me = 9.109 × 10^-31 kg.

1. Choose the correct option.

i) In which of the following systems will the radius of the first orbit of the electron be the smallest?
(A) hydrogen
(B) singly ionized helium
(C) deuteron
(D) tritium
Answer:
(D) tritium

ii) The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(B) 4

iii) In the spectrum of hydrogen atom which transition will yield longest wavelength?
(A) n = 2 to n = 1
(B) n = 5 to n = 4
(C) n = 7 to n = 6
(D) n = 8 to n = 7
Answer:
(D) n = 8 to n = 7

iv) Which of the following properties of a nucleus does not depend on its mass number?
(A) radius
(B) mass
(C) volume
(D) density
Answer:
(D) density

v) If the number of nuclei in a radioactive sample at a given time is N, what will be the number at the end of two half-lives?
(A) \(\frac{N}{2}\)
(B) \(\frac{N}{4}\)
(C) \(\frac{3N}{4}\)
(D) \(\frac{N}{8}\)
Answer:
(B) \(\frac{N}{4}\)

2. Answer in brief.

i) State the postulates of Bohr’s atomic model.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :

1. The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
2. The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
3. Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

ii) State the difficulties faced by Rutherford’s atomic model.
Answer:
(1) According to Rutherford, the electrons revolve in circular orbits around the atomic nucleus. The circular motion is an accelerated motion. According to the classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion, should go on radiating energy. Due to the loss of energy, the radius of its orbit should go on decreasing. Therefore, the electron should move along a spiral path and finally fall into the nucleus in a very short time, of the order of 10^-16 s in the case of a hydrogen atom. Thus, the atom should be unstable. We exist because atoms are stable.

(2) If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution of the electron would go on increasing. The electron, therefore, would emit radiation of continuously changing frequency, and hence give rise to a con-tinuous spectrum. However, atomic spectrum is a line spectrum.

iii) What are alpha, beta and gamma decays?
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.
Answer:
(1) Excitation energy of an electron in an atom : The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state.

(2) Binding energy of an electron in an atom is defined as the minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero.

(3) Ionization energy of an electron in an atom is defined as the minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero.

v) Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.
Answer:
For the first line in the Lyman series,
\(\frac{1}{\lambda_{\mathrm{L} 1}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}\)
∴ v_L1 = \(\frac{c}{\lambda_{\mathrm{L} 1}}=\frac{3 R_{c}}{4}\), where v denotes the frequency,
c the speed of light in free space and R the Rydberg constant.
For the limit of the Lyman series,

Hence, the result.

Question 3.
State the postulates of Bohr’s atomic model and derive the expression for the energy of an electron in the atom.
Answer:
The postulates of Bohr’s atomic model (for the hydrogen atom) :
(1) The electron revolves with a constant speed in acircular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
(2) The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2π, where h is Planck’s constant.
(3) Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of elec-tromagnetic radiation, i.e., the photon, emitted is equal to the energy difference of the two states.

Consider the electron revolving in the nth orbit around the nucleus of an atom with the atomic number Z. Let m and e be the mass and the charge of the electron, r the radius of the orbit and v the linear speed of the electron.

According to Bohr’s first postulate, centripetal force on the electron = electrostatic force of attraction exerted on the electron by the nucleus
∴ \(\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Z e^{2}}{r^{2}}\) ……………. (1)
where ε₀ is the permittivity of free space.
∴ Kinetic energy (KE) of the electron
= \(\frac{1}{2} m v^{2}=\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (2)
The electric potential due to the nucleus of charge +Ze at a point at a distance r from it is
V = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Ze}}{r}\)
∴ Potential energy (PE) of the electron
= charge on the electron × electric potential
= – e × \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e}{r}=-\frac{Z e^{2}}{4 \pi \varepsilon_{0} r}\) …………….. (3)
Hence, the total energy of the electron in the nth orbit is
E = KE + PE = \(\frac{-Z e^{2}}{4 \pi \varepsilon_{0} r}+\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\)
∴ E = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0} r}\) ………….. (4)
This shows that the total energy of the electron in the nth orbit of the atom is inversely proportional to the radius of the orbit as Z, ε₀ and e are constants. The radius of the nth orbit of the electron is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) …………….. (5)
where h is Planck’s constant.
From Eqs. (4) and (5), we get,
E_n = \(-\frac{Z e^{2}}{8 \pi \varepsilon_{0}}\left(\frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}\right)=-\frac{m Z^{2} e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\) ……………… (6)
This gives the expression for the energy of the electron in the nth Bohr orbit. The minus sign in the expression shows that the electron is bound to the nucleus by the electrostatic force of attraction.
As m, Z, e, ε₀ and h are constant, we get
E_n ∝ \(\frac{1}{n^{2}}\)
i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.
[ Note : Energy levels are most conveniently expressed in electronvolt. Hence, substituting the values of m, e, £0 and h, and dividing by the conversion factor 1.6 × 10^-19 J/eV,
E_n ≅ \(-\frac{13.6 Z^{2}}{n^{2}}\) (in eV)
For hydrogen, Z = 1
∴ E_n ≅ \(-\frac{13.6}{n^{2}}\) (in eV).

Question 4.
Starting from the formula for energy of an electron in the n^th orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.
Answer:
According to Bohr’s third postulate for the model of the hydrogen atom, an atom radiates energy only when an electron jumps from a higher energy state to a lower energy state and the energy of the
quantum of electromagnetic radiation emitted in this process is equal to the energy difference between the two states of the electron. This emission of radiation gives rise to a spectral line.

The energy of the electron in a hydrogen atom,
when it is in an orbit with the principal quantum
number n, is
E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\)
where m = mass of electron, e = electronic charge, h = Planck’s constant and = permittivity of free space.

Let E_m be the energy of the electron in a hydrogen atom when it is in an orbit with the principal quantum number m and E, its energy in an orbit with the principal quantum number n, n < m. Then
E_m = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\) and E_n = \(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}\)
Therefore, the energy radiated when the electron jumps from the higher energy state to the lower energy state is
E_m – E_n = \(\frac{-m e^{4}}{8 \varepsilon_{0}^{2} h^{2} m^{2}}-\left(-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}\right)\)
= \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
This energy is emitted in the form of a quantum of radiation (photon) with energy hv, where V is the frequency of the radiation.
∴ E_m – E_n = hv
∴ v = \(\frac{E_{m}-E_{n}}{h}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3}}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
The wavelength of the radiation is λ = \(\frac{c}{v^{\prime}}\)
where c is the speed of radiation in free space.
The wave number, \(\bar{v}=\frac{1}{\lambda}=\frac{v}{c}\)
\(\bar{v}=\frac{1}{\lambda}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\)
where \(R\left(=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3} c}\right)\) is a constant called the Ryd berg constant.

This expression gives the wave number of the radiation emitted and hence that of a line in hydrogen spectrum.

For the Lyman series, n = 1,m = 2, 3, 4, ………… ∞
∴ \(\frac{1}{\lambda_{\mathrm{L}}}=R\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line m this series, \(\frac{1}{\lambda_{\mathrm{Ls}}}=R\left(\frac{1}{1^{2}}\right)\) as m = ∞.
For the Balmer series, n = 2, m = 3, 4, 5, … ∞.
∴ \(\frac{1}{\lambda_{\mathrm{B}}}=R\left(\frac{1}{4}-\frac{1}{m^{2}}\right)\) and for the shortest wavelength line in this series, \(\frac{1}{\lambda_{\mathrm{Bs}}}=R\left(\frac{1}{4}\right)\) as m = ∞
[Note: Johannes Rydberg (1854—1919), Swedish spectroscopist, studied atomic emission spectra and introduced the idea of wave number. The empirical formula \(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)\) where m and n are simple integers, is due to Rydberg. When we consider the finite mass of the nucleus, we find that R varies slightly from element to element.]

Question 5.
Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
Answer:
The radius of the ,ith Bohr orbit is
r = \(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\) ………….. (1)
and the linear speed of an electron in this orbit is
ν = \(\frac{Z e^{2}}{2 \varepsilon_{0} n h}\) …………… (2)
where ε ₀ permittivity of free space, h ≡ Planck’s constant, n ≡ principal quantum number, m ≡ electron mass, e electronic charge and Z ≡ atomic number of the atom.
Since angular speed ω = \(\frac{v}{r}\), then from Eqs. (1) and (2), we get,
ω = \(\frac{v}{r}=\frac{Z e^{2}}{2 \varepsilon_{0} n h} \cdot \frac{\pi m Z e^{2}}{\varepsilon_{0} h^{2} n^{2}}=\frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}\) ………………. (3)
which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Eq. (3), the frequency of revolution of the electron,
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \times \frac{\pi m Z^{2} e^{4}}{2 \varepsilon_{0}^{2} h^{3} n^{3}}=\frac{m Z^{2} e^{4}}{4 \varepsilon_{0}^{2} h^{3} n^{3}}\) …………….. (4)
as required.
[Note : From Eq. (4), the period of revolution of the electron, T = \(\frac{1}{f}=\frac{4 \varepsilon_{0}^{2} h^{3} n^{3}}{m Z e^{4}}\). Hence, f ∝ \(\frac{1}{n^{3}}\) and T ∝ n³].

Obtain the formula for ω and continue as follows :

This is required quantity.

Question 6.
Determine the series limit of Balmer, Paschen and Bracket series, given the limit for Lyman series is 912 Å.
Answer:
Data : λ_L∞ = 912 Å

as n = 5 and m = ∞
From Eqs. (1) and (2), we get,
\(\frac{\lambda_{\mathrm{Pa} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 9}\) = 9
∴ λ_Pa∞ = 9λ_L∞ = (9) (912) = 8202 Å
\(\frac{\lambda_{\mathrm{Pf} \infty}}{\lambda_{\mathrm{L} \infty}}=\frac{R_{\mathrm{H}}}{R_{\mathrm{H}} / 25}\) = 25
∴ λ_Pf∞ = 25λ_L∞ = (25) (912) = 22800 Å
This is the series limit of the pfund series.

Question 7.
Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.
Answer:
(a) A radioactive transformation in which an α-particle is emitted is called α-decay.
In an α-decay, the atomic number of the nucleus decreases by 2 and the mass number decreases by 4.
Example : \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \alpha\)
Q = [m_u – m_Th – m_α]c²

(b) A radioactive transformation in which a β-particle is emitted is called β-decay.
In a β^–-decay, the atomic number of the nucleus increases by 1 and the mass number remains unchanged.
Example : \({ }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+{ }_{-1}^{0} e+\bar{v}_{\mathrm{e}}\)
where \(\bar{v}_{\mathrm{e}}\) is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_u – m_Th – m_α]c²
In a β⁺-decay, the atomic number of the nucleus decreases by 1 and the mass number remains unchanged.
Example : \(\begin{aligned}
&30 \\
&15
\end{aligned} P \rightarrow{ }_{14}^{30} \mathrm{Si}+{ }_{+1}^{0} e+v_{\mathrm{e}}\)
where v_e is the neutrino emitted to conserve the momentum, energy and spin.
Q = [m_P – m_Si – m_e]c²
[Note : The term fi particle refers to the electron (or positron) emitted by a nucleus.]

A given nucleus does not emit α and β-particles simultaneously. However, on emission of α or β-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a γ-ray photon in a transition to the lower energy state. Hence, α and β-particle emissions are often accompanied by γ-rays.

Question 8.
Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.
Answer:
Nuclear fission is a nuclear reaction in which a heavy nucleus of an atom, such as that of uranium, splits into two or more fragments of comparable size, either spontaneously or as a result of bombardment of a neutron on the nucleus (induced fission). It is followed by emission of two or three neutrons.
The mass of the original nucleus is more than the sum of the masses of the fragments. This mass difference is released as energy, which can be enormous as in the fission of ²³⁵U.
Nuclear fission was discovered by Lise Meitner, Otto Frisch, Otto Hahn and Fritz Strassmann in 1938.

The products of the fission of ²³⁵U by thermal neutrons are not unique. A variety of fission fragments are produced with mass number A ranging from about 72 to about 138, subject to the conservation of mass-energy, momentum, number of protons (Z) and number of neutrons (N). A few typical fission equations are

A type of nuclear reaction in which lighter atomic nuclei (of low atomic number) fuse to form a heavier nucleus (of higher atomic number) with the’ release of enormous amount of energy is called nuclear fusion.

Very high temperatures, of about 107 K to 108 K, are required to carry out nuclear fusion. Hence, such a reaction is also called a thermonuclear reaction.

Example : The D-T reaction, being used in experimental fusion reactors, fuses a deuteron and a triton nuclei at temperatures of about 10⁸ K.

(2) The value of the energy released in the fusion of two deuteron nuclei and the temperature at which the reaction occurs mentioned in the textbook are probably misprints.]

Question 9.
Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?
Answer:
In a nuclear reactor fuel rods are used to provide a suitable fissionable material such as \(\begin{gathered}
236 \\
92
\end{gathered}\)U. Control rods are used to start or stop the reactor. Moderators are used to slow down the fast neutrons ejected in a nuclear fission to the appropriate lower speeds. Material used as a coolant removes the energy released in the nuclear reaction by converting it into thermal energy for production of electricity.

In a nuclear reactor, a nuclear fission chain reaction is used in a controlled manner, while in a nuclear bomb, the nuclear fission chain reaction is not controlled, releasing tremendous energy in a very short time interval.
[Note : The first nuclear bomb (atomic bomb) was dropped on Hiroshima in Japan on 06 August 1945. The second bomb was dropped on Nagasaki in Japan on 9 August 1945.]

Question 10.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
Answer:
Data : M = 4.00151 u, = 1.00728 u,
m_n = 1.00866 u, 1 u = 931.5 MeV/c²
The binding energy of an alpha particle
(Zm_p + N_n -M)c²
=(2m_p + 2m_n -M)c²
= [(2)(1.00728u) + 2(1.00866 u) – 4.00151 u]c²
= (2.01456 + 2.01732 – 4.00151)(931.5) MeV
= 28.289655 MeV
= 28.289655 × 10⁶ eV × 1.602 × 10^-19 J
= 4.532002731 × 10^-12 J

Question 11.
An electron in hydrogen atom stays in its second orbit for 10^-8 s. How many revolutions will it make around the nucleus in that time?
Answer:
Data : z = 1, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, ε₀ = 8.85 × 10^-12 C² / N.m², h = 6.63 × 10 ^-34 J.s, n = 2, t = 10^-8 s
The periodic time of the electron in a hydrogen atom,

Let N be the number of revolutions made by the electron in time t. Then, t = NT.
∴ N = \(\frac{t}{T}=\frac{10^{-8}}{3.898 \times 10^{-16}}\) = 2.565 × ⁷

Question 12.
Determine the binding energy per nucleon of the americium isotope \(_{95}^{244} \mathrm{Am}\) , given the mass of \({ }_{95}^{244} \mathrm{Am}\) to be 244.06428 u.
Answer:
Data : Z = 95, N = 244  – 95 = 149,
m_p = 1.00728 u, m_n = 1.00866 u,
M = 244.06428 u, 1 u = 931.5 MeV/c²
The binding energy per nucleon,

= 7.3209 MeV/nucleon

Question 13.
Calculate the energy released in the nuclear reaction \({ }_{3}^{7} \mathrm{Li}\) + p → 2α given mass of \({ }_{3}^{7} \mathrm{Li}\) atom and of helium atom to be 7.016 u and 4.0026 u respectively.
Answer:
Data: M₁ (\({ }_{3}^{7} \mathrm{Li}\) Li atom)= 7.016 u, M₂ (He atom)
= 4.0026 u, m_p = 1.00728 u, 1 u = 931.5 MeV/c²
∆M = M₁ + m_p – 2M₂
= [7.016 + 1.00728 – 2(4.0026)]u
= 0.01808 u = (0.01808)(931.5) MeV/c²
= 16.84152 MeV/c²
Therefore, the energy released in the nuclear reaction = (∆M) c² = 16.84152 MeV

Question 14.
Complete the following equations describing nuclear decays.

Answer:
(a) \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \alpha+{ }_{86}^{222} \mathrm{Em}\)
Em (Emanation) ≡ Rn (Radon)
Here, α particle is emitted and radon is formed.

(b) \({ }_{8}^{19} \mathrm{O} \rightarrow e^{-}+{ }_{9}^{19} \mathrm{~F}\)
Here, e^– ≡ \({ }_{-1}^{0} \beta\) is emitted and fluorine is formed.

(c) \(\underset{90}{228} \mathrm{Th} \rightarrow{ }_{2}^{4} \alpha+{ }_{88}^{224} \mathrm{Ra}\)
Here, α particle is emitted and radium is formed.

(d) \({ }_{7}^{12} \mathrm{~N} \rightarrow{ }_{6}^{12} \mathrm{C}+{ }_{1}^{0} \beta\)
\({ }_{1}^{0} \beta\) is e⁺ (positron)
Here, β⁺ is emItted and carbon is formed.

Question 15.
Calculate the energy released in the following reactions, given the masses to be \({ }_{88}^{223} \mathrm{Ra}\) : 223.0185 u, \({ }_{82}^{209} \mathrm{~Pb}\) : 208.9811, \({ }_{6}^{14} C\) : 14.00324, \({ }_{92}^{236} \mathrm{U}\) : 236.0456, \({ }_{56}^{140} \mathrm{Ba}\) : 139.9106, \({ }_{36}^{94} \mathrm{Kr}\) : 93.9341, \({ }_{6}^{11} \mathrm{C}\) : 11.01143, \({ }_{5}^{11} \mathrm{~B}\) : 11.0093. Ignore neutrino energy.

Answer:

(a) \({ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}\)
The energy released in this reaction = (∆M) c²
= [223.0185 – (208.9811 + 14.00324)j(931 .5) MeV
= 31.820004 MeV

(b) \({ }_{92}^{236} \mathrm{U} \rightarrow{ }_{56}^{140} \mathrm{Ba}+{ }_{36}^{94} \mathrm{Kr}+2 \mathrm{n}\)
The energy released in this reaction =
(∆M) c² = [236.0456 – (139.9106 + 93.9341 + (2)(1 .00866)1(93 1 .5)MeV
= 171.00477 MeV

(c) \({ }_{6}^{11} \cdot \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}\) + neutrino
The energy released in this reaction = (∆M) c²
= [11.01143 – (11.0093 + O.00055)](931.5) MeV
= 1.47177 MeV

Question 16.
Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be 3.839 × 10^-12 per second.
Answer:
Data: 15.3 decays per gram per minute (living organism), 12.3 disintegrations per gram per minute (very old charcoal). Hence, we have,
\(\frac{A(t)}{A_{0}}=\frac{12.3}{15.3}\), λ = 3.839 × 10^-12 per second

Question 17.
The half-life of \({ }_{38}^{90} \mathrm{Sr}\) is 28 years. Determine the disintegration rate of its 5 mg sample.
Answer:
Data: T_1/2 = 28 years = 28 × 3.156 × 10⁷ s
=8.837 × 10⁸s, M = 5 mg =5 × 10^-3g
90 grams of \({ }_{38}^{90} \mathrm{Sr}\) contain 6.02 × 10²³ atoms
Hence, here, N = \(\frac{\left(6.02 \times 10^{23}\right)\left(5 \times 10^{-3}\right)}{90}\)
= 3.344 × 10¹⁹ atoms
∴ The disintegration rate = Nλ = N\(\frac{0.693}{T_{1 / 2}}\)
= \(\frac{\left(3.344 \times 10^{19}\right)(0.693)}{8.837 \times 10^{8}}\)
= 2.622 × 10¹⁰ disintegrations per second

Question 18.
What is the amount of \({ }_{27}^{60} \mathrm{Co}\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Answer:
Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci
= (10.0 × 10^-3)(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸ dis/s
T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷) s
= 1.673 × 10⁸ s
Decay constant, λ = \(\frac{0.693}{T_{1 / 2}}=\frac{0.693}{1.673 \times 10^{8}} \mathrm{~s}^{-1}\)
=4.142 × 10^-9 s^-1
Activity = Nλ
∴ N = \(\frac{\text { activity }}{\lambda}=\frac{3.7 \times 10^{8}}{4.142 \times 10^{-9}} \text { atoms }\)
= 8.933 × 10¹⁶ atoms
=60 grams of \({ }_{27}^{60} \mathrm{Co}\) contain 6.02 × 10²³ atoms
Mass of 8.933 × 10¹⁶ atoms of \({ }_{27}^{60} \mathrm{Co}\)
= \(\frac{8.933 \times 10^{16}}{6.02 \times 10^{23}} \times 60 \mathrm{~g}\)
= 8.903 × 10^-6 g = 8.903 µg
This is the required amount.

Question 19.
Disintegration rate of a sample is 10¹⁰ per hour at 20 hrs from the start. It reduces to 6.3 × 10⁹ per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.
Answer:
Data : A (t₁) = 10¹⁰ per hour, where t₁ = 20 h,
A (t₂) = 6.3 × 10⁹ per hour, where t₂ = 30 h
A(t) = A₀e^-λt ∴ A(t₁) = A₀e^-λt₁ and A(t₂) = A_oe^-λt₂

∴ 1.587 e^10λ ∴ 10λ =2.3031og₁₀(1.587)
∴ λ = (0.2303)(0.2007) = 0.04622 per hour
The half life of the material, T_1/2 = \(\frac{0.693}{\lambda}=\frac{0.693}{0.04622}\)
= 14.99 hours
Now, A₀ = A (t₁)e^λt₁ = 10¹⁰e^{(0.04622)(20)}
= 10¹⁰ e^0.9244
Let x = e^0.9244 ∴ 2.3031og₁₀x = 0.9244
∴ 1og₁₀x = \(\frac{0.9244}{2.303}\) = 0.4014
∴ x = antilog 0.4014 = 2.52
∴ A₀ = 2.52 × 10¹⁰ per hour
Now A₀ = N₀λ ∴ N₀ = \(\frac{A_{0}}{\lambda}=\frac{2.52 \times 10^{10}}{0.04622}\)
= 5.452 × 10¹¹
This is the initial number of radioactive atoms in the sample.

Question 20.
The isotope ⁵⁷Co decays by electron capture to ⁵⁷Fe with a half-life of 272 d. The ⁵⁷Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
(a) Find the mean lifetime and decay constant for ⁵⁷Co.
(b) If the activity of a radiation source ⁵⁷Co is 2.0 µCi now, how many ⁵⁷Co nuclei does the source contain?
(c) What will be the activity after one year?
Answer:
Data: T_1/2 = 272d = 272 × 24 × 60 × 60s = 2.35 × 10⁷ s,
A₀ = 2.0uCi = 2.0 × 10^-6 × 3.7 × 10¹⁰
= 7.4 × 10⁴ dis/s
t = 1 year = 3.156 × 10⁷ s
(a) T_1/2 = \(\frac{0.693}{\lambda}\) = 0.693 τ ∴ The mean lifetime for
⁵⁷Co = τ = \(\frac{T_{1 / 2}}{0.693}=\frac{2.35 \times 10^{7}}{0.693}\) = 3391 × 10⁷ s
The decay constant for ⁵⁷Co = λ = \(\frac{1}{\tau}\)
= \(\frac{1}{3.391 \times 10^{7} \mathrm{~s}}\)
= 2949 × 10^-8 s^-1

(b)A₀ = N₀A ∴ N₀ = \(\frac{A_{0}}{\lambda}\) = A₀τ
= (7.4 × 10⁴)(3.391 × 10⁷)
= 2.509 × 10¹² nuclei
This is the required number.

(c) A(t) = A₀e^-λt = 2e^{-(2.949 × 10-8)(3.156 × 107)}
= 2e^-0.9307 = 2 / e^0.9307
Let x = e^0.9307 ∴ Iog_ex = 0.9307
∴ 2.303log₁₀x = 0.9307
∴ log₁₀x = \(\frac{0.9307}{2.303}\) = 0.4041
∴ x = antilog 0.4041 = 2.536
∴ A (t) = \(\frac{2}{2.536}\) μCi = 0.7886 μCi

Question 21.
A source contains two species of phosphorous nuclei, \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 14.3 d) and \({ }_{15}^{32} \mathrm{P}\) (T_1/2 = 25.3 d). At time t = 0, 90% of the decays are from \({ }_{15}^{32} \mathrm{P}\) . How much time has to elapse for only 15% of the decays to be from \({ }_{15}^{32} \mathrm{P}\) ?
Answer:

∴ (0.04846 – 0.02739) t = 2.303 (2.1847 – 0.4771)
∴ t = \(\frac{(2.303)(1.7076)}{0.02107}\) = 186.6 days
This is the required time.

Question 22.
Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of ¹⁴C averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were ¹⁴C? (b) An archaeological specimen containing 500 mg of carbon, shows 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when the specimen died was equal to the average value of the air? Half-life of 14C is 5730 years?
Answer:
0.693
Data: T_1/2 = 5730y ∴ λ = \(\frac{0.693}{5730 \times 3.156 \times 10^{7}} \mathrm{~s}^{-1}\)
= 3.832 × 10^-12 s^-1, A = 0.255 Bq per gram of carbon in part (a); M = 500 mg = 500 × 10^-3 g,
174 decays in one hour \(\frac{174}{3600}\) dis/s = 0.04833 dis/s in part (b) (per 500 mg].

(a) A = Nλ ∴ N = \(\frac{A}{\lambda}=\frac{0.255}{3.832 \times 10^{-12}}\)
= 6.654 × 10¹⁰
Number of atoms in 1 g of carbon = \(\frac{6.02 \times 10^{23}}{12}\)
=5.017 × 10²²
\(\frac{5.017 \times 10^{22}}{6.654 \times 10^{10}}\) = 0.7539 × 10¹²
∴ 1 ¹⁴C atom per 0.7539 × 10¹² atoms of carbon
∴ 4 ¹⁴C atoms per 3 × 10¹² atoms of carbon

(b) Present activity per gram = \(\)
= 0.09666 dis/s per gram
A₀ = 0.255 dis/s per gram
Now, A(t) = A₀e^-λt

This is the required quantity.

Question 23.
How much mass of ²³⁵U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV
Answer:
Data: Power = 3000 MW = 3 × 10⁹ J/s
∴ Energy to be produced each day
=3 × 10⁹ × 86400 J each day
= 2.592 × 10¹⁴ J each day
Energy per fission = 202.79 MeV
= 202.79 × 10⁶ × 1.6 × 10^-19 J = 3,245 × 10^-11 J
∴ Number of fissions each day
= \(\frac{2.592 \times 10^{14}}{3.245 \times 10^{-11}}\) × 10²⁴ each day
0.235 kg of ²³⁵U contains 6.02 × 10²³ atoms
7988 x 1024
∴ M = \(\left(\frac{7.988 \times 10^{24}}{6.02 \times 10^{23}}\right)\) (o.235) = 3.118 kg
This is the required quantity.

Question 24.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504 u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584 u) and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259 u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
[Answer: 9.3% and 11.7%]
Answer:
Data : Average atomic mass of magnesium =

12th Physics Digest Chapter 15 Structure of Atoms and Nuclei Intext Questions and Answers

Use your brainpower (Textbook Page No. 336)

Question 1.
Why don’t heavy nuclei decay by emitting a single proton or a single neutron?
Answer:
According to quantum mechanics, the probability of these emissions is extremely low.

Maharashtra State Board Class 12 Textbook Solutions

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