Class 12 Chemistry Suggestion 2023-2024 | Best Notes for TBSE, CBSE, ICSE Board Exams.

Class 12 Chemistry based on NCERT Syllabus 

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1. Chapter 1 The Solid State
2. Chapter 2 Solutions
3. Chapter 3 Electro chemistry
4. Chapter 4 Chemical Kinetics
5. Chapter 5 Surface Chemistry
6. Chapter 6 General Principles and Processes of Isolation of Elements
7. Chapter 7 The p Block Elements
8. Chapter 8 The d and f Block Elements
9. Chapter 9 Coordination Compounds
10. Chapter 10 Haloalkanes and Haloarenes
11. Chapter 11 Alcohols Phenols and Ethers
12. Chapter 12 Aldehydes Ketones and Carboxylic Acids
13. Chapter 13 Amines
14. Chapter 14 Biomolecules
15. Chapter 15 Polymers
16. Chapter 16 Chemistry in Everyday Life
    
    
    

Class 12 Chemistry Chapter 8 The d and f Block Elements:

Section Name	Topic Name
8	The d – and f – Block Elements
8.1	Position in the Periodic Table
8.2	Electronic Configurations of the d-Block Elements
8.3	General Properties of the Transition Elements (d-Block)
8.4	Some Important Compounds of Transition Elements
8.5	The Lanthanoids
8.6	The Actinoids
8.7	Some Applications of d – and f -Block Elements

8.1 Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?

Ans: The outer electronic configuration of Ag (Z=47) is 4d105s1. It shows+1 and + 2 O.S. (in AgO and AgF2). And in + 2 O.S., the electronic configuration is d9 i.e, d-subshell is incompletely filled. Hence, it is a transition element.

8.2. In the series Sc(Z = 21) to (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol-1. Why?

Ans: The enthalpy of atomisation is directly linked with the stability of the crystal lattice and also the strength of the metallic bond. In case of zinc (3d104s2 configuration), no electrons from the 3d-orbitals are involved in the formation of metallic bonds since all the orbitals are filled. However, in all other elements belonging to 3d series one or more d-electrons are involved in the metallic bonds. This means that the metallic bonds are quite weak in zinc and it has therefore, lowest enthalpy of atomisation in the 3d series.

8.3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Ans: Manganese (Z = 25) shows maximum number of O.S. This is because its outer EC is 3d54s2. As 3d and 4s are close in energy, it has maximum number of e-1 s to loose or share. Hence, it shows O.S. from +2 to +7 which is the maximum number.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.7.Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Ans: Cr2+ is a stronger reducing agent than Fe2+. This is because E°(Cr3+/Cr2+) is negative (- 0.41V) whereas E°(Fe3+/Fe2+) is positive (+ 0.77 V). Thus, Cr2+ is easily oxidised to Fe3+ but Fe2+ cannot be easily oxidised to Fe3+.

8.8.Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).

Ans:

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

j

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+  towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+  is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1  = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9. In what way are the electronic configuration of the transition elements different from non-transition elements?

Sol: Electronic configuration of transition elements : (n – 1)d1-10 ns1-2. Electronic configuration of non-transition elements : ns1-2 or ns2np1-6. From comparison, it is quite evident that the transition elements have incomplete d-orbitals (s- orbitals in some cases) while the non-transition elements have no d-orbitals present in the valence shells of their atoms. This is responsible for the difference in the characteristics of the elements belonging to these classess of elements.

8.10. What are the different oxidation states exhibited by the lanthanoids?

Sol: Lanthanides exhibits + 2, + 3 and + 4 oxidation states. The most common oxidation state of lanthanoids is +3.

8.11. Explain giving reasons:

(i)Transition metals and many of their compounds show paramagnetic behaviour.

(ii)The enthalpies of atomisation of the transition metals are high.

(iii)The transition metals generally form coloured compounds.

(iv)Transition metals and their many compounds act as good catalyst

Sol: (i) Magnetic properties: Transition elements and many of their compounds are paramagnetic, i.e., they are weakly attracted by a magnetic field. This is due to the presence of unpaired electrons in atoms, ions or molecules. The paramagnetic character increases as the number of . unpaired electrons increases. The paramagnetic character is measured in terms of magnetic moment and is given by

 where n – number of unpaired electrons.

(ii) Because of large number of unpaired electrons in d-orbitals of their atoms they have stronger interatomic intefactions and hence stronger metallic bonding between atoms resulting in higher enthalpies of atomisation.

(iii) Formation of coloured compounds (both in solid state as well as in aqueous solution) is another very common characteristics of transition metals. This is due to absorption of some radiation from visible light to cause d-d transition of electrons in transition metal atom. The d-orbitals do not have same energy and under the influence of ligands, the d-orbitals split into two sets of orbitals having different energies; transition of electrons can take place from one set of d-orbitals to another set within the same sub-shell. Such transitions are called d-d transitions. The energy difference for these d-d transitions fall in the visible region. When white light is incident on compounds of transition metals, they absorb a particular frequency and remaining colours are emitted imparting a characteristic colour to the complex. Zn2+ and Ti4+ salts are white because they do not absorb any radiation in visible region.

(iv)Catalytic properties: Many of transition metals and their compounds act as catalyst in variety of reactions, e.g., finely divided iron in manufacture of NH3 by Haber’s process, V2O5 or Pt in manufacture of H2S04 by Contact process, etc.). The catalytic activity is due to following two reasons.

(a)The ability of transition metal ion to pass ” easily from one oxidation state to another

and thus providing a new path to reaction with lower activation energy.

(b)The surface of transition metal acts as very good adsorbent and thus provides increased concentration of reactants on their surface causing the reaction to occur.

8.12. What are interstitial compounds? Why are such compounds well known for transition metals?

Sol: Transition metals form large number of interstitial compounds. They are able to entrap small atoms of elements like H, G, N, B, etc., in their crystal lattice and even can make weak bonds with them.

Due to formation of interstitial compounds, their malleability and ductility decreases and tensile . strength increases. Steel and cast iron are hard in comparison to wrought iron due to the presence of trapped carbon atoms in interstitial spaces.

8.13. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Sol: The transition metals show a number of variable oxidation states due to the participation of (n – 1) d electrons in addition to ns electrons in the bond formation. They therefore, exhibit a large number of variable oxidation states. On the other hand, the non-transition metals generally belonging to s-block do not show variable oxidation states because by the loss of valence s-electrons, they acquire the configuration of the nearest noble gas elements.

In the p-block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not.

8.14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Sol: Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr2O3) with sodium or potassium carbonate in free excess of air. The reaction with sodium carbonate occurs as follows:

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr,07.2H20 can be crystallised.

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.

Orange crystals of potassium dichromate crystallise out. The chromates and dichromates depending upon pH of the solution. If pH of potassium dichromate is increased it is converted to yellow potassium chromate.

8.15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i)iodide

(ii)iron (II) solution and

(iii)H2S

Sol: K2Gr207is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from+6 to + 3. The oxidising action can be represented as follows:

8.16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) S02 and (iii) oxalic acid? Write the ionic, equations for the reactions.

Sol: Potassium permanganate (KMn04) is prepared by the fusion of a mixture of pyrolusite (Mn02),potassiufn hydroxide and oxygen, first green coloured potassium manganate is formed. 2MnO2 + 4KOH + 02 —> 2K2Mn04+2H20 The potassium manganate is extracted by water, which then undergoes disproportionation in neutral or acidic solution to give potassium permanganate.

8.17. For M2+/M and M3+/M2+ systems the E° values

for some metals are as follows:

Cr2+/Cr   –> -0.9 V

Mn2+/Mn  –> -1.2V

Fe2+/Fe     –> -0.4 V

Cr3+/Cr2+  –> -0.4 V

Mn3+/Mn2+   –>+ 1.5V

Fe3+/Fe2+   –>+ 0.8V

(ii)the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Sol: (i) Cr3+/Cr2+ has negative reduction potential. Hence, Cr3+ cannot be reduced to Cr2+. Mn3+/Mn2+ has a large positive reduction potential. Hence, Mn3+ can be easily reduced to Mn2+. Fe3+/Fe2+ has small positive reduction potential. Hence, Fe3+ is more stable than Mn3+ but less stable than Cr3+.

(ii)From the E° values, the order of oxidation of the metal to the divalent cation is : Mn > Cr > Fe.

8.18. Predict which of the following will be coloured in aqueous solution?

Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+, Co2+.

Sol: Only those ions will be coloured which have incomplete d-orbitals. The ions with either empty or filled d-orbitals are colourless. Keeping this in view, the coloured ions among the given list are :

Ti3+(3d1), V3+(3d2), Mn2+(3d5), Fe3+(3d5), Co2+ (3d7)

Sc3+ (3d°) and Cu+ (3d10) ions are colourless.

8.19. Compare the stability of +2 oxidation state for the elements of the first transition series.

Sol: In general, the stability of +2 oxidation state in first transition series decreases from left to right due to increase in the sum of first and second ionisation energies. However Mn2+ is more stable due to half filled d-orbitals (3d5) and Zn2+ is more stable due to completely filled d-orbitals (3d10).

8.20. Compare the chemistry of actinoids with that of the lanthanoids with special reference to

(i)electronic configuration,

(ii)atomic and ionic sizes and

(iii)oxidation state

(iv)chemical reactivity.

Sol: (i) Electronic configuration: The general electronic configuration of lanthanoids is [Xe]54 4f1-14 5d0-1  6s2 and that of actinoids is [Rn]86 5f0-14 6d0-1  7s2, lanthanoids . belong to 4 f series whereas actinoids belong to 5f-series.

(ii) Atomic and ionic sizes: Both lanthanoids and actinoids show decrease in size of their atoms or ions in + 3 oxidation state as we go from left to right. In lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoid contraction. The contratibn is greater from element to element in actinodes due to poorer shielding by 5f electrons.

(iii)Oxidation state: Lanthanoids show limited oxidation states (+ 2, + 3, + 4) out of which + 3 is most common whereas actinoids show +3, +4, +5, +6, +7 oxidation states.This is because of large energy gap between 4f 5d and 6s orbitals. However, actinoids show a large number of oxidation states because of small energy ap- between 5f 6d and Is orbitals.

(iv) Chemical reactivity: The earlier members

of the lanthanoids series are quite reactive similar to calcium but, with increase in atomic number, they behave more like aluminium. The metals combine with hydrogen when . gently heated in the gas. Carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated with carbon. They liberate hydrogen from dilute acid and burn in halogens to form halides. They form oxides M203 and hydroxides M(OH)3.

Actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them gives a mixture of oxide and hydride and combination with most non-metals take place at moderate temperatures. HCl attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers, alkalis have no action. Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.

8.21. How would you account for the following:

(i) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidizing.

(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.

(iii) The d1 configuration is very unstable in ions.

Sol: (i) E° value for Cr3+/Cr2+ is negative (-0-41 V) whereas E° values for Mn3+/Mn2+is positive (+1.57 V). Hence, Cr2+ ion can easily undergo oxidation to give Cr3+ ion and, therefore, act as strong reducing agent whereas Mn3+ can easily undergo’ reduction to give Mn2+ and hence act as an oxidizing agent.

(ii) Co (III) has .greater tendency to form coordination complexes than Co (II). Hence, in the presence of ligands, Co (II) changes to Co (III), i.e., is easily oxidized.

(iii) The ions with dx configuration have the tendency to lose the only electron present in d-subshell to acquire stable d° configuration. Hence, they are unstable and undergo oxidation or disproportionation.

8.22. What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution

Sol: Disproportionation reactions are those in which the same substance undergoes oxidation as well as reduction, i.e., oxidation number of an element increases as well as decreases to form two different products.

8.23. Which metal in the first transition metal series exhibits +1 oxidation state most frequently and why?

Sol: Cu with configuration [Ar] 4s13d10 exhibits +1 oxidation state and forms Cu+ ion because by losing one electron, the cation or positive ion acquires a stable configuration of d-orbitals (3d10).

8.24. Calculate the number of unpaired electrons in the following gaseous ions : Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution.

Sol: Mn3+ = 3d1 = 4 unpaired electrons, Cr3+ = 3d3 = 3 electrons,V3+ = 3d2 = 2 electrons, Ti3+=3d1 = l electron.Out of these, Cr3+ is most stable in aqueous solution because of half-filled t2g level.

8.25. Give examples and suggest reasons for the following features of the transition metal chemistry:

(i) The lowest oxide of transition metal is basic the highest is amphoteric/ acidic.

(ii) A transition metal exhibits highest oxidation state ih oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Sol: (i) The lower oxide of transition metal is basic because the metal atom has low oxidation state whereas higher once are acidic due to high oxidation state. For example, MnO is basic whereas Mn2O7is acidic. Oxides in lower oxidation state are ionic hence basic. Oxides in higher oxidation state are covalent hence acidic

(ii) A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are highly electronegative elements, small in size and strongest oxidising agents. For example, osmium shows an oxidation states of + 6 in O5F6and vanadium shows an oxidation states of + 5 in V2O5.

(iii) Oxo metal anions have highest oxidation state, e.g., Cr in Cr2072- has an. oxidation state of + 6 whereas Mn in Mn04– has an oxidation state of + 7. This is again due to the combination of the metal with oxygen, which is highly electronegative and oxidizing agent.

8.26. Indicate the steps in the preparation of:

(i)K2Cr207from chromite ore

(ii)KMn04 from pyrolusite ore.

Sol:

8.27. What are alloys? Name an alloy which contains some lanthanoid metals. Mention its uses.

Sol: An alloy is a homogeneous mixture of different metals or metals and non-metals.

Misch metal is an alloy of cerium (Ce). lanthanum (La), neodymium (Nd), iron (Fe) and traces of carbon, sulphur, aluminium etc. It is used in making parts of jet engines.

8.28. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29,59,74,95,102,104.

Sol: The f-block elements in which the. last electron enters into f-sub shell-are called inner-transition elements. These include lanthanoids (Z=58 to 71) and actinoids (Z=90 to 103). Thus, the elements with atomic numbers 59,95 and 102 are the? inner transition elements.

8.29. The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

Sol: Lanthanoids show limited number of oxidation state, viz, + 2, + 3 and + 4 (out of which + 3 is most common). This is because of large energy gap between 4f 5d and 6s subshells. The dominant oxidation state of actinoids is also + 3 but they show a number of other oxidation states also. For example, uranium (Z=92) and plutonium (Z – 94), show + 3, + 4, + 5 and + 6, neptunium (Z = 94) shows + 3, +4, + 5 and + 7, etc. This is because of the small energy difference between. 5f, 6d and 7s orbitals of the actinoids.

8.30. Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element

Sol: Last actinoid=Lawrencium (Z = 103)

Electronic configuration = [Rn]86 5f14 6d1 7s2 Possible oxidation state = + 3.

8.31 Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Sol.

8.32. Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to co-relate this type of behaviour with the electronic configuration of these elements.

Sol: +4 oxidation state : 58Ce, 59Pr, 65Tb

+ 2 oxidation state : 60Nd, 62Sm, 63Eu, 69Tm, 70Yb.

In general +2 oxidation state is exhibited by the elements with configuration 5d06s2 so that two electrons may be easily lost. Similarly +4 oxidation state is shown by the elements which after losing four electrons acquire configuration either close to 4f0 or 4f7.

8.33. Compare the chemistry of actinoids with that of lanthanoids with reference to:

(i)Electronic configuration

(ii)Oxidation states

(iii)Chemical reactivity

Sol: (i)Electronic configuration : In lanthanoids 4f- orbitals are progressively filled whereas in actinoids 5f-orbitals are progressively filled.

(ii)Oxidation states : Lanthanoids shows +3 oxidation state. Some elements shows +2 and +4 oxidation state also. Actinoids shows +3, +4, +5 +6, +7 oxidation states. Although +3 and +4 are most common.

(iii)Chemical reactivity : Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.

8.34. Write the electronic configurations of the elements with the atomic numbers 61,91,101 and 109.

Sol: Z=61 (Promethium, Pm) [Xe]544f55d0 6s2

Z = 91 (Protactinium, Pa) => [Rn]86 5f2 6d1 7s2

Z = 101 (Mendelevium, Md)=> [Rn]86 5f13 6d0 7s2

Z = 109 (Meitnerium, Mt) [Rn]86 5f14 6d7 7s2

8.35. Com pare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(i)electronic configurations

(ii)oxidation states

(iii)ionisation enthalpies and

(iv)atomic sizes

Sol: (i) Electronic configuration: The elements in the same vertical column generally have similar electronic configuration. First transition series shows only two exceptions, i.e., Cr = 3d5  4s1  and Cu = 3d10 4s1. But second transition series shows more exceptions, i.e., Y = 4d1 5s2, Nb = 4d1 , 5s1 , Mo=4d5  5s1 , Ru=4d1  5s1 , Rh=4d8  5s1 , Pd , =4d10 5s°, Ag=4d10 5s1 . In third transition, there are two exceptions, i.e„ Pt = 5d9  6s1  and Au = 5d10 6s1 .

Thus in the same vertical column, in a number of cases, the electronic configuration of the elements of three series are not similar.

(ii) Oxidation states: The elements in the same vertical column generally show similar oxidation states. The number of oxidation states shown by the elements in the middle of each series is maximum and minimum at the extreme ends.

(iii)Ionization enthalpies: The first ionization enthalpies in each series generally increases gradually as we more from left to right though some exceptions are observed in each series. The first ionization enthalpies of some elements in the second (4d) series are higher while some of them have lower value than the elements of 3d series in the same vertical column. However, the first ionization enthalpies of third (5d) series are higher than those of 3d and Ad series. This is because of weak shielding of nucleus by 4f-electrons in the 5d series.

(iv)Atomic sizes: In general, ions of the same charge or atoms in a given series show progressively decrease in radius with increasing atomic number though the decrease is quite small. But the size of the atoms of the Ad series is larger then the corresponding elements of the 3d series whereas size of elements of the 5d-series nearly the same as those of Ad series because of lanthanoid contraction.

8.36. Write down the number of 3d electrons in each of the following ions:Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe2+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Sol:

8.37. Comment on the statement that elements of first transition series possess many properties different from those of the heavier transition elements.

Sol: The heavier transition elements belong to fourth (Ad) and fifth (5d) and sixth (6d) transition series. Their properties are expected to be different from the elements belonging to the first (3d) series due to the following reasons :

(i) The atomic radii of the elements belonging to Ad and 5d series are more due to greater number of electron shells. However, the difference in Ad and 5d transition elements are comparatively less because of lanthanoid contraction.

(ii) Because of stronger inter atomic bonding, the m.p. and b.p. of the elements of Ad and 5d series are higher.

(iii) Ionisation enthalpies are expected to decrease as we move from one series to the other. However, the values for the elements of 5d series are higher as compared to the elements belonging to the other two series due to lanthanoid contraction.

Actually atomic size decreases on account of it and effective nuclear charge increases. As a result, there is an increase in ionisation energy in case of 3d elements.

8.38. What can be inferred from the magnetic moment values of the following complex species?

Sol:

বিশেষ নোট: এই প্রশ্নগুলো ছাড়াও টেক্সট বইয়ের ভেতর থেকে প্রশ্ন আসার সম্ভাবনা বেশি ।

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Chapter 12 Aldehydes Ketones and Carboxylic Acids

Section Name	Topic Name
12	Aldehydes, Ketones and Carboxylic Acids
12.1	Nomenclature and Structure of Carbonyl Group
12.2	Preparation of Aldehydes and Ketones
12.3	Physical Properties
12.4	Chemical Reactions
12.5	Uses of Aldehydes and Ketones
12.6	Nomenclature and Structure of Carboxyl Group
12.7	Methods of Preparation of Carboxylic Acids
12.8	Physical Properties
12.9	Chemical Reactions
12.10	Uses of Carboxylic Acids

NCERT INTEXT QUESTION

12.1. Write the structures of the following compounds:

(i) α-Methoxypropionaldehyde

(ii) 3-Hydroxybutanal

(iii) 2-Hydroxycyclopentane carbaldehyde

(iv) 4-OxopentanaI

(v) Di-sec.butylketone

(vi) 4-fluoroaeetophenone

Ans:

12.2. Write the structures of the products of the following reactions:

Ans:

12.3. Arrange the following compounds in increasing order of their boiling points:

CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3

Ans: The order is : CH3CH2CH3 < CH3OCH3 < CH3CHO <CH3CH2OH

All these compounds have comparable molecular masses CH3CH2OH undergoes extensive intermolecular Il-bonding and thus its b.pt. is the highest. CH3CHO is more pdlar than CH3OCH3 so that dipole-dipoie interactions in CH3CHO are greater than in CH3OCH3. Thus, b.pt. of CH3CHO > CH3OCH3. CH3CH2CH3 has only weak van der waals forces between its molecules and hence has the lowest b.pt.

12.4. Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reactions :

(a) Ethanal, propanal, propanone, butanone

(b) Benzaldehyde, p-tolualdehyde, p-nitrobenzaldehyde, acetophenone

Ans: (a) The increasing order of reactivity of the carbonyl compounds towards nucleophilic addition reactions is :

butanone < propanone < propanal < ethanal

The reactivity is based upon two factors. These are: steric factors and electronic factors. 

(b) The increasing order of reactivity is :

acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde

Explanation: Acetophenone being a ketone is the least reactive towards nucleophilic addition. All others are aldehydes. Among them, p-tolualdehyde is less reactive than benzaldehyde because CH3 group present at the para position w.r.t.  -CHO group will increase the electron density on the carbonyl carbon atom due to hyper conjugation effect. As a result, the nucleophile attack occurs to lesser extent as compared to benzaldehyde.

In p-nitrobenzaldehyde, the nitro group has an opposing effect. It is electron withdrawing in nature due to -I effect as well as -R effect. The electron density on the carbonyl carbon atom decreases and this favours the nucleophile attack.

12.5. Predict the products of the following reactions:

Ans:

12.6. Give the 1UPAC names of the following compounds:

(i) PhCH2CH2COOH

(ii) (CH3)2 C=CHCOOH

Ans: (i) 3 – Phenylpropanoic acid

(ii) 3 – Methylbut-2-enoic acid

(iii) 2-Methylcyclohexanecarboxylic acid

(iv) 2,4,6 – Trinitrobenzoic acid

12.7. Show how each of the following compounds can be converted into benzoic acid.

(i) Ethylbenzene

(ii) Acetophenone

(iii) Bromobenzene

(iv) Phenylethene (styrene)

Ans:

12.8. Which acid from each of the following pairs would you expect to be a stronger acid?

(i) CH3COOH or CH2FCOOH

(ii) CH2FCOOH or CH2ClCOOH

(iii) CH2FCH2CH2COOH or CH3CHFCH2COOH

Ans:

Explanation: CH3 group with +I effect increases the electron density on the oxygen atom in O – H bond in the carboxyl group and cleavage of bond becomes diffcult. It therefore, decreases the acidic strength. The F atom has very strong -I effect, i.e., electron withdrawing influence. It decreases the electron density on the oxygen atom and cleavage of bond becomes easy. The acidic character therefore, increases. It is further related to the

1. No. of F atoms present in the molecule.
2.  Relative position of the F atom in the carbon atom chain.

In the light of the above discussion.

(i) CH2FCOOH is a stronger acid.

(ii) CH2FCOOH is a stronger acid.

(iii) CH3CHFCH2COOH is a stronger acid.

NCERT EXERCISES

12.1. What is meant by the following terms? Give an example of the reaction in each case.

(i) Cyanohydrin

(ii) Acetal

(iii) Semicarbazone

(iv) Aldol

(v) Hemiacetal

(vi) Oxime

(vii) Ketal

(viii) Imine

(ix) 2,4-DNP derivative

(x) Schiff’s base.

Ans: (i) Cyanohydrin: gem-Hydroxynitriles, i.e., compounds possessing hydroxyl and cyano groups on the same carbon atom are called cyanohydrins. These are produced by addition of HCN to aldehydes or ketones in a weakly basic medium.

(ii) gem – Dialkoxy compounds in which the two alkoxy groups are present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in presence of dry HCl gas.

These are easily hydrolysed by dilute mineral acids to regenerate the original aldehydes. Therefore, these are used for the protection of aldehyde group in organic synthesis.

(iii) Semicarbazones are derivatives of aldehydes and ketones and are produced by action of semicarbazide on them in acidic medium.

(iv) Aldols are P-hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones in presence of a dilute aqueous base. For example,

(v) gem – Alkoxyalcohols are called hemiacetals. These are produced by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.

(vi) Oximes are produced when aldehydes or ketones react with hydroxyl amine in weakly acidic medium.

(vii) Ketals are produced when a ketone is heated with dihydric alcohols like ethylene glycol in presence of dry HCl gas or /3-toluene sulphonic acid (PTS).

These are easily hydrolysed by dilute mineral acids to regenerate the original ketones. Therefore, ketals are used for protecting keto groups in organic synthesis.

(viii) Compounds containing -C = N – group are called imines. These are produced when aldehydes and ketones react with ammonia derivatives.

(ix)2, 4-Dinitrophenyl hydrazone (i.e., 2,4-DNP derivatives) are produced when aldehydes or ketones react with 2,4-dinitrophenyl hydrazine in weakly acidic medium.

(x) Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or SchifFs bases.

12.2. Name the following compounds according to IUPAC system of nomenclature:

(i) CH3CH (CH3)—CH2 CH2—CHO

(ii) CH3CH2COCH(C2H5)CH2CH2Cl

(iii) CH3CH=CHCHO

(iv) CH3COCH2COCH3

(v) CH3CH(CH3)CH2C(CH3)2COCH3

(vi) (CH3)3CCH2COOH.

(vii) OHCC6H4CHO-p

Ans: (i) 4-Methyl pentanal

(ii) 6-Chloro-4-ethylhexan-3-one

(iii) But-2-en-l-al

(iv) Pentane-2,4-dione

(v) 3,3,5-Trimethyl-hexan-2-one

(vi) 3,3-Dimethyl butanoic acid

(vii) Benzene-1,4-dicarbaldehyde

12.3. Draw the structures of the following compounds.

(i) 3-Methylbutanal

(ii) p-Methylbenzaldehyde

(iii) 4-Chloropentan-2-one

(iv) p, p’-Dihydroxybenzophenone

(v) p-Nitropropiophenone

(vi) 4-Methylpent-3-en-2-one.

(vii) 3-Bromo-4-phenylpentanoic acid

(viii) Hex-2-en-4-ynoic acid

Ans:

12.4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

(i) CH3CO(CH2)4CH3

(ii) CH3CH2CH BrCH2CH(CH3)CHO

(iii) CH3(CH2)5CHO

(iv) Ph—CH=CH—CHO

Ans:

12.5. Draw structures of the following derivatives:

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

(ii) Cydopropanone oxime

(iii) Acetaldehydedimethylacetal

(iv) The semicarbazone of cyclobutanone

(v) The ethylene ketal of hexan-3-one

(vi) The methyl hemiacetal of formaldehyde

Ans:

12.6. Predict the product when cyclohexanecarbaldehyde reacts with following reagents :

(i) C6H5MgBr followed by H30+

(ii) Tollen’s reagent

(iii) Semicarbazide in the weakly acidic medium

(iv) Excess of ethanol in the presence of acid

(v) Zinc amalgam and Cyclohexanecarbaldehyde Semicarbazide

Ans:

12.7. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

(i) Methanal

(ii) 2-Methylpentanal

(iii) Benzaldehyde.

(iv) Benzophenone

(v) Cyclohexanone

(vi) 1-Phenylpropanone

(vii) Phenylacetaldehyde

(viii) Butan-l-ol 1

(ix) 2,2-Dimethylbutanal

Ans: 2-Methylpertfanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde contain one or more a-hydrogen and hence undergo aldol condensation. The reactions and the structures of the expected products are given below:

12.8. How will you convert ethanal into the following compounds?

(i) Butane-1,3-diol

(ii) But-2-enal

(iii) But-2-enoic acid

Ans:

12.9. Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Ans:

12.10. An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. 

Ans: Since the given compound with molecular formula C9H10O forms a 2,4-DNP derivative and reduces Tollen’s reagent, it must be an aldehyde. Since it undergoes Cannizzaro reaction, therefore, CHO group is directly attached to die benzene ring.

Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore, it must be an ortho- substituted benzaldehyde. The only o-substituted aromatic aldehyde having molecular formula C9H10O is o-ethyl benzaldehyde. Ail the reactions can now be explained on the basis of this structure.

12.11. An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B} and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (Q on dehydration gives but-l-ene. Write equations for the reactions involved.

Ans: Since an ester A with molecular formula C8H16O2 upon hydrolysis gives carboxylic acid B and the alcohol C and oxidation of C with chromic acid produces the acid B, therefore, both the carboxylic acid B and alcohol C must contain the same number of carbon atoms.

Further, since ester A contains eight carbon atoms, therefore, both the carboxylic acid B and the alcohol C must contain four carbon atoms each.

Since the alcohol C on dehydration gives but-l-ene, therefore, C must be a straight chain alcohol, i.e., butan-l-ol.

If C is butan-l-ol, then the acid B must be butanoic acid and the ester A must be butyl butanoate.The chemical equations are as follows:

12.12. Arrange the following in increasing order of the property indicated :

(i) Acetaldehyde, Acetone, Di tert. butyl ketone, Methyl tert. butyl ketone (reactivity towards HCN). (C.B.S.E. Sample Paper 2011, 2015, C.B.S.E. Delhi 2012)

(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) (C.B.S.E. Delhi2008)

(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 5-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) (C.B.S.E. Sample Paper 2011, 2015; C.B.S.E. Delhi 2012, C.B.S.E. Outside Delhi 2015, Rajasthan Board 2015)

Ans: (i) Cyanohydrin derivatives are formed as a result of the reaction in which the nucleophile (CN– ion) attacks the carbon atom of the carbonyl group. The order of reactivity

• decreases with increase in +I effect of the alkyl group.
• decreases with increase in steric hindrance due to the size as well as number of the alkyl groups. In the light of the above information, the decreasing order of reactivity is :
  

(ii) We know that alkyl group with +I effect decreases the acidic strength. The +I effect of isopropyl group is more than that of n-propyl group. Similarly, bromine (Br) with -I-effect increases the acidic strength. Closer its position in the carbon atom chain w.r.t., carboxyl (COOH) group, more will be its -I-effect and stronger will be the acid. In the light of this, the increasing order of acidic strength is :

(CH3)2CHCOOH< CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br) COOH

(iii) We have learnt that the electron donating group (OCH3) decreases the acidic strength of the benzoic acid. At the same time, the electron withdrawing group (N02) increases the same. Keeping this in mind, the increasing order of acidic strength is:

12.13. Give simple chemical tests to distinguish between the following pairs of compounds.

(i) PropanalandPropanone

(ii) Acetophenone and Benzophenone

(iii) Phenol and Benzoic acid

(iv) Benzoic acid and Ethyl benzoate

(v) Pentan-2-one and Pentan-3-one

(vi) Benzaldehyde and Acetophenone.

(vii) EthanalandPropanal

Ans:

12.14. Row will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.

(i) Methyl benzoate

(ii) m-nitrobenzoic acid

(iii) p-nitrobenzoic acid

(iv) Phenylaceticacid

(v) p-nitrobenzaldehyde

Ans:

12.15. How will you bring about the following conversions in not more than two steps?

(i) PropanonetoPropene

(ii) Benzoic acid to Benzaldehyde

(iii) Ethanol to 3-Hydroxybutanal

(iv) Benzene to m-Nitroacetophenone

(v) Benzaldehyde to Benzophenone –

(vi) Bromobenzeneto 1-PhenylethanoL

(vii) Benzaldehyde to 3-Phenylpropan-1-ol.

(viii) Benzaldehyde to α Hydroxyphenylacetk acid

(ix) Benzoic acid to m-Nitrobenzy 1 alcohol

Ans:

12.16. Describe the following:

(i) Acetylation

(ii) Cannizzaro reaction

(iii) Cross aldol condensation

(iv) Decarboxylation

Ans: (i) Acetylation refers to the process of introducing an acetyl group into a compound namely, the substitution of an acetyl group for an active hydrogen atom. Acetylation is usually carried out in presence of a base such as pyridine, dimethylanitine, etc.

(ii) Cannizzaro reaction : Aldehydes which do not contain an a-hydrogen atom, when treated with concentrated alkali solution undergo disproportionation, i.e., self oxidation reduction. As a result, one molecule of the aldehyde is reduced to the corresponding alcohol at the cost of the other which is oxidised to the corresponding carboxylic acid. This reaction is called Cannizzaro reaction.

(iii) Cross aldol condensation: Aldol condensation between two different aldehydes is called cross aldol condensation.If both aldehydes contain a-hydrogens, It gives a mixture of four products.

(iv) Decarboxylation: The process of removal of a molecule of CO2 from a carboxylic acid is called decarboxylation. Sodium salts of carboxylic acids when heated with soda-lime undergoes decarboxylation to yield alkanes.

12.17. Complete each synthesis by giving missing starting material, reagent or products.

Ans:

12.18. Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2, fctrimethylcyclohexanone does not

(ii) There are two – NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(iii)During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Ans:

The yield of second reaction is very low because of the presence of three methyl groups at ex-positions with respect to the C = O, the nucleophilic attack by the CN– ion does not occur due to steric hinderance. Since there is no such steric hindrance in cyclohexanone, therefore, nucleophilic attack by the CN– ion occurs readily and hence cyclohexanone cyanohydrin is obtained in good yield.

Although semicarbazide has two – NH2 groups but one of them (i.e., which is directly attached to C = O) is involved in resonance as shown above. As a result, electron density on N of this -NH2 group decreases and hence it does not act as a nucleophile. In contrast, the other -NH2 group (i.e.. attached to NH) is not involved in resonance and hence lone pair of electrons present on N atom of this -NH2 group is available for nucleophilic attack on the C = O group of aldehydes and ketones.’

(iii) The formation of esters from a carboxylic acid and an alcohol in presence of an acid catalyst is a reversible reaction.

Thus to shift the equilibrium in the forward direction, the water or the ester formed should be removed as fast as it is formed.

12.19. An organic compound contains 69-77% carbon, 11-63 % hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tottens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Ans:

Since the compound form sodium hydrogen sulphite addition product, therefore, it must be either an – aldehyde or methyl/ cyclic ketone. Since the compound does not reduce Tollens’ reagent therefore, it cannot be an aldehyde. Since the compound gives positive iodoform test, therefore, the given compound is a methyl ketone. Since the given compound on vigorous oxidation gives a mixture ofethanoic acid and propanoic acid, therefore, the methyl ketone is pentan-2-one, i.e.,

12.20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than on phenol. Why?

Ans: Consider the resonating structures of carboxylate ion and phenoxide ion.

In case of phenoxide ion, structures (V – VII) carry a negative charge on the less electronegative carbon atom.Therefore, their contribution towards the resonance stabilization of phenoxide ion is very small.

In structures I and II, (carboxylate ion), the negative charge is delocalized over two oxygen atoms while in structures III and IV, the negative charge on the oxygen atom remains localized only the electrons of the benzene ring are delocalized. Since delocalization of benzene electrons contributes little towards the stability of phenoxide ion therefore, carboxylate ion is much more resonance stabilized than phenoxide ion. Thus, the release of a proton from carboxylic acids is much easier than from phenols. In other words, carboxylic acids are stronger acids than phenols.

Class 12 Chemistry Chapter 3 Electrochemistry

Section Name	Topic Name
3	Electrochemistry
3.1	Electrochemical Cells
3.2	Galvanic Cells
3.3	Nernst Equation
3.4	Conductance of Electrolytic Solutions
3.5	Electrolytic Cells and Electrolysis
3.6	Batteries
3.7	Fuel Cells
3.8	Corrosion

3.1. How would you determine the standard electrode potential of the system Mg2+1 Mg?

Ans: A cell will be set up consisting of Mg/MgSO4 (1 M) as one electrode and standard hydrogen electrode Pt, H, (1 atm)H+/(l M) as second electrode, measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that e-1 s flow from mg electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as follows :

3.2. Can you store copper sulphate solutions in a zinc pot?

Ans: Zn being more reactive than Cu, displaces Cu from CuSO4 solution as follows:Zn (s) + CuSO4 (aq) –> ZnSO4(ag)+Cu (s)

3.3. Consult the table on standard electrode potentials and suggest three substances that can oxidise Fe2+ ions under suitable conditions.

Ans. The oxidation of Fe2+ ions to Fe3+ ions proceeds as follows :

Fe2+ → 3+ + e– ; E∘OX = – 0·77 V

Only those substances can oxidise Fe2+ ions to Fe3+ ions which can accept electrons released during oxidation or are placed above iron in electrochemical series. These are : Cl2(g), Br2(g) and Cr2O2−7 ions (in the acidic medium).

3.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Ans. For hydrogen electrode,H+ + e– —>1/2 H2,

3.5. Calculate the emf of the cell in which the following reaction takes place:

Ni(s)+2Ag+ (0.002 M) -> Ni2+ (0.160 M)+2Ag(s) Given that E(-)(cell) = 1.05 V .

Ans:

3.6. The cell in which the following reaction occurs: 2Fe3+ (aq) + 2I– (aq) —> 2Fe2+ (aq) +I2 (s) has E°cell=0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Ans:

3.7. Why does the conductivity of a solution decrease with dilution?

Ans: The conductivity of a solution is linked with the number of ions present per unit volume. With dilution, these decrease and the corresponding conductivity or specific conductance of the solution decreases.

3.8.  Suggest a way to determine the value of water.

Ans:

3.9. The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation and dissociation constant Given λ°(H+)=349.6 S cm2 mol-1 andλ°(HCOO-) = 54.6 S cm2 mol-1

Ans:

3.10. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

Ans:

3.11. Suggest a list of metals which can be extracted electrolytically.

Ans: The highly reactive metals having large -ve E° values, which can themselves act as powerful reducing agents can be extracted electrolytically. The process is known as electrolytic reduction. For details, consult Unit-6. For example, sodium, potassium, calcium, magnesium etc.

3.12. Consider the reaction: Cr2O72--+ 14H+ + 6e- -> 2Cr3+ + 7H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72- ?

Ans:

3.13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Ans: A lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide (PbO2) and 38% H2SO4 solution as electrolyte. When the battery is in use, the reaction taking place are:

On charging the battery, the reverse reaction takes place, i.e., PbSO4 deposited on electrodes is converted back to Pb and PbO2 and H2SO4 is regenerated.

3.14. Suggest two materials other than hydrogen that can be used as fuels in the fuel cells.

Ans: Methane (CH4) and methanol (CH3OH) can also be used as fuels in place of hydrogen in the fuel cells.

3.15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Ans: The water present on the surface of iron dissolves acidic oxides of air like CO2 , SO2 , etc. to form acids which dissociate to give H+ ions :

Thus, an electrochemical cell is set up on the surface.

Ferrous ions are further oxidised by atmospheric oxygen to ferric ions which combine with water to form hydrated ferric oxide, Fe2O3. xH2O, which is rust.

NCERT EXERCISES

3.1. Arrange the following metals in the order in which they displace each other from their salts.

Al, Cu, Fe, Mg and Zn

Sol: Mg, Al, Zn, Fe, Cu.

3.2. Given the standard electrode potentials, K+/K=-2. 93 V, Ag+/Ag = 0.80 V, Hg2+/Hg =0.79V, Mg2+/Mg=-2.37V, Cr3+/Cr=0.74V.

Arrange these metals in their increasing order of reducing power.

Sol: Higher the oxidation potential more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag<Hg<Cr<Mg<K.

3.3. Depict the galvanic cell in which the reaction

Zn(s) + 2Ag+(aq) —-> 7M2+(aq) + 2Ag (s) takes place. Further show:

(i) Which of the electrode is negatively charged?

(ii) The carriers of the current in the cell.

(iii) Individual reaction at each electrode.

Sol. The set-up will be similar to as shown below,

(i) Anode, i. e, zinc electrode will be negatively charged.

(ii) The current will flow from silver to copper in the external circuit.

(iii) At anode: Zn(s) ——–> Zn2+(aq) + 2e–

At cathode: 2Ag+(aq) + 2e– ——–> 2Ag(s)

3.4. Calculate the standard cell potentials of the galvanic cells in which the following reactions take place.

Also calculate ∆G° and equilibrium constant for the reaction. (C.B.S.E. Outside Delhi 2008)

Sol:

3.5. Write the Nernst equation and emf of the following cells at 298 K:

Sol:

3.6. In the button cells widely used in watches and other devices the following reaction takes place:

Sol:

3.7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Sol: The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by K (kappa). Thus, if K is the specific conductance and G is the conductance of the solution, then

Now, if I = 1 cm and A = 1 sq.cm, then K = G.

Hence, conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte.

Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by ∆m.

Variation of conductivity and molar conductivity with concentration: Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume.

3.8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.

Sol:

3.9. The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 x 10-3 S cm-1?

Sol:

3.10. The conductivity of NaCl at 298 K has been determined at different concentrations and the results are given below:

Sol:

3.11. Conductivity of 0.00241 M acetic acid is 7.896 x 10-5 S cm-1. Calculate its molar conductivity. If Λm0, for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?

Sol:

3.12. How much charge is required for the following reductions:

(i) 1 mol of Al3+ to Al?

(ii) 1 mol of Cu2+ to Cu ?

(iii) 1 mol of Mn04- to Mn2+?

Sol: (i) The electrode reaction is Al3+ + 3e ——> Al

∴ Quantity of charge required for reduction of 1 mol of Al3+=3F=3 x 96500C=289500C.

(ii) The electrode reaction is Cu2+ + 2e– ——–> Cu

∴ Quantity of charge required for reduction of 1 mol of Cu2+=2F=2 x 96500=193000 C.

(iii) The electrode reaction is Mn04- ———-> Mn2+.

i.e., Mn7+ + 5e–——-> Mn2+.

∴ Quantity of charge required = 5F

=5 x 96500 C=4825000.

3.13. How much electricity in terms of Faraday is required to produce :

(i) 20·0 g of Ca from molten CaCl2

(ii) 40·0 g of Al from molten Al2O3 ?

Sol:

3.14. How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to 02 (ii) 1 mol of FeO to Fe203

Sol:

3.15. A solution of Ni(N03)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Sol:

3.16. Three electrolytic cells A, B, C containing solutions of ZnS04, AgNO3 and CuS04, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 45 g of silver deposited at the cathode of call B. How long did the current flow? What mass of copper and zinc were deposited?

Sol:

3.17. Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible.

(a) Fe3+(aq) and I–(aq)

(b) Ag+(aq) and Cu(s)

(c) Fe3+(aq) and Br–(aq)

(d) Ag(s) and Fe3+(aq)

(e) Br2(aq) and Fe2+(aq).

Sol:

A particular reaction can be feasible if e.m.f. of the cell based on the E° values is positive. Keeping this in mind, let us predict the feasibility of the reactions.

3.18. Predict the products of electrolysis in each of the following.

(i) An aqueous solution of AgNO3 with silver electrodes.

(ii) An aqueous solution of AgNO3 with platinum electrodes.

(iii) A dilute solution of H2S04 with platinum electrodes.

(iv) An aqueous solution of CuCl2 with platinum electrodes.

Sol:

Chapter 1 The Solid State

Section Name	Topic Name
1	The Solid State
1.1	General Characteristics of Solid State
1.2	Amorphous and Crystalline Solids
1.3	Classification of Crystalline Solids
1.4	Crystal Lattices and Unit Cells
1.5	Number of Atoms in a Unit Cell
1.6	Close Packed Structures
1.7	Packing Efficiency
1.8	Calculations Involving Unit Cell Dimensions
1.9	Imperfections in Solids
1.10	Electrical Properties
1.11	Magnetic Properties

1.1. Why are solids rigid?

Ans: The constituent particles in solids have fixed positions and can oscillate about their mean positions. Hence, they are rigid.

1.2. Why do solids have definite volume?

Ans: Solids keep their volume because of rigidity in their structure. The interparticle forces are very strong. Moreover, the interparticle spaces are very few and small as well. As a result, their volumes cannot change by applying pressure.

1.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fibreglass, copper

Ans: Crystalline solids: Benzoic acid, potassium nitrate, copper Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibreglass

1.4. Why is glass considered as super cooled liquid ? (C.B.S.E. Delhi 2013)

Ans: Glass is considered to be super cooled liquid because it shows some of the characteristics of liquids, though it is an amorphous solid. For example, it is slightly thicker at the bottom. This can be possible only if it has flown like liquid, though very slowly.

1.5. Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?

Ans: As the solid has same value of refractive index along all directions, it is isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.

1.6. Classify the following solids in different categories based on the nature of the intermolecular forces: sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubedium, argon, silicon carbide.

Ans: Ionic, metallic, molecular, molecular, molecular (hydrogen-bonded), molecular (hydrogen-bonded), ionic, covalent, metallic, molecular, covalent (network).

1.7. Solid A is a very hard electrical insulator in. solid as well as in molten state and melts at extremely high temperature. What type of solid is it?

Ans: It is a covalent or network solid.

1.8. Why are ionic solids conducting in the molten state and not in the solid-state?

Ans: In the ionic solids, the electrical conductivity is due to the movement of the ions. Since the ionic mobility is negligible in the solid state, these are non-conducting in this state. Upon melting, the ions present acquire some mobility. Therefore, the ionic solids become conducting

1.9. What type of solids are electrical conductors, malleable and ductile?

Ans: Metallic solids

 1.10. Give the significance of a lattice point.

Ans: The lattice point denotes the position of a particular constituent in the crystal lattice. It may be atom, ion or a molecule. The arrangement of the lattice points in space is responsible for the shape of a particular crystalline solid.

1.11. Name the parameters that characterise a unit cell.

Ans: A unit cell is characterised by the following parameters:

(i)the dimensions of unit cell along three edges: a, b and c.

(ii)the angles between the edges: α (between b and c); β (between a and c) and γ (between a and b)

1.12. Distinguish between :

(i) Hexagonal and monoclinic unit cells

(ii) Face-centred and end-centred unit cells.

Ans:

(i) In a hexagonal unit cell :

a = b # c; α = β = 90° and γ = 120°

In a monoclinic unit cell :

a # b # c and α = γ = 90° and β # 90°

(ii) In a face-centered unit cell, constituent particles are located at all the corners as well as at the centres of all the faces.

In end-centered unit cell, constituent particles are located at all the corners as well as at the centres of two opposite faces. (C.B.S.E Foreign 2015)

1.13. Explain how many portions of an atom located at

(i)corner and (ii)body centre of a cubic unit cell is part of its neighbouring unit cell.

Ans: (i) An atom at the comer is shared by eight adjacent unit cells. Hence, portion of the atom at the comer that belongs to one unit cell=1/8.

(ii)An atom at the body centre is not shared by any other unit cell. Hence, it belongs fully to unit cell.

1.14. What is the two-dimensional coordination number of a molecule in a square close-packed layer?

Ans: In the two-dimensional square close-packed layer, a particular molecule is in contact with four molecules. Hence, the coordination number of the molecule is four.

1.15. A compound forms hexagonal close-packed. structure. What is the total number of voids in 0. 5 mol of it? How many of these are tetrahedral voids?

Ans:

No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 1023 =3.011 x 1023

No. of octahedral voids = No. of atoms in packing =3.011 x 1023

No. of tetrahedral voids = 2 x No. of atoms in packing

= 2 x 3.011 x 1023 = 6.022 x 1023

Total no. of voids = 3.011 x 1023 + 6.022 x 1023

= 9.033 x 1023

1.16. A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound? (C.B.S.E. Foreign 2015)

Ans: Let us suppose that,

the no. of atoms of N present in ccp = x

Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,

the no. of tetrahedral voids occupied = 2x/3

The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2

∴ The formula of the compound = N3M2 or M2N3

1.17. Wh ich of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centered cubic and (iii) hexagonal close-packed lattice?

Ans: Packing efficiency of:

Simple cubic = 52.4% bcc = 68% hcp = 74%

hcp lattice has the highest packing efficiency.

1.18. An element with molar mass 2:7 x 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2:7 x 103 kg m-3, what is the nature of the cubic unit cell ? (C.B.S.E. Delhi 2015)

Ans: 

Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc) or cubic close packed (ccp).

1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

Ans: When a solid is heated, vacancy defect is produced in the crystal. On heating, some atoms or ions leave the lattice site completely, i.e., lattice sites become vacant. As a result-of this defect, density of the substances decreases.

1.20. What types of stoichiometric defects are shown by (C.B.S.E. Delhi 2013)

(i) ZnS

(ii) AgBr?

Ans:

(i) ZnS crystals may show Frenkel defects since the cationic size is smaller as compared to anionic size.

(ii) AgBr crystals may show both Frenkel and Schottky defects.

1.21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

Ans: Let us take an example NaCl doped with SrCl, impurity when SrCl2 is added to NaCl solid as an impurity, two Na+ ions will be replaced and one of their sites will be occupied by Sr21- while the other will remain vacant. Thus, we can say that when a cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

1.22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.

Ans: Let us take an example of NaCl. When NaCl crystal is heated in presence of Na vapour, some Cl–ions leave their lattice sites to combine with Na to form NaCl. The e-1 s lost by Na to form Na+ (Na+ + Cl–—> NaCl) then diffuse into the crystal to occupy the anion vacancies. These sites are called F-centres. These e-s absorb energy from visible light, get excited to higher energy level and when they fall back to ground state, they impart yellow colour to NaCl crystal.

1.23. A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?

Ans: Impurity from group 15 should be added to get n-type semiconductor.

1.24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

Ans: Ferromagnetic substances make better permanent magnets. This is because when placed in magnetic field, their domains get oriented in the directions of magnetic field and a strong magnetic field is produced. This ordering of domains persists even when external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

1.1. Define the term ‘amorphous’. Give a few examples of amorphous solids.

Sol. Amorphous solids are those substances, in which there is no regular arrangement of its constituent particles, (i.e., ions, atoms or molecules). The arrangement of the constituting particles has only short-range order, i.e., a regular and periodically repeating pattern is observed over short distances only, e.g., glass, rubber, and plastics.

1.2. What makes glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Sol. Glass is a supercooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO2) in which tetrahedral units SiO4 are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz can be converted into glass by melting it and cooling the melt very rapidly. In the glass, SiO4 tetrahedra are joined in a random manner.

1.3 Classify each of the following solids as ionic, metallic, modular, network (covalent), or amorphous:

(i) Tetra phosphorus decoxide (P4O10) (ii) Ammonium phosphate, (NH4)3PO4 (iii) SiC (iv) I2 (v) P4 (vii) Graphite (viii), Brass (ix) Rb (x) LiBr (xi) Si

Sol.

1.4 (i) What is meant by the term ‘coordination number’?

(ii) What is the coordination number of atom

(a) in a cubic close-packed structure?

(b) in a body centred cubic structure?

Sol. (i) The number of nearest neighbours of a particle are called its coordination number.

(ii) (a) 12 (b) 8

1.5. How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer. (C.B.S.E. Outside Delhi 2011)

Sol.

1.6 ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Sol. Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following. Water = 273 K, Ethyl alcohol = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.

The intermoleoilar forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol. Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal’s forces (or London dispersion forces).

1.7. How will you distinguish between the following pairs of terms :

(a) Hexagonal close packing and cubic close packing

(b) Crystal lattice and unit cell

(c) Tetrahedral void and octahedral void.

Sol.

(a) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer

(ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCABC…..type).

(b) Crystal lattice: It deplicts the actual shape as well as size of the constituent particles in the crystal. It is therefore, called space lattice or crystal lattice.

Unit cell: Each bricks represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.

(c) Tetrahedral void: A tetrahedral void is formed when triangular void made by three spheres of a particular layer and touching each other.

Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over anothet set of spheres.

1.8 How many lattice points are there is one unit cell of each of the following lattices?

(i) Face centred cubic (if) Face centred tetragonal (iii) Body centred cubic

Sol.

1.9 Explain:

(i) The basis of similarities and differences between metallic and ionic crystals.

(ii) Ionic solids are hard and brittle.

Sol. (i) Metallic and ionic crystals

Similarities:

(a) There is electrostatic force of attraction in both metallic and ionic crystals.

(b) Both have high melting points.

(c) Bonds are non-directional in both the cases.

Differences:

(a) Ionic crystals are bad conductors of electricity in solids state as ions are not free to move. They can conduct electricity only in die molten state or in aqueous solution. Metallic crystals are good conductors of electricity in solid state as electrons are free to move.

(b) Ionic bond is strong due to strong electrostatic forces of attraction.

Metallic bond may be strong or weak depending upon the number of valence electrons and the size of the kernels.

(ii) Ionic solids are hard and brittle.Ionic solids are hard due to the presence of strong electrostatic forces of attraction. The brittleness in ionic crystals is due to the non- directional bonds in them.

1.10 Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic, (ii) body centred cubic, and (iii) face centred cubic (with the assumptions that atoms are touching each other).

Sol. Packing efficiency: It is the percentage of total space filled by the particles.

1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10-8 cm and density is 10.5 g cm-3, calculate the atomic mass of silver.

Sol.

1.12. A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q?

Sol. Contribution by atoms Q present at the eight corners of the cube = 18= x 8 = 1

Contribution by atom P present at the body centre = 1

Thus, P and Q are present in the ratio 1:1.

∴ Formula of the compound is PQ.

Co-ordination number of atoms P and Q = 8.

1.13 Niobium crystallises in a body centred cubic structure. If density is 8.55 g cm-3, calculate atomic radius of niobium, using its atomic mass 93u.

Sol.

1.14 If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between rand R.

Sol. A sphere is fitted into the octahedral void as shown in the diagram.

1.15 Copper crystallises into a fee lattice with edge length 3.61 x 10-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm-3.

Sol.

This calculated value of density is closely in agreement with its measured value of 8.92 g cm3.

Question 16.

Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Solution:

98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni2+ = x

Then Ni present as Ni3+ = 98 – x

Total charge on x Ni2+ and (98 – x) Ni3+ should

be equal to charge on 100 O2- ions.

Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94

∴ Fraction of Ni present as Ni2+ = 9498 × 100 = 96%

Fraction of Ni present as Ni3+ = 498 × 100 = 4%

Question 17.

What are semi-conductors? Describe the two main types of semiconductors and contrast their conduction mechanisms.

Solution:

Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two

main types of semiconductors are n-type and p-type.

(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 element like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor: When a silicon or germanium is doped with group 13 element like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of missing fourth electron. Here, this hole moves throughout the crystal like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-type semiconductor, p stands for positive hole, since it is the positive hole that is responsible for conduction.

Question 18.

Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Solution:

The ratio less than 2 : 1 in Cu20 shows cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. For maintaining electrical neutrality, every two Cu+ ions will be replaced by one Cu2+ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor.

Question 19.

Ferric oxide crystallises in a hexagonal dose- packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Solution:

Suppose the number of oxide ions (O2-) in the packing = 90

∴ Number of octahedral voids = 90

As 2/3rd of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions 2 present = 23 × 90 = 60

∴ Ratio of Fe3+ : O2- = 60 : 90 = 2 : 3

Hence, the formula of ferric oxide is Fe2O3.

Question 20.

Classify each of the following as being either a p-type or n-type semiconductor :

1. Ge doped with In
2. B doped with Si.

Solution:

1. Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
2. B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.

Question 21.

Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?

Solution:

For a face centred cubic unit cell (fcc)

Edge length (a) = 22–√r = 2 x 1.4142 x 0.144 mm = 0.407 nm

Question 22.

In terms of band theory, what is the difference

1. between a conductor and an insulator
2. between a conductor and a semiconductor?

Solution:

In most of the solids and in many insulating solids conduction takes place due to migration of electrons under the influence of electric field. However, in ionic solids, it is the ions that are responsible for the conducting behaviour due to their movement.

(i) In metals, conductivity strongly depends upon the number of valence electrons available in an atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other, as to form a band. If this band is partially filled or it overlaps with the higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal behaves as a conductor.

If the gap between valence band and next higher unoccupied conduction band is large, electrons cannot jump into it and such a substance behaves as insulator.

(ii) If the gap between the valence band and conduction band is small, some electrons may jump from valence band to the conduction band. Such a substance shows some conductivity and it behaves as a semiconductor. Electrical conductivity of semiconductors increases with increase in temperature, since more electrons can jump to the conduction band. Silicon and germanium show this type of behaviour and are called intrinsic semiconductors. Conductors have no forbidden band.

Question 23.

Explain the following terms with suitable examples :

1. Schottky defect
2. Frenkel defect
3. Interstitial defect
4. F-centres.

Solution:

(i) Schottky defect : In Schottky defect a pair of vacancies or holes exist in the crystal lattice due to the absence of equal number of cations and anions from their lattice points. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent.

(ii) Frenkel defect : This defect arises when some of the ions in the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation, e.g., AgBr, ZnS, etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal.

(iii) Interstitial defect : When some constituent particles (atoms or molecules) occupy an interstitial site of the crystal, it is said to have interstitial defect. Due to this defect the density of the substance increases.

(iv) F-Centres : These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.

Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.

1. What is the length of the side of the unit cell?
2. How many unit cells are there in 1.00 cm3 of aluminium?

Solution:

(i) For an fee unit cell, r = a22√ (given, r = 125 pm)

a = 2√2 r = 2√2 × 125 pm

= 353.55 pm

≅354 pm

(ii) Volume of one unit cell = a3 = (354 pm)3

= 4.4 × 107 pm3

= 4.4 × 107 × 10-30cm3

= 4.4 × 10-23 cm3

Therefore, number of unit cells in 1.00 cm3 = 

= 2.27 × 1022

Question 25.

If NaCI is doped with 10-3 mol % SrCl2, what is the concentration of cation vacancies?

Solution:

Let moles of NaCI = 100

∴ Moles of SrCl2 doped = 10-3

Each Sr2+ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.

∴ Moles of cation vacancy in 100 moles NaCI = 10-3

Moles of cation vacancy in one mole

NaCI = 10-3 × 10-2 = 10-5

∴ Number of cation vacancies

= 10-5 × 6.022 × 1023 = 6.022 × 1018 mol-1

Question 26.

Explain the following with suitable example:

1. Ferromagnetism
2. Paramagnetism
3. Ferrimagnetism
4. Antiferromagnetism
5. 12-16 and 13-15 group compounds.

Solution:

(i) Ferromagnetic substances : Substances which are attracted very strongly by a magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co and CrO2 show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetic moments are due to unpaired electrons in the same direction.

The ferromagnetic material, CrO2, is used to make magnetic tapes used for audio recording.

(ii) Paramagnetic substances : Substances which are weakly attracted by the external magnetic field are called paramagnetic substances. The property thus exhibited is called paramagnetism. They are magnetised in the same direction as that of the applied field. This property is shown by those substances whose atoms, ions or molecules contain unpaired electrons, e.g., O2, Cu2+, Fe3+, etc. These substances, however, lose their magnetism in the absence of the magnetic field.

(iii) Ferrimagnetic substances : Substances which are expected to possess large magnetism on the basis of the unpaired electrons but actually have small net magnetic moment are called ferrimagnetic substances, e.g., Fe3O4, ferrites of the formula M2+Fe2O4 where M = Mg, Cu, Zn, etc. Ferrimagnetism arises due to the unequal number of magnetic moments in opposite direction resulting in some net magnetic moment.

(iv) Antiferromagnetic substances : Substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but actually they possess zero net magnetic moment are called antiferromagnetic substances, e.g., MnO. Antiferromagnetism is due to the presence of equal number of magnetic moments in the opposite directions

(v) 13-15 group compounds : When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc.

12-16 group compounds : Combination of elements of groups 12 and 16 yield some solid compounds which are referred to as 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have ionic character.

Class 12 Chemistry Chapter 8 The d and f Block Elements 

Section Name	Topic Name
8	The d – and f – Block Elements
8.1	Position in the Periodic Table
8.2	Electronic Configurations of the d-Block Elements
8.3	General Properties of the Transition Elements (d-Block)
8.4	Some Important Compounds of Transition Elements
8.5	The Lanthanoids
8.6	The Actinoids
8.7	Some Applications of d – and f -Block Elements

8.1. Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element?

Ans: The outer electronic configuration of Ag (Z=47) is 4d105s1. It shows+1 and + 2 O.S. (in AgO and AgF2). And in + 2 O.S., the electronic configuration is d9 i.e, d-subshell is incompletely filled. Hence, it is a transition element.

8.2. In the series Sc(Z = 21) to (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol-1. Why?

Ans: The enthalpy of atomisation is directly linked with the stability of the crystal lattice and also the strength of the metallic bond. In case of zinc (3d104s2 configuration), no electrons from the 3d-orbitals are involved in the formation of metallic bonds since all the orbitals are filled. However, in all other elements belonging to 3d series one or more d-electrons are involved in the metallic bonds. This means that the metallic bonds are quite weak in zinc and it has therefore, lowest enthalpy of atomisation in the 3d series.

8.3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Ans: Manganese (Z = 25) shows maximum number of O.S. This is because its outer EC is 3d54s2. As 3d and 4s are close in energy, it has maximum number of e-1 s to loose or share. Hence, it shows O.S. from +2 to +7 which is the maximum number.

8.4.

Ans.

8.5. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Ans:  There is a irregularity in the IE’s of 3d-series due to alternation of energies of 4s and 3d orbitals when an e-1 is removed. Thus, there is a reorganisation energy accompanying ionization. This results into release of exchange energy which increases as the number of e-1 s increases in the dn configuration. Cr has low 1st IE because loss of 1 e- gives stable EC (3d6). Zn has very high IE because e~ has to be removed from 4s orbital of the stable configuration (3d10 4s2) After the loss of one e–, removal of 2nd e–, becomes difficult. Hence, 2nd IE’s are higher and in general, increase from left to right. However, Cr and Cu show much higher values because 2nd e– has to be removed from stable configuration of Cr+ (3d5) and Cu+ (3d10)

8.6. Why is the highest oxidation state of a metal exhibited by its fluoride and oxide only? (C.B.S.E. Delhi 2010)

Ans: Both fluorine and oxygen have very high electronegativity values. They can oxidise the metals to the highest oxidation state. As a result, the highest oxidation states are shown by the fluorides and oxides of the metals; transition metals in particular.

8.7.Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Ans: Cr2+ is a stronger reducing agent than Fe2+. This is because E°(Cr3+/Cr2+) is negative (- 0.41V) whereas E°(Fe3+/Fe2+) is positive (+ 0.77 V). Thus, Cr2+ is easily oxidised to Fe3+ but Fe2+ cannot be easily oxidised to Fe3+.

8.8.Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).

Ans:

8.9.Explain why Cu+ ion is not stable in aqueous solutions?

Ans: Cu+ (aq) is not stable, while Cu2+ (aq) is stable. This is becuase ΔhydH of Cu2+(aq) is much higher than that of Cu+(aq) and hence it compensates for the 2nd IE of Cu. Thus, many Cu(I) compounds are unstable in aqueous solution and undergo disproportionation as follows :

2 Cu+ —–> Cu2+ + Cu

8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (C.B.S.E. Sample Paper 2011, Jharkhand Board 2010)

Ans: The decrease or contraction in atomic radii, as well as ionic radii in actinoid elements (actinoid contraction), is more as compared to lanthanoid contraction because 5/ electrons have more poor shielding effect as compared to 4f electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

8.1. Write down the electronic configuration of (i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+(v) Co2+ (vi) Lu2+(vii) Mn2+ (viii) Th4+.

Sol: (i) Cr3+ = [Ar]183d3

(ii)Pm3+ = [Xe]54 4f4

(iii)Cu+ = [Ar]18 3d10

(iv)Ce4+ = [Xe]54

(v)Co2+ = [Ar]18 3d7

(vi)Lu2+ = [Xe]54 4f14 5d1

(vii) Mn2+ = [Ar]18 3d5 (viii)Th4+= [Rn]86

8.2. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their+3 state?

Sol: Electronic configuration of Mn2+ is 3d5. This is a half-filled configuration and hence stable. Therefore, third ionization enthalpy is’very high, i. e., third electron cannot be lost easily. Electronic configuration of Fe2+ is 3d6. It can lose one electron easily to achieve a stable configuration 3d5.

8.3. Explain briefly how+2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol: Here after losing 2 electrons from j-orbitals, the 3d-orbital gets gradually occupied with increase in atomic number. Since the number of unpaired electrons in 3d orbital increases, the stability of the cations (M2+) increases from Sc2+ to Mn2+.

8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Sol: In the first series of transition elements, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, Mn (Z = 25) has electronic configuration [Ar] 3d5 4 s2. It shows oxidation states + 2 to + 7 but Mn (II) is most stable because of half-filled configuration [Ar] 3d5. Similarly Sc3+ is more stable then Sc+ and Fe3+ is more stable than Fe2+ due to half filled it f-orbitals.

8.5. What may be the stable oxidation state of the transition element with the following delectron configurations in the ground state of their atoms: 3d3,3d5, 3d8 and 3d4?

Sol: (a) 3d3 4s1 = + 5.

(b) 3d5 4s2 = + 2, + 7,3d5 4s1 =+6.

(c)3d84s2 = + 2.

(d)3d44s2 = 3d5 4s1 = + 6(and + 3).

8.6. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol: Cr2072- and Cr042- (Group number = Oxidation state of Cr = 6).

Mn04–  (Group number = Oxidation state of Mn = 7).

8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol: Lanthanoid Contraction : In the lanthanoids , the electrons are getting filled in the 4f-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the 4 f- electrons However,

the f-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.

Consequences of lanthanoid Contraction

(a)Separation Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.

(b)Variation in basic strength of hydroxides: Due to lanthanoid contraction, size of M3+ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. Thus basic character of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3.

(c)Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of Y and La is quite, large as compared to the difference in the value of Zr and Hf. This is because of the lanthanoid contraction.

(d)Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.

M3+ (aq) + 3e– —–> 4 M(aq)

(e)Variation in physical properties like melting point, boiling point, hardness etc.

8.8. What are the characteristics of the transition . elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Sol: General characteristics of transition elements.

(i)Electronic configuration – (n -1) d1-10 ns1-2

(ii)Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.

(iii)Atomic and ionic size-ions of same charge in a given series show progressive decrease in radius with increasing atomic number.

(iv)Oxidation state-Variable; ranging from+2 to +7.

(v)Paramagnetism – The ions with unpaired electrons are paramagnetic.

(vi)Ionisation enthalpy – Increases with increase in charge.

Formation of coloured ions – Due to presence of unpaired electrons.

(viii) Formation of complex compounds – Due to small size and high charge density of metal ions.

(ix)They possess catalj^c properties – Due to

their ability to adopt multiple oxidation states. .

(x)Formation of interstitial compounds.

(xi)Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or in any stable oxidation state and they are placed between s and p- block elements. Zn, Cd and Hg have fully filled d- orbitals in their ground state hence may not be regarded as the transition elements.

8.9. In what way are the electronic configuration of the transition elements different from non-transition elements?

Sol: Electronic configuration of transition elements : (n – 1)d1-10 ns1-2. Electronic configuration of non-transition elements : ns1-2 or ns2np1-6. From comparison, it is quite evident that the transition elements have incomplete d-orbitals (s- orbitals in some cases) while the non-transition elements have no d-orbitals present in the valence shells of their atoms. This is responsible for the difference in the characteristics of the elements belonging to these classess of elements.

8.10. What are the different oxidation states exhibited by the lanthanoids?

Sol: Lanthanides exhibits + 2, + 3 and + 4 oxidation states. The most common oxidation state of lanthanoids is +3.

8.11. Explain giving reasons:

(i)Transition metals and many of their compounds show paramagnetic behaviour.

(ii)The enthalpies of atomisation of the transition metals are high.

(iii)The transition metals generally form coloured compounds.

(iv)Transition metals and their many compounds act as good catalyst

Sol: (i) Magnetic properties: Transition elements and many of their compounds are paramagnetic, i.e., they are weakly attracted by a magnetic field. This is due to the presence of unpaired electrons in atoms, ions or molecules. The paramagnetic character increases as the number of . unpaired electrons increases. The paramagnetic character is measured in terms of magnetic moment and is given by

μ=

n(n+2)−−−−−−−

√

 where n – number of unpaired electrons.

(ii) Because of large number of unpaired electrons in d-orbitals of their atoms they have stronger interatomic intefactions and hence stronger metallic bonding between atoms resulting in higher enthalpies of atomisation.

(iii) Formation of coloured compounds (both in solid state as well as in aqueous solution) is another very common characteristics of transition metals. This is due to absorption of some radiation from visible light to cause d-d transition of electrons in transition metal atom. The d-orbitals do not have same energy and under the influence of ligands, the d-orbitals split into two sets of orbitals having different energies; transition of electrons can take place from one set of d-orbitals to another set within the same sub-shell. Such transitions are called d-d transitions. The energy difference for these d-d transitions fall in the visible region. When white light is incident on compounds of transition metals, they absorb a particular frequency and remaining colours are emitted imparting a characteristic colour to the complex. Zn2+ and Ti4+ salts are white because they do not absorb any radiation in visible region.

(iv)Catalytic properties: Many of transition metals and their compounds act as catalyst in variety of reactions, e.g., finely divided iron in manufacture of NH3 by Haber’s process, V2O5 or Pt in manufacture of H2S04 by Contact process, etc.). The catalytic activity is due to following two reasons.

(a)The ability of transition metal ion to pass ” easily from one oxidation state to another

and thus providing a new path to reaction with lower activation energy.

(b)The surface of transition metal acts as very good adsorbent and thus provides increased concentration of reactants on their surface causing the reaction to occur.

8.12. What are interstitial compounds? Why are such compounds well known for transition metals?

Sol: Transition metals form large number of interstitial compounds. They are able to entrap small atoms of elements like H, G, N, B, etc., in their crystal lattice and even can make weak bonds with them.

Due to formation of interstitial compounds, their malleability and ductility decreases and tensile . strength increases. Steel and cast iron are hard in comparison to wrought iron due to the presence of trapped carbon atoms in interstitial spaces.

8.13. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Sol: The transition metals show a number of variable oxidation states due to the participation of (n – 1) d electrons in addition to ns electrons in the bond formation. They therefore, exhibit a large number of variable oxidation states. On the other hand, the non-transition metals generally belonging to s-block do not show variable oxidation states because by the loss of valence s-electrons, they acquire the configuration of the nearest noble gas elements.

In the p-block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO3 and WO3 are not.

8.14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Sol: Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr2O3) with sodium or potassium carbonate in free excess of air. The reaction with sodium carbonate occurs as follows:

The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr,07.2H20 can be crystallised.

Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.

Orange crystals of potassium dichromate crystallise out. The chromates and dichromates depending upon pH of the solution. If pH of potassium dichromate is increased it is converted to yellow potassium chromate.

8.15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i)iodide

(ii)iron (II) solution and

(iii)H2S

Sol: K2Gr207is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from+6 to + 3. The oxidising action can be represented as follows:

8.16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (II) ions (ii) S02 and (iii) oxalic acid? Write the ionic, equations for the reactions.

Sol: Potassium permanganate (KMn04) is prepared by the fusion of a mixture of pyrolusite (Mn02),potassiufn hydroxide and oxygen, first green coloured potassium manganate is formed. 2MnO2 + 4KOH + 02 —> 2K2Mn04+2H20 The potassium manganate is extracted by water, which then undergoes disproportionation in neutral or acidic solution to give potassium permanganate.

8.17. For M2+/M and M3+/M2+ systems the E° values

for some metals are as follows:

Cr2+/Cr   –> -0.9 V

Mn2+/Mn  –> -1.2V

Fe2+/Fe     –> -0.4 V

Cr3+/Cr2+  –> -0.4 V

Mn3+/Mn2+   –>+ 1.5V

Fe3+/Fe2+   –>+ 0.8V

(ii)the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Sol: (i) Cr3+/Cr2+ has negative reduction potential. Hence, Cr3+ cannot be reduced to Cr2+. Mn3+/Mn2+ has a large positive reduction potential. Hence, Mn3+ can be easily reduced to Mn2+. Fe3+/Fe2+ has small positive reduction potential. Hence, Fe3+ is more stable than Mn3+ but less stable than Cr3+.

(ii)From the E° values, the order of oxidation of the metal to the divalent cation is : Mn > Cr > Fe.

8.18. Predict which of the following will be coloured in aqueous solution?

Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+, Co2+.

Sol: Only those ions will be coloured which have incomplete d-orbitals. The ions with either empty or filled d-orbitals are colourless. Keeping this in view, the coloured ions among the given list are :