CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. How would you account for the following ?
Many of the transition elements are known to form interstitial compounds. [CBSE Delhi 2012]
Answer:
In the crystal lattice, transition elements have interstitial vacant space into which small sized non-metal atoms such as H, B, C, or N are trapped. These compounds are non-stoichiometric, neither typically ionic nor covalent. e.g., TiC, MH, Fe₃H etc.

Teacher's Note:
1. Interstitial compounds are harder and have higher melting points than the pure metal, but retain metallic conductivity.

Question 2. When Cu²⁺ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation. [NCERT Exemplar]
Answer:
2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂
(Cu₂I₂ is the white ppt.)
Cu²⁺ gets reduced to Cu⁺ while I⁻ oxidizes to I₂.

Teacher's Note:
1. This is a redox reaction: the white Cu₂I₂ often looks brown at first because of the I₂ liberated with it.

Teacher's Note:
1. Ce⁴⁺ (4f⁰) and Eu²⁺ (4f⁷) are the classic exceptions to the +3 state; Dy (vi) is the distractor.

Question 4. How would you account for the following? Zr (Z = 40) and Hf (Z = 72) have almost identical radii. [CBSE Delhi 2013]
Answer:
Due to lanthanoid contraction, the atomic radii of 4d and 5d transition series elements are almost same.

Teacher's Note:
1. The 14 lanthanoids (4f filling) come between Zr's period and Hf's period, and their poor shielding cancels the size increase expected for the extra shell.

Question 5. Explain the following observation: The members of the actinoid series exhibit a large number of oxidation states than the corresponding members of the lanthanoid series. [CBSE OD 2012]
Answer:
Due to the comparable energies of 5f, 6d and 7s levels members of actinoid series exhibit a large number of oxidation states.

Teacher's Note:
1. In lanthanoids the 4f electrons are buried deep and are not easily used, so +3 dominates; actinoids show states up to +7 (e.g. Np, Pu).

Question 6. What are the different oxidation states exhibited by the lanthanoids? [NCERT]
Answer:
In the lanthanide series, +3 oxidation state is most common that is Ln (III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

Teacher's Note:
1. Give an example of each extra state for full marks: Eu²⁺, Yb²⁺ (+2) and Ce⁴⁺, Tb⁴⁺ (+4).

Question 7. Explain CuSO₄.5H₂O is blue while CuSO₄ is colourless.
Answer:
Because water molecules act as ligands and result in crystal field splitting of d-orbitals of Cu²⁺ ion.

Teacher's Note:
1. Without ligands the d-orbitals are not split, so no d–d transition in the visible region is possible.

Question 8. Which element among 3d series exhibits highest oxidation state?
Answer:
Mn

Teacher's Note:
1. Mn shows +7 (in MnO₄⁻), using all its 3d and 4s electrons.

Question 9. In 3d series (Sc to Zn), the enthalpy of atomisation of Zn is low. Why?
Answer:
Due to absence of unpaired electrons.

Teacher's Note:
1. Zn (3d¹⁰4s²) has no unpaired electrons, so metallic bonding is weak.

Question 10. Which element among 3d series exhibits only one oxidation state?
Answer:
Sc

Teacher's Note:
1. Sc shows only +3; Zn also shows only one state (+2), so either is accepted if justified by configuration.

Question 11. Why is the 3rd ionisation energy of Mn (Z = 25) unexpectedly high?
Answer:
Due to half-filled electronic configuration (3d⁵) of Mn²⁺.

Teacher's Note:
1. The third electron has to be removed from the stable 3d⁵ configuration of Mn²⁺.

Short Answer Questions (2 Marks)

Question 12. Identify the following:
(i) Transition metal of 3d series that exhibits the maximum number of oxidation states.
(ii) An alloy consisting of approximately 95% lanthanoid metal used to produce bullet, shell and lighter flint. [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) Mn
(ii) Mischmetal

Teacher's Note:
1. Mn (3d⁵4s²) shows every state from +2 to +7; mischmetal is about 95% lanthanoid, 5% iron).

Question 13. Give reasons for the following:
(i) E° values of Mn, Ni and Zn are more negative than expected.
(ii) [Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless. [HOTS]
OR
Ti³⁺ is coloured whereas Sc³⁺ is colourless in aqueous solution. [CBSE 2020 (56/5/1)]
Answer:
(i) Negative E° values for Mn²⁺ and Zn²⁺ are related to stabilities of half-filled and fully filled configurations, respectively. But for Ni²⁺, E° value is related to the highest negative enthalpy of hydration.
(ii) This is due to d-d transition of electron in [Ti(H₂O)₆]³⁺ complex. Ti³⁺ has one electron in d-orbital (3d¹) which absorbs energy corresponding to blue-green region and jumps from t_2g to e_g set of d-orbitals (t_2g¹e_g⁰ → t_2g⁰e_g¹). But Sc³⁺ has no electron in the d-orbital.

Teacher's Note:
1. Ti³⁺ absorbs blue-green light, so the solution looks purple (the complementary colour).

Question 14. How would you account for the following:
(i) Mn (III) undergoes disproportionation reaction easily.
(ii) Co (II) is easily oxidised in the presence of strong ligands. [CBSE (F) 2011]
Answer:
(i) Mn³⁺ is less stable and changes to Mn²⁺ which is more stable due to half filled d-orbital configuration. That is why, Mn³⁺ undergoes disproportionation reaction.
(ii) Co (II) has electronic configuration 3d⁷4s⁰, i.e., it has three unpaired electrons. In the presence of strong ligands, two unpaired electrons in 3d-subshell pair-up and third unpaired electron shifts to higher energy subshell from where it can be easily lost and hence oxidised to Co(III).

Teacher's Note:
1. The disproportionation is 2Mn³⁺ + 2H₂O → Mn²⁺ + MnO₂ + 4H⁺; writing it earns credit.

Question 15. When FeCr₂O₄ is fused with Na₂CO₃ in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured (C). An acidified solution of compound (C) oxidises Na₂SO₃ to (D). Identify (A), (B), and (D.) [CBSE 2019 (56/2/1)]
Answer:
A = Na₂CrO₄ (Sodium chromate), B = Na₂Cr₂O₇ (Sodium dichromate), C = K₂Cr₂O₇ (Potassium dichromate), D = Na₂SO₄ (Sodium sulphate)

Teacher's Note:
1. Learn the three steps of K₂Cr₂O₇ preparation: chromite → chromate (yellow) → dichromate (orange) → potassium dichromate.

Question 16. Complete the following chemical reaction equations:
(i) MnO₄⁻(aq) + C₂O₄²⁻(aq) + H⁺(aq) →
(ii) Cr₂O₇²⁻(aq) + Fe²⁺(aq) + H⁺(aq) →
Answer:
(i) [MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] × 2
[C₂O₄²⁻ → 2CO₂ + 2e⁻] × 5
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
(ii) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
[Fe²⁺ → Fe³⁺ + e⁻] × 6
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Teacher's Note:
1. Check every ionic equation for both atoms and total charge: in (i) the charge is +4 on each side.

Question 17. Complete the following chemical equations:
(i) 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O →
(ii) Cr₂O₇²⁻ + 3Sn²⁺ + 14H⁺ → [CBSE Delhi 2016]
Answer:
(i) [MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻] × 8
[S₂O₃²⁻ + 10OH⁻ → 2SO₄²⁻ + 5H₂O + 8e⁻] × 3
8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻
(ii) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
[Sn²⁺ → Sn⁴⁺ + 2e⁻] × 3
Cr₂O₇²⁻ + 3Sn²⁺ + 14H⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Teacher's Note:
1. Water as a reactant with no H⁺ signals a neutral medium, where MnO₄⁻ gives MnO₂ (3 electrons), not Mn²⁺.

Question 18. Complete and balance the following equations:
(i) MnO₄⁻ + I⁻ + H⁺ →
(ii) Na₂Cr₂O₇ + KCl → [CBSE 2019 (56/5/2)]
Answer:
(i) 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
(ii) Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Teacher's Note:
1. Reaction (ii) works because K₂Cr₂O₇ is less soluble than Na₂Cr₂O₇ and crystallises out.

Question 19. Transition metals can act as catalysts because these can change their oxidation state. How does Fe(III) catalyse the reaction between iodide and persulphate ions? [NCERT Exemplar]
Answer:
Reaction between iodide and persulphate ions is:
2I⁻ + S₂O₈²⁻ \( \xrightarrow{\text{Fe(III)}} \) I₂ + 2SO₄²⁻
Role of Fe(III) ions:
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻

Teacher's Note:
1. The catalyst is regenerated in the second step; adding the two steps gives back the overall reaction.

Question 20. Explain each of the following observations:
(i) Mn²⁺ is much more resistant than Fe²⁺ towards oxidation. [CBSE Delhi 2012]
(ii) Among lanthanoids, Ln(III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained. [CBSE (AI) 2012]
Answer:
(i) Mn³⁺(d⁴) is less stable than Mn²⁺ (d⁵, half filled) while Fe³⁺ (d⁵, half filled) is more stable than Fe²⁺ (d⁶). That is why Mn²⁺ is more resistance than Fe²⁺ towards oxidation.
(ii) Lanthanoid metals show +2 and +4 oxidation states to attain extra stable f⁰ and f⁷ configurations.

Teacher's Note:
1. Examples help in (ii): Ce⁴⁺ (f⁰), Eu²⁺ and Tb⁴⁺ (f⁷), Yb²⁺ (f¹⁴).

Question 21. Explain each of the following observations:
(i) Actinoids exhibit a much larger number of oxidation states than the lanthanoids. [CBSE 2019 (56/2/1)]
(ii) There is hardly any increase in atomic size with increasing atomic numbers in a series of transition metals. [CBSE (F) 2012]
Answer:
(i) This is due to small energy gap between 5f, 6d and 7s subshells in actinoids.
(ii) This is because with increase in atomic number in a series, the increased nuclear charge is partly cancelled by the increased shielding effect of electrons in the d-orbitals of penultimate shell.

Teacher's Note:
1. The new electron enters the inner (n − 1)d shell, so the outer shell, which decides size, hardly changes.

Question 22. Suggest reasons for the following features of transition metal chemistry :
(i) The transition metals and their compounds are usually paramagnetic.
(ii) The transition metals exhibit variable oxidation states. [CBSE Comptt. Delhi 2015]
Answer:
(i) Due to presence of unpaired electrons in d-orbitals.
(ii) Due to incomplete filling of d-orbitals. Due to very small energy difference in between (n−1)d and n s-orbitals.

Teacher's Note:
1. Paramagnetism increases with the number of unpaired electrons; species with all electrons paired (e.g. Zn²⁺, Sc³⁺) are diamagnetic.

Question 23. Complete the following chemical equations :
(i) MnO₄⁻(aq) + S₂O₃²⁻(aq) + H₂O(l) →
(ii) Cr₂O₇²⁻(aq) + Fe²⁺(aq) + H⁺(aq) → [CBSE Delhi 2012]
Answer:
(i) 8MnO₄⁻(aq) + 3S₂O₃²⁻(aq) + H₂O(l) → 8MnO₂(s) + 6SO₄²⁻(aq) + 2OH⁻(aq)
(ii) Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6Fe²⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)

Teacher's Note:
1. In neutral or faintly alkaline medium permanganate is reduced only to MnO₂ (3 electrons), so thiosulphate is oxidised to sulphate.
2. Always check atoms and charges: in (ii) the charge is +24 on both sides.

Short Answer Questions (3 Marks)

Question 24. Give reasons:
(a) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
(b) Iron has higher enthalpy of atomisation than that of copper.
(c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured. [CBSE Delhi/OD 2018]
Answer:
(a) Because Mn²⁺ is more stable than Mn³⁺ due to half-filled d⁵ configuration whereas Fe²⁺ becomes unstable after loosing an electron from half-filled orbital.
(b) Due to presence of higher number of unpaired electrons in iron, they have stronger metallic bonding. Hence, the enthalpy of atomisation is more of iron than that of copper.
(c) Sc³⁺ is colourless as it does not contain unpaired electrons to undergo d-d transition while Ti³⁺ is coloured as it contains unpaired electrons to undergo d-d transition by absorbing light from visible region and radiate complementary colour.

Teacher's Note:
1. In (a) compare the configurations: Mn²⁺ 3d⁵ → Mn³⁺ 3d⁴ breaks a half-filled set, while Fe²⁺ 3d⁶ → Fe³⁺ 3d⁵ creates one.

Question 25. Give reasons for the following:
(a) Transition metals show variable oxidation states.
(b) E° value for (Zn²⁺/Zn) is negative while that of (Cu²⁺/Cu) is positive.
(c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7. [CBSE OD Set-2 2019]
Answer:
(a) Because of comparable energies of (n − 1)d and ns orbitals / Incomplete filling of d-orbital.
(b) Because of stable 3d¹⁰ configuration of Zn²⁺ whereas due to low hydration enthalpy and high enthalpy of atomization of Cu²⁺.
(c) Due to the ability of oxygen to form multiple bonds with metal.
Detailed Answer:
(a) Because of availability of partially filled orbitals and comparable energies of ns and (n − 1)d orbitals.
(b) E° value for (Zn²⁺/Zn) is negative due to stable completely filled d¹⁰ configuration in Zn²⁺. The positive value of (Cu²⁺/Cu) accounts for its inability to liberate H₂ from acids due to its high enthalpy of atomization and low hydration energy.
(c) Mn can form multiple bonds with oxygen by using 2p orbital of oxygen and 3d orbital of Mn because of which it shows highest oxidation state of +7 with fluorine, Mn cannot form multiple bonds thus shows an oxidation state of +4.

Teacher's Note:
1. A positive E°(M²⁺/M) means the metal cannot displace H₂ from dilute acids; copper is the only 3d metal of this kind.

Question 26. Give reasons for the following:
(i) Transition elements act as catalysts.
(ii) It is difficult to obtain oxidation state greater than two for copper.
(iii) Cr₂O₇²⁻ is a strong oxidising agent in acidic medium whereas WO₃ and MoO₃ are not.
Answer:
(i) Due to large surface area and ability to show variable oxidation states.
(ii) Due to high value of third ionisation enthalpy.
(iii) Mo(VI) and W(VI) are more stable than Cr(VI).

Teacher's Note:
1. Down a group of d-block elements, higher oxidation states become more stable (Cr(VI) < Mo(VI) < W(VI)), the opposite of the trend in p-block groups.

Question 27. Observed and calculated values for the standard electrode potentials of elements from Ti to Zn in the first reactivity series are depicted in figure (1):
Explain the following observations:
(i) The general trend towards less negative E° values across the series
(ii) The unique behaviour of Copper
(iii) More negative E° values of Mn and Zn [CBSE SQP 2021]

Answer:
(i) The general trend towards less negative E° values across the series is related to the general increase in the sum of the first and second ionisation enthalpies.
(ii) The high energy to transform Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.
(iii) The stability of the half-filled d sub-shell in Mn²⁺ and the completely filled d¹⁰ configuration in Zn²⁺ are related to their more negative E° values.

Teacher's Note:
1. When a graph shows a dip below the trend line (Mn, Zn), look for an extra-stable configuration of the M²⁺ ion.

Question 28. Following ions are given:
Cr²⁺, Cu²⁺, Cu⁺, Fe²⁺, Fe³⁺, Mn³⁺
Identify the ion which is
(i) a strong reducing agent.
(ii) unstable in aqueous solution.
(iii) a strong oxidising agent.
Give suitable reason in each. [CBSE OD Set-1 2020]
Answer:
(i) Cr²⁺, because its configuration changes from d⁴ to d³ and having a half-filled t_2g level.
(ii) Cu⁺ in an aqueous medium energy is required to remove one electron from Cu⁺ to Cu²⁺, high hydration energy of Cu²⁺ compensates for it. Therefore Cu⁺ ion in an aqueous solution is unstable.
2Cu⁺ → Cu²⁺(aq) + Cu(s)
(iii) Mn³⁺, because its configuration changes from Mn³⁺ to Mn²⁺ results in the half filled d⁵ configuration, which has extra stability.

Teacher's Note:
1. Cr²⁺ and Mn³⁺ are both d⁴; the direction each one moves (to d³ or to d⁵) decides whether it reduces or oxidises.

Question 29. Account for the following:
(i) Eu²⁺ is a strong reducing agent.
(ii) Orange colour of dichromate ion changes to yellow in alkaline medium.
(iii) E°(M²⁺/M) values for transition metals show irregular variation. [CBSE Foreign Set-2 2017]
Answer:
(i) Eu²⁺ is a strong reducing agent because Eu³⁺ is more stable than Eu²⁺.
(ii) Dichromate ion changes to chromate ion/OH⁻
Cr₂O₇²⁻ (orange) → CrO₄²⁻ (yellow)
(iii) Due to the irregular variation in ionisation enthalpies (sum of 1st and 2nd ionisation enthalpies), heat of sublimation and enthalpy of hydration/due to irregular electronic configurations from left to right in a period, which changes the ionisation potential.
Detailed Answer:
(i) Electronic configuration of Eu = 4f⁷6s². On oxidation, the evolution of the electrons takes place. After the removal of 2 electrons it achieves stable half filled electronic configuration (Eu²⁺, 4f⁷), but it readily loses one more electron to attain the common and more stable +3 oxidation state, so Eu²⁺ acts as a strong reducing agent.
(ii) Cr₂O₇²⁻ + H₂O ⇄ 2CrO₄²⁻ + 2H⁺
(Orange)      (yellow)
When an alkali is added to an orange solution of dichromate, a yellow solution is obtained due to the formation of chromate ions.

Teacher's Note:
1. +3 is the characteristic oxidation state of all lanthanoids; any lanthanoid ion in +2 (Eu²⁺, Yb²⁺) tends to be oxidised to +3 and so acts as a reducing agent.

Question 30. Explain the following:
(a) Out of Sc³⁺, Co²⁺ and Cr³⁺ ions, only Sc³⁺ is colourless in aqueous solutions. (Atomic no. Co = 27; Sc = 21 and Cr = 24)
(b) The \( E^{\circ}_{Cu^{2+}/Cu} \) for copper metal is positive (+0.34), unlike the remaining members of the first transition series.
(c) La(OH)₃ is more basic than Lu(OH)₃. [CBSE SQP 2018-2019]
Answer:
(a) Co²⁺ : [Ar]3d⁷ Sc³⁺ : [Ar]3d⁰ Cr³⁺ : [Ar]3d³
Co²⁺ and Cr³⁺ have unpaired electrons. Thus, they are coloured in aqueous solution. Sc³⁺ has no unpaired electron. Thus it is colourless.
(b) Metal copper has high enthalpy of atomisation and enthalpy of ionisation. Therefore the high energy required to convert Cu(s) to Cu²⁺(aq) is not balanced by its hydration enthalpy.
(c) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with increase in atomic size. Thus covalent character between lanthanoid ion and OH⁻ increases from La³⁺ to Lu³⁺. Thus the basic character of hydroxides decreases from La(OH)₃ to Lu(OH)₃.

Teacher's Note:
1. Smaller cation → greater polarising power → more covalent M–OH bond → weaker base; this chain explains (c).

Question 31. Give the reasons for following:
(i) Transition elements and their compounds act as catalysts.
(ii) E° value for (Mn²⁺|Mn) is negative whereas for (Cu²⁺|Cu) is positive.
(iii) Actinoids show irregularities in their electronic configuration. [CBSE Delhi Set 1 2019]
Answer:
(i) Due to variable oxidation state.
(ii) Mn²⁺ is stable due to exactly half filled 3d⁵ configuration/ Due to high Δ_aH° and low Δ_hydH° for Cu²⁺/Cu is positive.
(iii) Due to comparable energies of 5f, 6d and 7s orbitals.
Detailed Answer:
(i) Transition metals have the ability to adsorb many other substances on their surface and activate them as a result of chemisorption. Transition metals also exhibit a variety of oxidation states, and can change oxidation states relatively easily. This makes transition metals and their compounds good catalysts.
(ii) E° values are indicative of the stability of the oxidized form of the element. The lower the E° value, more stable the oxidized form of the element. Mn²⁺ with a half filled d-subshell (d⁵) is stable, so Mn is easily oxidized to Mn²⁺, making the E° value negative. Cu²⁺ with a partially filled d subshell (d⁹) is not stable, and is relatively easily reduced to element form. This makes its E° value positive.
(iii) The electronic configurations of actinoids show irregularities because the energies of their 5f, 6d, and 7s orbitals are close to each other. Electrons can easily move between these subshells.

Teacher's Note:
1. For catalysis give both reasons: variable oxidation states (forming unstable intermediates) and a surface that adsorbs reactants.

Question 32. Give reasons for the following:
(i) Transition metals form alloys.
(ii) Mn₂O₃ is basic whereas Mn₂O₇ is acidic.
(iii) Eu²⁺ is a strong reducing agent. [CBSE Delhi Set 2 2019]
Answer:
(i) Due to comparable radii / comparable size.
(ii) In Mn₂O₃, Mn is in +3 (lower) oxidation state while in Mn₂O₇, Mn is in higher oxidation state (+7)
(iii) Because its stable oxidation state is +3.
Detailed Answer:
(i) Transition metals easily form alloys with other transition metals because they have almost similar size. So, they can easily replace each other in the crystal lattice.
(ii) The transition metal oxides in the lower oxidation state of metals are basic in nature and in higher oxidation state they are acidic in nature. The oxidation state of Mn in Mn₂O₃ is +3 and Mn₂O₇ has +7. Therefore, Mn₂O₃ is basic and Mn₂O₇ is acidic.
(iii) The common oxidation state of lanthanide metals is +3. Eu²⁺ is formed by losing the two s electrons acquires half filled (4f⁷) configuration. But still, they oxidise to their common +3 state. So the Eu²⁺ loses one electron and is oxidized to Eu³⁺. So, Eu²⁺ acts as a strong reducing agent.

Teacher's Note:
1. Remember the series for manganese oxides: MnO (basic), Mn₂O₃, MnO₂ (amphoteric), Mn₂O₇ (acidic); acidity rises with oxidation state.

Question 33. Give reasons for the following:
(a) Transition metals have high enthalpies of atomization.
(b) Manganese has lower melting point even though it has a higher number of unpaired electrons for bonding.
(c) Ce⁴⁺ is a strong oxidising agent. [CBSE OD Set 3 2019]
Answer:
(a) Because of strong interatomic interactions / Strong metallic bonding between atoms.
(b) Due to stable 3d⁵ configuration, interatomic interaction is poor between unpaired electrons.
(c) Because Ce is more stable in +3 oxidation state.
Detailed Answer:
(a) The transition elements have high enthalpies of atomisation because they have large number of unpaired electrons in their atoms. This results in stronger interatomic interaction and stronger bonding between atoms.
(b) Mn has low melting point because of 3d⁵4s² configuration, which is highly stable and delocalised, so are not available for bonding, as a result interatomic forces becomes weaker. Hence Mn has low melting point bounded half-filled d orbital electrons with nucleus results in weak interatomic interaction.
(c) The formation of Ce⁴⁺ is promoted by its noble gas configuration reverting to the common +3 state. E° value for Ce⁴⁺/Ce³⁺ is + 1.74 V thus readily gains an electron and acts as a strong oxidising agent.

Teacher's Note:
1. Ce⁴⁺ (4f⁰) is stable enough to exist, but +3 is still more stable, so Ce⁴⁺ takes an electron readily; this is why it is used as an oxidant in titrations.

Question 34. (a) What happens when
(i) Manganate ions (MnO₄²⁻) undergoes disproportionation reaction in acidic medium?
(ii) Lanthanum is heated with sulphur?
(b) Explain the following trends in the properties of the members of the First series of transition elements:
(i) E°(M²⁺/M) value for copper is positive (+0.34 V) in contrast to the other members of the series.
(ii) Cr²⁺ is reducing while Mn³⁺ is oxidising, though both have d⁴ configuration.
(iii) The oxidising power in the series increases in the order VO₂⁺ < Cr₂O₇²⁻ < MnO₄⁻. [CBSE SQP 2020]
Answer:
(a) (i) MnO₄²⁻ ions disproportionate in acidic medium to give permanganate ions and manganese(IV) oxide.
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
(ii) Lanthanum sulphide is formed.
2La + 3S \( \xrightarrow{\text{heat}} \) La₂S₃
(Deduct overall ½ mark if equation not balanced/statements not written)
(b) (i) Copper has high enthalpy of atomisation and low enthalpy of hydration, thus the high energy is required to transform Cu(s) to Cu²⁺(aq) which is not balanced by hydration enthalpy, therefore E°(M²⁺/M) value for copper is positive (+0.34 V).
(ii) Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having more stable half filled t_2g level. On the other hand, the change from Mn³⁺ to Mn²⁺ results an extra stable d⁵ configuration.
(iii) This is due to the increasing stability of the species of lower oxidation state to which they are reduced.

Teacher's Note:
1. In the disproportionation of MnO₄²⁻, Mn(+6) goes up to +7 (MnO₄⁻) and down to +4 (MnO₂); check the electrons: 2 × 1 lost = 1 × 2 gained.

Teacher's Note:
1. E° depends on three terms: enthalpy of atomisation, ionisation enthalpy and hydration enthalpy; explain any anomaly using these.

Long Answer Questions (5 Marks)

Question 36. (i) (a) Which transition element in 3d series has positive \( E^{\circ}_{M^{2+}/M} \) value and why?
(b) Name a member of lanthanoid series which is well known to exhibit +4 oxidation state and why?
(ii) Account for the following
(a) The highest oxidation state is exhibited in oxoanions of transition metals.
(b) HCl is not used to acidify KMnO₄ solution.
(c) Transition metals have high enthalpy of atomisation. [CBSE Comptt. Delhi Set-1, 2, 3 2017]
Answer:
(i) (a) Copper; Due to high Δ_aH° and low Δ_hydH°
(b) Cerium; Due to stable 4f⁰ configuration/Tb; Due to stable 4f⁷ configuration
(ii) (a) Due to ability of oxygen to form multiple bonds to metal
(b) HCl is oxidized to chlorine
(c) Due to strong interatomic bonding
Detailed Answer:
(ii) (a) Due to high electronegativity and small size, oxygen acts as a strong oxidising agent. This results in oxygen’s ability to oxidise the metal to attain highest oxidation state.
(b) As KMnO₄ is a very strong oxidising agent, it oxidizes HCl resulting in evolution of chlorine gas. Therefore, HCl is not used to acidify KMnO₄ solution.

Teacher's Note:
1. Ce⁴⁺ is 4f⁰ (noble gas core) and Tb⁴⁺ is 4f⁷; both +4 states are linked to an empty or half-filled 4f set.

Question 37. (a) Give reasons:
(i) Transition metals and their compounds show catalytic activities.
(ii) Separation of a mixture of Lanthanoid elements is difficult.
(iii) Zn, Cd and Hg are soft and have low melting point.
(b) Account for the following:
(i) Ti³⁺ is coloured whereas Sc³⁺ is colourless in aqueous solution.
(ii) Cr²⁺ is a strong reducing agent.
Answer:
(a) (i) The catalytic activities of transition metals and their compounds is due to the ability of adopt variable oxidation states and to form complexes. It can also provide a large surface area for the reactants to be adsorbed. (Any one)
(ii) Separation of lanthanoid elements is difficult because all lanthanoid elements have almost similar physical as well as chemical properties. Due to the lanthanoid contraction the change in the atomic or ionic radii is very small.
(iii) Zn, Cd and Hg are soft and have low melting point because no d-orbitals are available for metallic bond formation and bonds formed are very weak.
(b) (i) Ti³⁺ has incomplete d (3d¹) orbital whereas Sc³⁺ has empty (3d⁰) d-orbital.
(ii) Cr²⁺ ion can lose electron to form Cr³⁺, so acts as a strong reducing agent.

Teacher's Note:
1. For (b)(ii), always add the reason for the change: Cr³⁺ (d³, half-filled t_2g) is more stable than Cr²⁺ (d⁴).

Question 38. (i) The elements of 3d transition series are given as: Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(a) Which element has the highest m.p. and why?
(b) Which element is a strong oxidising agent in +3 oxidation state and why?
(c) Which element is soft and why?
(ii) Write the equations involved in the preparation of potassium dichromate from sodium chromate (Na₂CrO₄). [CBSE (F) 2016]
Answer:
(i) (a) Cr, the highest melting point of Cr is attributed to the involvement of greater number of electrons (5) from 3d in addition to 4s electrons in interatomic metallic bonding.
(b) Mn, because the change from Mn³⁺ (d⁴) to Mn²⁺ (d⁵) results in the half filled configuration which has extra stability.
(c) Zn, in Zn (3d¹⁰4s²) all the electrons present in d-orbitals are paired and hence metallic bonds present in it are weak. That is why, it is soft.
(ii) Sodium chromate is acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na₂Cr₂O₇.2H₂O can be crystallised.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Teacher's Note:
1. Co³⁺ is also a strong oxidising agent (E° = +1.97 V), but Mn³⁺ is the expected answer because of the stable d⁵ product.

Question 39. (i) Is the variability in oxidation number of transition elements different from that of non-transition elements? Illustrate with examples.
(ii) Give reasons:
(a) d-block elements exhibit more oxidation states than f-block elements.
(b) Orange solution of potassium dichromate turns yellow on adding sodium hydroxide to it.
(c) Zirconium (Z = 40) and Hafnium (Z = 72) have almost similar atomic radii. [CBSE Sample Paper 2017]
Answer:
(i) In transition elements, the oxidation states differ from each other by unity, e.g., Fe³⁺ and Fe²⁺ etc., while in non-transition elements (p-block elements), the oxidation states differ by two, e.g., Pb⁴⁺ and Pb²⁺, etc.
In transition elements the higher oxidation states are more stable for the heavier elements in a group, e.g., Mo(VI) is more stable than Cr(VI) whereas in non-transition elements (p-block elements), the lower oxidation states are more stable for heavier elements due to inert pair effect, e.g., Pb(II) is more stable than Pb(IV).
(ii) (a) d–block elements exhibit more oxidation states because of less energy gap between d and s subshell whereas f-block elements have large energy gap between f and d subshell.
(b) On adding NaOH, pH of solution increases and the orange colour of the solution changes to yellow due to conversion of dichromate ion to chromate ion.
Cr₂O₇²⁻ (Dichromate ion, Orange) + 2OH⁻ ⇄ 2CrO₄²⁻ (Chromate ion, Yellow) + H₂O
(c) This is due to filling of 4f-orbitals which have poor shielding effect (Lanthanoid contraction).

Teacher's Note:
1. Part (ii)(a) is really about lanthanoids: actinoids show many oxidation states, so in exams compare d-block with the 4f series.

Question 40. (i) Describe the preparation of potassium permanganate from pyrolusite ore. Write balanced chemical equation for one reaction to show the oxidizing nature of potassium permanganate.
(ii) Draw the structures of chromate and dichromate ions. [CBSE Sample Paper 2017]
Answer:
(i) Conversion of pyrolusite (MnO₂) into potassium manganate (K₂MnO₄).
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
Electrolytic oxidation:
K₂MnO₄ ⇄ 2K⁺ + MnO₄²⁻
At anode: MnO₄²⁻ → MnO₄⁻ + e⁻
At cathode: H⁺ + e⁻ → H, 2H → H₂
It oxidises oxalate (C₂O₄²⁻) to carbon dioxide (CO₂) in acidic medium.
[MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O] × 2
[C₂O₄²⁻ → 2CO₂ + 2e⁻] × 5
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
(ii)

Teacher's Note:
1. Chromate is tetrahedral; dichromate is two tetrahedra sharing one corner oxygen, with a Cr–O–Cr bond angle of about 126°.