CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 08

Very Short Answer Questions (1 Mark)

Question 1. Zinc, cadmium and mercury are generally not considered as transition metals. Give reasons. (HP Board 2011, Assam Board 2013)
Answer:
These elements in their most common oxidation state of + 2 have completely filled d-orbitals. Hence, d-d transition cannot occur.

Teacher's Note:
1. Their atoms and their M²⁺ ions both have a full (n − 1)d¹⁰ subshell, so they do not meet the definition of a transition element.

Question 2. Name the element in the 3d series that
(i) shows maximum oxidation state
(ii) is diamagnetic (Karnataka Board 2012)
Answer:
(i) Manganese (ii) Zinc.

Teacher's Note:
1. Mn reaches +7 (in MnO₄⁻); Zn (3d¹⁰ 4s²) has all electrons paired.

Question 3. Why Zn²⁺ salts are white while Ni²⁺ salts or Cu²⁺ salts are blue? (HP Board 2011, Assam Board 2012, HP Board 2013)
Answer:
Zn²⁺ has completely filled d-orbitals (3d¹⁰) while Ni²⁺ has incompletely filled d-orbitals (3d⁸) or Cu²⁺ has 3d⁹. Hence, d-d transition can take place in Ni²⁺ salts or Cu²⁺ salts but not in Zn²⁺ salts.

Teacher's Note:
1. Strictly, aqueous Ni²⁺ salts are green and Cu²⁺ salts are blue; the reason (d–d transition) is the same for both.

Question 4. Calculate the magnetic moment of Ni²⁺. (Raj. Board 2012)
Answer:
For Ni²⁺ = 3d⁸ = ↑↓ ↑↓ ↑↓ ↑ ↑ . Hence, n = 2.
μ = \( \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \) = 2.84 BM.

Teacher's Note:
1. Fill the five d-orbitals singly first (Hund's rule) and then pair: 8 electrons give 3 pairs and 2 unpaired electrons.

Question 5. What happens when chromates are kept in acidic solution and dichromates in the alkaline solution ?
Answer:
Chromates in the acidic solution are converted into dichromates and thus the colour becomes orange whereas dichromates in the alkaline solution are converted into chromates and the colour becomes yellow.

Teacher's Note:
1. The interconversion is not a redox change: chromium stays +6 in both chromate and dichromate.

Question 6. Complete the following reaction :
(NH₄)₂Cr₂O₇ \( \xrightarrow{\text{Heat}} \)
Answer:
(NH₄)₂Cr₂O₇ (s) \( \xrightarrow{\text{Heat}} \) N₂ (g) + 4 H₂O (l) + Cr₂O₃ (s)

Teacher's Note:
1. This is an internal redox reaction: N goes from −3 to 0 and Cr from +6 to +3; it is used to prepare pure N₂ in the laboratory.

Question 7. Name one ore each of manganese and chromium. (Assam Board 2013)
Answer:
Mn = Pyrolusite, MnO₂ ;
Cr = Chromite/ferrochrome, FeO.Cr₂O₃.

Teacher's Note:
1. Chromite ore is written FeCr₂O₄ or FeO.Cr₂O₃; both formulas mean the same compound.

Question 8. What happens when KMnO₄ is heated with conc. H₂SO₄ or acidified oxalic acid or treated with acidified FeSO₄ ?
Answer:
(i) With conc. H₂SO₄, KMnO₄ forms manganese heptoxide, Mn₂O₇, which on warming decomposes to MnO₂ :
2KMnO₄ + 2H₂SO₄ (conc.) → Mn₂O₇ + 2KHSO₄ + H₂O ; 2Mn₂O₇ → 4MnO₂ + 3O₂
(ii) Acidified KMnO₄ oxidises oxalic acid to CO₂ and is itself decolourised :
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ \( \xrightarrow{333\,K} \) 2Mn²⁺ + 8H₂O + 10CO₂
(iii) Acidified KMnO₄ oxidises ferrous salt to ferric salt :
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Teacher's Note:
1. In acidic medium each MnO₄⁻ gains 5 electrons (Mn +7 to +2), so it oxidises 5 Fe²⁺ ions or 5/2 oxalate ions.

Question 9. Write one reaction using alkaline KMnO₄.
Answer:
CH₂=CH₂ (Ethylene) + H₂O + (O) \( \xrightarrow[\text{(Baeyer's reagent)}]{\text{Alkaline KMnO}_4} \) HOCH₂–CH₂OH (Ethylene glycol)

Teacher's Note:
1. Cold dilute alkaline KMnO₄ (Baeyer's reagent) is decolourised by alkenes, which makes it a test for unsaturation.

Question 10. Write ionic reaction for the reaction between MnO₄⁻ ions and oxalate ions at 333 K.
Answer:
2 MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ \( \xrightarrow{\text{Heat}} \) 2 Mn²⁺ + 8 H₂O + 10 CO₂ .

Teacher's Note:
1. Check the balance by charge: left side −2 + 16 − 10 = +4 and right side 2 × (+2) = +4.

Short Answer Questions (2 Marks)

Question 11. One among the lanthanides, Ce (III), can be easily oxidized to Ce (IV) (At. No. of Ce = 58). Explain why?
Or
Of the lanthanides, cerium (At. No. 58) forms a tetrapositive ion, Ce⁴⁺ in aqueous solution. Why?
Answer:
Ce (III) having the configuration 4f¹ 5d⁰ 6s⁰ can easily lose an electron to acquire the configuration 4f⁰ and form Ce(IV). In fact, this is the only (+ IV) lanthanide which exists in solution.

Teacher's Note:
1. Ce⁴⁺ has the noble gas (Xe) configuration, which is why it can exist, though it is still a good oxidising agent.

Question 12. The actinide contraction differs from lanthanide contraction? Explain.
Answer:
Lanthanide contraction refers to the gradual decrease in the size of the lanthanides and their trivalent ions. (Z = 58 to 71) whereas actinoid contraction refers to the gradual decrease in the size of the actinoids or their ions (M³⁺ or M⁴⁺) (Z = 90 to 103). They differ in the fact that in the actinoid series, the contraction is greater from element to element due to poor shielding by 5f electrons than by 4f electrons.

Teacher's Note:
1. 5f orbitals are even more diffuse than 4f, so they shield still worse and the contraction per element is larger.

Question 13. Which out of the two, La(OH)₃ and Lu(OH)₃, is more basic and why? [Manipur Board 2011]
Answer:
La(OH)₃ is more basic than Lu(OH)₃. As the size of the lanthanoid ions decreases from La³⁺ to Lu³⁺, the covalent character of the hydroxides increases (Fajan's rules). Hence, the basic strength decreases from La(OH)₃ to Lu(OH)₃.

Teacher's Note:
1. Smaller cation → more polarising → more covalent M–OH bond → weaker base.

Question 14. The ionization energies of the elements of first transition series do not vary much with increasing atomic number. Explain why.
Answer:
Along the series, the nuclear charge increases which tends to increase the ionization enthalpy. But at the same time, the electron is added to the inner (n − 1)d subshell, which shields the outer ns electrons from the nucleus. These two opposing effects nearly balance each other. Hence, the ionization enthalpies increase only slightly along the series.

Teacher's Note:
1. The same two opposing effects (nuclear charge and shielding by d-electrons) also explain why atomic radii stay nearly constant across the series.

Question 15. Transition metal compounds generally act as catalysts. Give reasons. (Hr. Board 2012, CBSE 2013)
Answer:
Transition metals and their compounds act as catalysts because :
(i) They can show variable oxidation states and so can form unstable intermediate compounds with the reactants, which then decompose to give products and regenerate the catalyst.
(ii) They provide a large surface area on which the reactants are adsorbed, so that bonds are weakened and the activation energy is lowered.
For example, V₂O₅ in the Contact process, finely divided Fe in Haber's process and Ni in hydrogenation of oils.

Teacher's Note:
1. Two reasons are expected for full marks: variable oxidation states (intermediate compound formation) and a surface on which reactants are held.

Question 16. Why do the transition elements form coloured compounds ? Explain. (CBSE 2008, MP Board 2011, Hr. Board 2012, Uttarakhand Board 2012, UP Board 2012)
Answer:
Transition metal ions have incompletely filled d-orbitals. In the presence of ligands, the five d-orbitals split into two sets of different energies. When visible light falls on the compound, an electron from the lower energy d-orbital absorbs a particular wavelength and jumps to the higher energy d-orbital (d-d transition). The colour observed is the complementary colour of the light absorbed.
For example, [Ti(H₂O)₆]³⁺ (3d¹) absorbs in the yellow-green region and appears purple. Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) configuration are colourless.

Teacher's Note:
1. The colour you see is the complementary colour of the light absorbed, not the absorbed colour itself.

Question 17. Explain why transition metal ions usually show paramagnetic behaviour. (Hr. Board 2012, UP Board 2012)
Answer:
Paramagnetism arises from the presence of unpaired electrons, each of which has a magnetic moment due to its spin and orbital motion. Most transition metal ions have incompletely filled d-orbitals and hence contain unpaired electrons. Therefore, they are paramagnetic. The magnetic moment increases with the number of unpaired electrons n, according to μ = \( \sqrt{n(n+2)} \) BM.

Teacher's Note:
1. Ions with all electrons paired (d⁰ or d¹⁰, e.g., Sc³⁺, Zn²⁺) are diamagnetic.

Question 18. Why do transition elements exhibit the tendency for complex formation ? Explain with suitable examples. (Raj. Board 2011)
Answer:
Transition metals form a large number of complexes because of :
(i) the comparatively small size of the metal atoms/ions,
(ii) their high ionic charges, and
(iii) the availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons from the ligands.
Examples : [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺, [Ni(CO)₄], [Co(NH₃)₆]³⁺.

Teacher's Note:
1. Small size and high charge give a high charge/size ratio, so the metal ion strongly attracts the lone pairs of ligands.

Question 19. What are interstitial compounds ? Why are such compounds well known for transition metals ? (Hr. Board 2011, 2012)
Answer:
Interstitial compounds are those which are formed when small atoms like H, C, N or B are trapped inside the crystal lattices (interstitial spaces) of metals, e.g., TiC, Mn₄N, Fe₃H, VH_0.56, TiH_1.7.
Transition metals form such compounds because their crystal lattices have a large number of interstitial voids of suitable size which can accommodate these small atoms.
These compounds are generally non-stoichiometric, have high melting points (higher than the pure metals), are very hard, retain metallic conductivity and are chemically inert.

Teacher's Note:
1. Interstitial compounds are neither typically ionic nor covalent; their formulas usually do not correspond to normal oxidation states.

Question 20. Mention the direct consequence of the following factors on the chemical behaviour of the transition elements :
(i) They have incompletely filled d-orbitals in the ground state or in one of oxidized states of their atoms.
(ii) They contribute more valence electrons per atom in the formation of metallic bonds.
Answer:
(i) They form complexes (and also show variable oxidation states, coloured ions and paramagnetism).
(ii) Strong metallic bonding; therefore they are hard, have high densities, high melting points and high enthalpies of atomisation.

Teacher's Note:
1. Link each structural cause to a property: partly filled d-orbitals give complexes and colour; more bonding electrons give hardness and high melting points.

Question 21. Write the structure of potassium dichromate. Give its uses also.
Answer:
The dichromate ion consists of two tetrahedra sharing one corner oxygen atom, the Cr–O–Cr bond angle being 126° :

Uses : (i) As an oxidising agent in volumetric analysis (primary standard). (ii) In leather tanning (chrome tanning). (iii) In the preparation of azo compounds and in dyeing and calico printing. (iv) In chromium plating and as a cleaning solution (chromic acid mixture).

Teacher's Note:
1. All six terminal Cr–O bonds are equivalent (163 pm); only the bridging Cr–O bonds are longer (179 pm).

Short Answer Questions (3 Marks)

Teacher's Note:
1. Compare the totals, not single values: add IE₁ to IE₄ for each metal before deciding which forms +4 compounds more easily.

Question 23. Explain the following about transition metals :
(i) Magnetic behaviour (ii) Metallic character (iii) Oxidation states. (Uttarakhand Board 2012)
Answer:
(i) Magnetic behaviour : Most transition metal ions are paramagnetic due to the presence of unpaired electrons in the d-orbitals. The magnetic moment is given by μ = \( \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons. Ions with d⁰ or d¹⁰ configuration are diamagnetic.
(ii) Metallic character : All transition elements are metals. They are hard, lustrous, malleable and ductile, have high tensile strength and are good conductors of heat and electricity. This is due to the strong metallic bonding involving both ns and (n − 1)d electrons.
(iii) Oxidation states : Transition elements show variable oxidation states because the energies of (n − 1)d and ns electrons are nearly the same, so both can take part in bond formation. The number of oxidation states is maximum in the middle of the series (Mn shows +2 to +7) and fewer at either end.

Teacher's Note:
1. For oxidation states, quote Mn (+2 to +7) as the element with the largest range in the 3d series.

Question 24. Discuss briefly the following physico-chemical properties of transition metals
(i) Ionisation potential (ii) Ionic radii (iii) Complex formation.
Answer:
(i) Ionisation potential : The ionisation enthalpies of transition elements are higher than those of s-block elements but lower than those of p-block elements. Along a series they increase only gradually because the increase in nuclear charge is largely balanced by the shielding of the (n − 1)d electrons.
(ii) Ionic radii : For ions of the same charge, the ionic radii decrease gradually with increase in atomic number along a series. For the same element, the ionic radius decreases as the oxidation state increases (e.g., Fe²⁺ > Fe³⁺).
(iii) Complex formation : Transition metals form a large number of complexes because of the small size and high charge of their ions and the availability of vacant d-orbitals to accept lone pairs from ligands, e.g., [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺.

Teacher's Note:
1. Higher charge on the same metal means fewer electrons for the same nuclear pull, so the ion is smaller.

Question 25. Explain the following about the transition elements.
(i) Oxidation states are variable
(ii) Exhibit good catalytic properties
(iii) Form interstitial compounds (HP Board 2011)
Answer:
(i) The energies of (n − 1)d and ns orbitals are very close. Hence, electrons from both can be lost, giving rise to variable oxidation states, e.g., Fe shows +2 and +3; Mn shows +2 to +7.
(ii) Due to their variable oxidation states they can form unstable intermediate compounds with reactants, and they also provide a large surface area for adsorption of the reactants. Both lower the activation energy, e.g., Fe in Haber's process, V₂O₅ in Contact process.
(iii) The crystal lattices of transition metals have interstitial voids in which small atoms like H, C, N and B can be trapped, e.g., TiC, Mn₄N, Fe₃H.

Teacher's Note:
1. Unlike s-block elements, transition elements can change their oxidation state in steps of one unit (e.g., Fe²⁺/Fe³⁺).

Question 26. Explain : (i) compounds of transition metals are often coloured. (HP Board 2013)
(ii) copper (I) is diamagnetic whereas copper (II) is paramagnetic.
(iii) A transition metal forms alloys with other transition metals easily. (MP Board 2012)
Answer:
(i) Transition metal ions have partly filled d-orbitals. In the presence of ligands, the d-orbitals split and an electron can be excited from a lower to a higher d-orbital by absorbing visible light (d-d transition). The colour seen is the complementary colour of the light absorbed.
(ii) Cu⁺ = [Ar] 3d¹⁰ has no unpaired electron, so it is diamagnetic. Cu²⁺ = [Ar] 3d⁹ has one unpaired electron, so it is paramagnetic.
(iii) Transition metals have very similar atomic sizes (within about 15%), so in the molten state one metal atom can easily take the place of another in the crystal lattice. On cooling, they form alloys (substitutional solid solutions).

Teacher's Note:
1. For alloy formation, the key point is similar atomic radii, which lets atoms of one metal replace atoms of another in the lattice.

Question 27. Account for the following :
(i) Transition metals and their compounds show catalytic activities.
(ii) Zn, Cd and Hg are non-transition elements.
(iii) Zr and Hf are of almost identical atomic radii. (CBSE 2022)
Answer:
(i) Transition metals show variable oxidation states and can form unstable intermediate compounds; they also provide a large surface area for adsorption of reactants. Thus they provide a new path of lower activation energy.
(ii) Zn, Cd and Hg have completely filled d-orbitals [(n − 1)d¹⁰ ns²] in their ground state as well as in their common oxidation state (+2). Hence, they are not regarded as transition elements.
(iii) This is due to lanthanoid contraction. The filling of 4f orbitals before 5d causes a steady decrease in size, which cancels the expected increase in size from Zr (4d) to Hf (5d). Hence, Zr (160 pm) and Hf (159 pm) have almost identical radii.

Teacher's Note:
1. Lanthanoid contraction is the standard reason for Zr/Hf, Nb/Ta and Mo/W having nearly equal radii.

Question 28. How is potassium dichromate prepared from chromite ore ? Give its three oxidizing properties. Write the chemical equations of the reactions involved. (J & K Board 2011, 2012 Hr. Board 2012, Kerala Board 2012)
Answer:
Preparation : (i) Chromite ore is fused with sodium carbonate (or NaOH) in free access of air :
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
(ii) The yellow solution of sodium chromate is filtered and acidified with sulphuric acid :
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
(iii) Sodium dichromate solution is treated with potassium chloride :
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. Orange crystals of K₂Cr₂O₇ crystallise out.
Oxidising properties (in acidic medium, Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O) :
(a) It oxidises iodide to iodine : Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
(b) It oxidises ferrous salts to ferric salts : Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(c) It oxidises H₂S to sulphur : Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O

Teacher's Note:
1. KCl is used in the last step because K₂Cr₂O₇ is much less soluble than Na₂Cr₂O₇, so it crystallises out.

Question 29. What happens when potassium dichromate reacts with (i) acidified solution of ferrous sulphate (ii) acidified solution of potassium iodide (iii) sodium chloride in the presence of conc. H₂SO₄ (iv) when sulphur dioxide is passed through its acidified solution (v) K₂Cr₂O₇ is heated. (J & K Board 2012)
or
What happens when acidified K₂Cr₂O₇ reacts with
(i) H₂S (ii) FeSO₄ (iii) KI (Hr. Board 2012)
Answer:
(i) Ferrous salt is oxidised to ferric salt : Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) Iodine is liberated : Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
(iii) Red vapours of chromyl chloride are formed : K₂Cr₂O₇ + 4NaCl + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O
(iv) SO₂ is oxidised to sulphate and the solution turns green : Cr₂O₇²⁻ + 2H⁺ + 3SO₂ → 2Cr³⁺ + 3SO₄²⁻ + H₂O
(v) On heating, it decomposes : 4K₂Cr₂O₇ \( \xrightarrow{\text{Heat}} \) 4K₂CrO₄ + 2Cr₂O₃ + 3O₂
or (i) Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O (ii) and (iii) as in (i) and (ii) above.

Teacher's Note:
1. In every acidic reaction the orange dichromate is reduced to green Cr³⁺; that colour change is a useful observation to mention.

Question 30. Name the gas evolved when conc HCl is added to powdered potassium dichromate. On passing the evolved gas through acidified KBr solution, the solution turns brown. Write the balanced equations for the reactions involved. (West Bengal Board 2013)
Answer:
The gas evolved is Cl₂.
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂
Cl₂ + 2KBr → 2KCl + Br₂ (brown colour)

Teacher's Note:
1. Chlorine is a stronger oxidising agent than bromine, so it displaces Br₂ from bromide, giving the brown colour.

Question 31. Assign reasons for the following observations :
(i) Mn²⁺ compounds are more stable than Fe²⁺ compounds towards oxidation to their +3 state.
(ii) Interstitial compounds are well known for transition elements.
(iii) An aqueous solution of potassium chromate is yellow but changes its colour on decreasing the pH of the solution.
Answer:
(i) Mn²⁺ has the stable half-filled 3d⁵ configuration; oxidation to Mn³⁺ (3d⁴) would destroy it. Fe²⁺ is 3d⁶; on oxidation it becomes Fe³⁺ (3d⁵), which is the stable half-filled configuration. Hence, Fe²⁺ is easily oxidised but Mn²⁺ is not.
(ii) Transition metals have interstitial voids in their crystal lattices which can easily accommodate small atoms like H, C, N, B.
(iii) On decreasing the pH (adding acid), yellow chromate changes to orange dichromate : 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O

Teacher's Note:
1. Moving to or away from a half-filled d⁵ configuration decides the relative stability of Mn²⁺/Fe²⁺ towards oxidation.

Question 32. (a) Write the balanced chemical equations involved in the preparation of KMnO₄ from pyrolusite ore (MnO₂). (CBSE 2020)
(b) Using balanced chemical equations, illustrate the oxidising properties of potassium permanganate in acidic and basic media.
Answer:
(a) 2MnO₂ + 4KOH + O₂ \( \xrightarrow{\text{Fuse}} \) 2K₂MnO₄ + 2H₂O
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (or by electrolytic oxidation : MnO₄²⁻ → MnO₄⁻ + e⁻)
(b) In acidic medium : MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
e.g., 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O ; MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
In neutral or faintly alkaline medium : MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
e.g., 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻ ; 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻

Teacher's Note:
1. Note the change in products with medium: iodide gives I₂ in acid but iodate (IO₃⁻) in neutral/alkaline solution.

Long Answer Questions (5 Marks)

Question 33. (a) Describe the preparation of potassium dichromate from chromite (FeCr₂O₄) ore. Write the chemical equations of the reactions involved. What is the effect of increasing pH of a solution of potassium dichromate ? (Hr. Board 2011)
(b) Represent the reaction of acidified potassium dichromate with (i) KI solution and (ii) FeSO₄ solution with the help of chemical equations.
(c) Draw the structure of chromate ion.
Answer:
(a) (i) Chromite ore is fused with sodium carbonate in free access of air to form sodium chromate :
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
(ii) The yellow solution of sodium chromate is filtered and acidified with H₂SO₄ to give sodium dichromate :
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
(iii) Sodium dichromate is treated with KCl, when less soluble potassium dichromate crystallises out :
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
Effect of increasing pH : On increasing pH (adding alkali), the orange dichromate solution changes to yellow due to formation of chromate ions :
Cr₂O₇²⁻ (orange) + 2OH⁻ → 2CrO₄²⁻ (yellow) + H₂O
(b) (i) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
(ii) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(c) The chromate ion is tetrahedral :

Teacher's Note:
1. Chromium stays +6 in both chromate and dichromate; changing pH only shifts the equilibrium, it is not a redox reaction.

Question 34. (a) Describe how potassium permanganate is made from pyrolusite. Write the chemical equations for the reactions involved.
(b) What happens when :
(i) Acidified solution of KMnO₄ is heated with oxalic acid.
(ii) K₂Cr₂O₇ is heated with sodium chloride and conc. H₂SO₄ ?
Answer:
(a) (i) Pyrolusite (MnO₂) is fused with KOH in the presence of air (or an oxidising agent like KNO₃) to give green potassium manganate :
2MnO₂ + 4KOH + O₂ \( \xrightarrow{\text{Fuse}} \) 2K₂MnO₄ + 2H₂O
(ii) The manganate is extracted with water and oxidised electrolytically (or by Cl₂/O₃) to permanganate :
MnO₄²⁻ \( \xrightarrow{\text{Electrolytic oxidation}} \) MnO₄⁻ + e⁻
In neutral or acidic solution, manganate also disproportionates : 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
The purple solution is concentrated when dark purple crystals of KMnO₄ separate out.
(b) (i) Oxalic acid is oxidised to CO₂ and the purple colour of KMnO₄ disappears :
2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ \( \xrightarrow{333\,K} \) 2Mn²⁺ + 8H₂O + 10CO₂
(ii) Red vapours of chromyl chloride are formed :
K₂Cr₂O₇ + 4NaCl + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O

Teacher's Note:
1. In the first step Mn is oxidised from +4 to +6 (green manganate); the second step takes it from +6 to +7 (purple permanganate).

Question 35. Give reasons : (a) Transition metals have high enthalpies of atomization.
(b) Among the lanthanoids, Ce(III) is easily oxidised to Ce(IV).
(c) Fe³⁺ | Fe²⁺ redox couple has less positive electrode potential than Mn³⁺ | Mn²⁺ couple.
(d) Copper (I) has d¹⁰ configuration, while copper (II) has d⁹ configuration, still copper (II) is more stable in aqueous solution than copper (I).
(e) The second and third transition series elements have almost similar atomic radii.
(f) Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal.
Answer:
(a) Transition metals have a large number of unpaired electrons which take part in strong interatomic metallic bonding. Hence, they have high enthalpies of atomisation.
(b) Ce(III) has the configuration 4f¹. On losing this electron, it acquires the stable noble gas configuration 4f⁰ of Ce(IV). Hence, Ce(III) is easily oxidised to Ce(IV).
(c) Mn²⁺ has the stable half-filled 3d⁵ configuration, so Mn³⁺ (3d⁴) is easily reduced to Mn²⁺ and E° (Mn³⁺/Mn²⁺) is highly positive (+1.57 V). Fe³⁺ itself has the stable 3d⁵ configuration and is not so easily reduced to Fe²⁺ (3d⁶). Hence, E° (Fe³⁺/Fe²⁺) is less positive (+0.77 V).
(d) The stability of an ion in aqueous solution depends on its hydration enthalpy as well as ionisation enthalpy. Cu²⁺ (aq) has a much more negative hydration enthalpy than Cu⁺ (aq), which more than compensates for the second ionisation enthalpy of copper. Hence, Cu²⁺ is more stable in aqueous solution and Cu⁺ disproportionates : 2Cu⁺ → Cu²⁺ + Cu.
(e) This is due to lanthanoid contraction. The 14 lanthanoids, in which poorly shielding 4f electrons are filled, come before the third series; the resulting decrease in size cancels the expected increase in size from the second to the third series.
(f) Oxygen is highly electronegative and small in size and it can form multiple bonds with the metal (M=O). Hence, oxygen can stabilise the highest oxidation state of the metal in oxoanions, e.g., Mn (+7) in MnO₄⁻, Cr (+6) in Cr₂O₇²⁻.

Teacher's Note:
1. For (c) and (d), always compare the d-configurations before and after the change; reaching d⁵ or losing it is usually the deciding factor.

Question 36. (a) Out of Ag₂SO₄, CuF₂, MgF₂ and CuCl, which compound will be coloured and why ?
(b) Explain : (i) CrO₄²⁻ is a strong oxidizing agent while MnO₄²⁻ is not. (ii) Zr and Hf have identical sizes.
(iii) The lowest oxidation state of manganese is basic while the highest is acidic.
(iv) Mn (II) shows maximum paramagnetic character amongst the divalent ions of the first transition series.
Answer:
(a) CuF₂ will be coloured because Cu²⁺ (3d⁹) has one unpaired electron and can show d-d transition. In Ag₂SO₄ (Ag⁺ = 4d¹⁰) and CuCl (Cu⁺ = 3d¹⁰) the d-orbitals are completely filled, and Mg²⁺ has no d-electrons; so these are colourless.
(b) (i) Cr in CrO₄²⁻ is in the highest oxidation state, i.e., +6 and can only be reduced, while Mn in MnO₄²⁻ is in oxidation state +6 and its most stable oxidation state is +7, so it tends to be oxidised rather than reduced.
(ii) Due to lanthanoid contraction.
(iii) In the lowest oxidation state (+2, MnO), Mn has most of its valence electrons and can give electrons (ionic, basic oxide). In the highest oxidation state (+7, Mn₂O₇), the bonds are covalent and the oxide reacts with water to form an acid (HMnO₄). Hence, MnO is basic and Mn₂O₇ is acidic.
(iv) Mn²⁺ has 3d⁵ configuration with five unpaired electrons, the maximum for a divalent ion of the first series. μ = \( \sqrt{5(5+2)} \) = 5.92 BM.

Teacher's Note:
1. As the oxidation state of a metal increases, its oxides change from basic through amphoteric to acidic: MnO (basic), Mn₂O₃/MnO₂ (amphoteric), Mn₂O₇ (acidic).

Question 37. (a) Complete the following chemical reaction equations :
(i) Cr₂O₇²⁻ (aq) + I⁻ (aq) + H⁺ (aq) →
(ii) MnO₄⁻ (aq) + Fe²⁺ (aq) + H⁺ (aq) →
(b) Explain the following observations :
(i) In general the atomic radii of transition elements decrease with atomic number in a given series
(ii) The E°_M²⁺/M for copper is positive (+ 0.34 V). It is the only metal in the first series of transition elements showing this type of behaviour
(iii) The E° value for Mn³⁺/Mn²⁺ couple is much more positive than for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ couple (CBSE 2009)
Answer:
(a) (i) Cr₂O₇²⁻ (aq) + 6I⁻ (aq) + 14H⁺ (aq) → 2Cr³⁺ (aq) + 3I₂ (s) + 7H₂O (l)
(ii) MnO₄⁻ (aq) + 5Fe²⁺ (aq) + 8H⁺ (aq) → Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H₂O (l)
(b) (i) As we move along a series, the nuclear charge increases by one unit at each step while the added electron goes into the inner (n − 1)d subshell. The shielding by d-electrons is not very effective, so the effective nuclear charge increases and the atomic radius decreases (the decrease becomes small in the middle of the series).
(ii) The high energy needed to convert Cu(s) to Cu²⁺(aq) (high enthalpy of atomisation + first and second ionisation enthalpies) is not balanced by its hydration enthalpy. Hence, E° (Cu²⁺/Cu) is positive and copper does not liberate H₂ from acids.
(iii) Mn³⁺ (3d⁴) on reduction gives Mn²⁺ with the stable half-filled 3d⁵ configuration; hence E° is highly positive (+1.57 V). Cr³⁺ has the stable t_2g³ (3d³) configuration and is not easily reduced (E° = −0.41 V), while Fe³⁺ (3d⁵) is itself stable, so E° (Fe³⁺/Fe²⁺) is only +0.77 V.

Teacher's Note:
1. A positive E°(M²⁺/M) means the metal is less reactive than hydrogen; copper is the only 3d metal of this kind.