RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
Transition metals give us coloured gemstones, industrial catalysts and the steel in bridges — all traceable to partly filled d orbitals. This chapter explains the trends in atomic size, oxidation state, colour and magnetism across the d-block, and then turns to the lanthanoids and actinoids, where the quiet shrinking known as the lanthanoid contraction reshapes the chemistry of the elements below them.
Which ion is expected to have a spin-only magnetic moment of 3.87 BM?
Answer
Cr³⁺ (d³). Using μ = √(n(n+2)), three unpaired electrons give μ = √(3 × 5) = √15 = 3.87 BM. For comparison, Ti³⁺ gives √3 = 1.73 BM, V³⁺ gives √8 = 2.83 BM and Mn²⁺ gives √35 = 5.92 BM.
Zirconium and hafnium have almost identical atomic radii. The best explanation is:
Answer
The lanthanoid contraction cancels the expected size increase from the fifth to the sixth period. Between Zr and Hf lie the fourteen lanthanoids, in which the added 4f electrons shield the nucleus poorly. The effective nuclear charge therefore rises steadily and pulls the outer shells inward, offsetting the extra shell Hf would otherwise gain. The two elements consequently have very similar sizes and remarkably similar chemistry.
An aqueous solution of potassium dichromate is treated with excess sodium hydroxide. The observed change is:
Answer
Orange to yellow, because dichromate is converted to chromate. The equilibrium 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O lies to the right in acid and to the left in alkali. Added OH⁻ removes H⁺, so by Le Chatelier's principle the orange Cr₂O₇²⁻ converts to yellow CrO₄²⁻. The oxidation state of chromium stays +6 throughout, so this is not a redox change.
Which of the following is NOT regarded as a transition element?
Answer
Zinc. Its atom is [Ar]3d¹⁰4s² and its only common ion, Zn²⁺, is [Ar]3d¹⁰, so neither the atom nor the ion has a partly filled d subshell. It is therefore a d-block element but not a transition element, and its compounds are white and diamagnetic. Copper qualifies because Cu²⁺ is 3d⁹.
Assertion (A): The E°(M³⁺/M²⁺) value for manganese is unusually high compared with its neighbours. Reason (R): Mn²⁺ has a half-filled 3d⁵ configuration, which gives it extra stability and makes further oxidation difficult.
Answer
Both A and R are true and R is the correct explanation of A. A high positive E°(M³⁺/M²⁺) means the +3 ion is a strong oxidising agent, i.e. it readily reverts to +2. Mn²⁺ is 3d⁵, a half-filled subshell with maximum exchange energy and spherical symmetry, so removing a sixth electron to reach Mn³⁺ is energetically costly. That extra stability of Mn²⁺ is precisely why the couple has a high potential.
Write the electronic configurations of Cr and Cu and explain why they do not follow the simple filling order.
Answer
Cr is [Ar]3d⁵4s¹ rather than 3d⁴4s², and Cu is [Ar]3d¹⁰4s¹ rather than 3d⁹4s². The 3d and 4s levels lie very close in energy, so a small stabilisation can decide the arrangement. A half-filled (d⁵) or completely filled (d¹⁰) subshell has all electrons with parallel spins available for exchange in the first case, and a symmetrical charge distribution in both, giving maximum exchange energy and lower total energy. Promoting one 4s electron into 3d therefore pays for itself in both metals.
Explain why transition metal ions are usually coloured while Sc³⁺ and Zn²⁺ are colourless. Predict the number of unpaired electrons and the spin-only magnetic moment of Fe³⁺.
Answer
Colour in transition metal ions comes from d–d transitions. Ligands split the five degenerate d orbitals into lower t₂g and upper e_g sets; the energy gap corresponds to visible light, so an electron absorbs one wavelength to jump between the sets and the complementary colour is seen. Sc³⁺ is 3d⁰ and has no electron to promote; Zn²⁺ is 3d¹⁰ and has no vacancy to promote an electron into. With no possible d–d transition, both are colourless. Fe³⁺ is [Ar]3d⁵, so in a weak field all five d electrons stay unpaired, n = 5. μ = √(n(n+2)) = √(5 × 7) = √35 = 5.92 BM.
Give two reasons why transition metals and their compounds are widely used as catalysts, with one example of each reason.
Answer
First, transition metals show variable oxidation states, so they can accept and release electrons to form unstable intermediates that offer a lower-energy route; vanadium(V) oxide catalyses the oxidation of SO₂ to SO₃ by cycling between V(V) and V(IV). Second, their surfaces have partly filled d orbitals that adsorb reactant molecules, holding them close together in the right orientation and weakening their bonds; finely divided iron adsorbs N₂ and H₂ in the synthesis of ammonia, and nickel adsorbs H₂ in the hydrogenation of oils.
(i) Describe the preparation of potassium permanganate from pyrolusite ore, writing the equations involved. (ii) Write the balanced ionic equations for the oxidation of (a) Fe²⁺ and (b) oxalate ion by permanganate in acidic medium. (iii) Explain why KMnO₄ titrations against oxalic acid are carried out warm and need no external indicator.
Answer
(i) Pyrolusite, MnO₂, is fused with potassium hydroxide in the presence of an oxidising agent such as air or potassium nitrate. This gives the dark green manganate ion: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O. The manganate is then oxidised to permanganate, either electrolytically or by disproportionation in acidic or neutral solution: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O. The purple solution is concentrated and crystallised to obtain dark purple KMnO₄ crystals. (ii) In acid, permanganate is reduced to Mn²⁺ by a five-electron change: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. (a) With iron(II): MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. (b) With oxalate: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. (iii) The oxalate reaction is slow to start at room temperature because two negative ions must approach each other, so the mixture is warmed to about 60 °C to give the reaction a workable rate; once Mn²⁺ forms it autocatalyses the reaction and it proceeds briskly. No indicator is needed because KMnO₄ is self-indicating: every drop added is decolourised while oxalate remains, and the first drop in excess leaves a permanent faint pink colour that marks the end point.
(i) What is lanthanoid contraction and what causes it? State three of its consequences. (ii) Compare the lanthanoids and actinoids with respect to oxidation states and radioactivity. (iii) Account for the fact that the +2 oxidation state becomes more stable across the first transition series from Ti to Zn.
Answer
(i) Lanthanoid contraction is the gradual decrease in atomic and ionic radii from lanthanum to lutetium as the 4f subshell fills. The cause is the shape of the 4f orbitals: they are diffuse and shield the nuclear charge very poorly, so each added proton is only partly offset by the added 4f electron. The effective nuclear charge acting on the outer electrons therefore rises steadily and the shells are pulled in. Three consequences: (1) the atomic radii of second and third transition-series congeners become almost equal, so Zr and Hf, and Nb and Ta, are extremely difficult to separate; (2) the size difference between successive lanthanoids is tiny, making their separation by fractional crystallisation or ion exchange very laborious; (3) the basic character of Ln(OH)₃ decreases steadily from La(OH)₃ to Lu(OH)₃, because the smaller, more highly polarising ion holds the hydroxide more covalently. (ii) Lanthanoids show a dominant +3 oxidation state, with +2 and +4 appearing only where they lead to a stable f⁰, f⁷ or f¹⁴ configuration (Ce⁴⁺, Eu²⁺). Actinoids show a far wider range, +3 to +7, because the 5f, 6d and 7s levels lie close in energy; uranium reaches +6 in UO₂²⁺ and neptunium +7. All actinoids are radioactive, and those beyond uranium are synthetic and increasingly short-lived, whereas only promethium among the lanthanoids is radioactive. (iii) Across the series the effective nuclear charge rises, so the third ionisation enthalpy climbs steeply and removing a third electron from the d subshell becomes progressively harder. Once the 4s electrons are lost, the remaining d electrons are held more and more tightly, so the ion is content to stop at +2. Titanium and vanadium readily go to +3 or +4, while at the far end zinc has a completely filled 3d¹⁰ core and forms only Zn²⁺.
A laboratory technician has four unlabelled aqueous solutions of first-row transition metal sulphates. Solution 1 is pale violet and its metal ion has one 3d electron. Solution 2 is bright blue. Solution 3 is very pale pink and its ion has a half-filled d subshell. Solution 4 is colourless and its ion is diamagnetic with a filled d subshell. (a) Identify the metal ion in each solution. (b) Calculate the spin-only magnetic moment of the ion in solution 1 and in solution 3. (c) Explain why solution 4 is colourless. (d) Explain in one or two sentences why solution 3 is only very faintly coloured despite having five d electrons.
Answer
(a) Solution 1 has a d¹ ion, so it is Ti³⁺ (titanium(III), pale violet). Solution 2, bright blue, is Cu²⁺ (d⁹). Solution 3 has a half-filled d⁵ subshell and is pale pink, so it is Mn²⁺. Solution 4 is colourless, diamagnetic and d¹⁰, so it is Zn²⁺. (b) Using μ = √(n(n+2)): for Ti³⁺, n = 1, so μ = √3 = 1.73 BM. For Mn²⁺, n = 5, so μ = √35 = 5.92 BM. (c) Zn²⁺ is 3d¹⁰. Every d orbital is full, so there is no vacant d orbital for an electron to be promoted into and no d–d transition can absorb visible light. With no absorption in the visible region, the solution appears colourless, and with no unpaired electrons it is also diamagnetic. (d) In Mn²⁺ all five d electrons have parallel spins in a half-filled subshell. Any d–d transition would have to pair two electrons and hence flip a spin, and spin-forbidden transitions have very low probability, so absorption is extremely weak and the colour is barely visible.
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