Class 12 Chemistry Chapter 4: The d- and f-Block Elements worksheet with answer key
A 20-mark worksheet on The d- and f-Block Elements for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (a) [Ar]3d⁶
Co is [Ar]3d⁷4s². On ionisation the two 4s electrons leave first, then one 3d electron: Co³⁺ is [Ar]3d⁶, with no 4s electrons.
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Answer: (a) Ag²⁺
A transition element has an incompletely filled d subshell in at least one of its common oxidation states. Zn²⁺, Cd²⁺ and Hg²⁺ are all d¹⁰; Ag²⁺ is 4d⁹, so silver, through Ag(II), qualifies while zinc, cadmium and mercury do not.
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Answer: (a) Zr and Hf
Because of the lanthanoid contraction the 4d and 5d members of a group have nearly the same radii: Zr 160 pm, Hf 159 pm. Down from 3d to 4d (Ti to Zr, Sc to Y, Cu to Ag) the radius increases.
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Answer: (d) Sc³⁺, Ti⁴⁺, Zn²⁺
Sc³⁺ and Ti⁴⁺ are 3d⁰ and Zn²⁺ is 3d¹⁰: none can have a d–d transition. Ti³⁺ (3d¹, purple), Fe³⁺ (3d⁵, yellow) and Cu²⁺ (3d⁹, blue) are coloured.
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Answer: (d) chromium (Z = 24)
3.87 BM = √(3 × 5), so n = 3. M³⁺ has (Z − 21) d electrons: V³⁺ d² (2 unpaired, 2.84), Cr³⁺ d³ (3 unpaired, 3.87), Mn³⁺ d⁴ (4, 4.90), Fe³⁺ d⁵ (5, 5.92). Only chromium fits.
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2Cr²⁺ + 2H⁺ → 2Cr³⁺ + H₂. Cr²⁺ is a strong reducing agent: on oxidation it goes from d⁴ to d³ Cr³⁺, which has a half-filled t₂g level and is stable in water.
- balanced ionic equation (1)
- Cr³⁺ d³, half-filled t₂g, so Cr²⁺ is readily oxidised (1)
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Colour in Ln³⁺ comes from the f electrons. La³⁺ is 4f⁰ and Lu³⁺ is 4f¹⁴, so neither can absorb by an f–f transition, and both are colourless. With no unpaired electrons, both are diamagnetic, not paramagnetic.
- f⁰ and f¹⁴, no f–f transition (1)
- no unpaired electrons, so not paramagnetic (1)
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Oxygen can form multiple bonds to the metal while fluorine forms only single bonds. In Mn₂O₇ each Mn is tetrahedral, bonded to terminal O atoms by multiple bonds and joined through an Mn–O–Mn bridge, which lets manganese reach +7; fluorine cannot satisfy +7 with single bonds.
- oxygen forms multiple (π) bonds to the metal, fluorine does not (1)
- so oxygen stabilises +7 (Mn₂O₇) while fluoride stops at +4 (1)
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(i) 2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O. (ii) 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O. (iii) 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻.
- nitrite to nitrate, balanced (1)
- iodide to iodine in acid, balanced (1)
- iodide to iodate, MnO₂ formed, in neutral solution (1)
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Eu: [Xe]4f⁷6s². Gd: [Xe]4f⁷5d¹6s². Eu loses its two 6s electrons to give Eu²⁺, 4f⁷, a half-filled f subshell, so +2 is reached (Eu²⁺ is, however, a strong reductant, returning to +3). Gd loses 6s² and 5d¹ to give Gd³⁺, also 4f⁷, half-filled, so its +3 state is particularly stable.
- both configurations correct (1)
- Eu²⁺ is 4f⁷ after losing 6s² (1)
- Gd³⁺ is 4f⁷, half-filled (1)
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(i) Copper: its E° is positive (+0.34 V), so it cannot reduce H⁺ to H₂; it dissolves only in oxidising acids. (ii) V > Fe > Ni, the order of increasingly less negative E°. (iii) E°(M²⁺/M) becomes less negative across the series, a decreasing tendency to form M²⁺, following the rising sum of the first and second ionisation enthalpies.
- copper, positive E° (1)
- V > Fe > Ni (1)
- less negative across the series, tracking IE₁ + IE₂ (1)
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