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The d- and f- Block Elements Set-2 MCQ Online Practice Test | Class 12 | Quiz Tweek

The d- and f- Block Elements Set-2

Practice Important MCQs for The d- and f- Block Elements — Chemistry, Class 12.

This chapter is central for both board exams and competitive tests because it links electronic configurations with measurable properties like magnetic moment, geometry, colour intensity, crystal-field splitting (CFSE), and Jahn–Teller effects. It also explains lanthanoid and actinoid trends (especially lanthanoid contraction) that directly affect chemical behaviour and separation of elements—frequently asked as reasoning-based MCQs.

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Q1. Calculate the spin-only magnetic moment $\mu_{so}$ (in $\mu_B$ ) for the $Cr^{3+}$ ion. Use $\mu_{so}=\sqrt{n(n+2)}$ , where $n$ is the number of unpaired electrons.

(A)

$1.73\ \mu_B$

(B)

$4.90\ \mu_B$

(C)

$3.87\ \mu_B$

(D)

$0.00\ \mu_B$

Q2. Which of the following best explains why $[Ni(CN)_4]^{2-}$ is square planar while $[NiCl_4]^{2-}$ is tetrahedral?

(A)

CN $^-$ is a strong-field ligand producing large $d$ -orbital splitting that leads to pairing in $Ni^{2+}$ ( $d^8$ ) and formation of a low-spin $dsp^2$ square-planar complex; Cl $^-$ is weak-field giving small splitting so the complex remains high-spin and tetrahedral.

(B)

The geometries arise mainly from steric effects: bulky CN $^-$ ligands hinder square-planar arrangement so $[Ni(CN)_4]^{2-}$ becomes tetrahedral, whereas smaller Cl $^-$ leads to square-planar $[NiCl_4]^{2-}$ .

(C)

CN $^-$ is actually a weak-field ligand, so $[Ni(CN)_4]^{2-}$ remains high-spin (tetrahedral), while Cl $^-$ provides a comparatively strong field that favours a low-spin square-planar $[NiCl_4]^{2-}$ .

(D)

The geometry is primarily decided by symmetry: absence or presence of inversion symmetry in the ligand arrangement determines whether the complex is tetrahedral or square planar, independent of ligand-field splitting.

Q3. Which of the following complexes is diamagnetic (has no unpaired electrons) under usual ligand-field behaviour?

(A)

$[V(H_2O)_6]^{2+}$

(B)

$[Co(H_2O)_6]^{2+}$

(C)

$[Fe(CN)_6]^{3-}$

(D)

$[Sc(H_2O)_6]^{3+}$

Q4. Statement: "Lanthanoid contraction causes the atomic and ionic radii of many 4d and 5d transition metals (for example Zr and Hf) to be very similar." Reason: "Poor shielding of the increasing nuclear charge by the 4f electrons across the lanthanoid series increases effective nuclear charge and leads to a steady decrease (contraction) in size." Which option correctly describes the two sentences?

(A)

Both the statement and the reason are false.

(B)

Both the statement and the reason are true, and the reason correctly explains the statement.

(C)

The statement is true but the reason is false.

(D)

The statement is false but the reason is true.

Q5. Which of the following gives the best explanation for why 3d transition-metal complexes are, in general, more labile (undergo faster ligand substitution) than their 4d and 5d analogues?

(A)

4d/5d metal ions have more spatially extended $d$ -orbitals, leading to stronger metal–ligand overlap, larger crystal-field splitting and stronger M–L bonds; this raises the activation energy for ligand substitution, making 4d/5d complexes kinetically more inert than 3d analogues.

(B)

4d/5d metal ions have larger and more diffuse $d$ -orbitals that overlap less with ligand orbitals, giving weaker M–L bonds; this lowers the activation energy for substitution, so 4d/5d complexes become more labile than 3d complexes.

(C)

Because of lanthanoid contraction, 4d and 5d ions have smaller radii, which reduces $\Delta_o$ and weakens M–L bonds; hence 4d/5d complexes are more labile.

(D)

3d metal complexes are generally less labile because they have higher M–L bond strength (smaller metal–ligand distances) than 4d/5d complexes.

Q6. For the octahedral complex $[Co(H_2O)_6]^{2+}$ (assume high-spin), determine the number of unpaired electrons and the spin-only magnetic moment $\mu_{\text{so}}$ using $\mu_{\text{so}}=\sqrt{n(n+2)}$ (in Bohr magnetons, BM).

(A)

1 unpaired; $\mu_{\text{so}}=1.73\ \mathrm{BM}$

(B)

3 unpaired; $\mu_{\text{so}}=\sqrt{3(3+2)}\approx 3.87\ \mathrm{BM}$

(C)

4 unpaired; $\mu_{\text{so}}=4.90\ \mathrm{BM}$

(D)

2 unpaired; $\mu_{\text{so}}=2.83\ \mathrm{BM}$

Q7. For an octahedral $d^6$ metal centre the CFSE values are: $\mathrm{CFSE_{low-spin}}=-2.4\Delta_o$ and $\mathrm{CFSE_{high-spin}}=-0.4\Delta_o$ . Low-spin configuration requires an extra pairing energy of $2P$ relative to high-spin. For Fe $^{2+}$ : $P=15000\ \mathrm{cm^{-1}}$ , $\Delta_o([Fe(CN)_6]^{4-})=21000\ \mathrm{cm^{-1}}$ and $\Delta_o([Fe(H_2O)_6]^{2+})=9000\ \mathrm{cm^{-1}}$ . Which statement follows from these data?

(A)

Both $[Fe(CN)_6]^{4-}$ and $[Fe(H_2O)_6]^{2+}$ will be low-spin.

(B)

Both $[Fe(CN)_6]^{4-}$ and $[Fe(H_2O)_6]^{2+}$ will be high-spin.

(C)

$[Fe(CN)_6]^{4-}$ will be high-spin and $[Fe(H_2O)_6]^{2+}$ will be low-spin.

(D)

$[Fe(CN)_6]^{4-}$ will be low-spin and $[Fe(H_2O)_6]^{2+}$ will be high-spin.

Q8. Which of the following pairs of octahedral complexes is expected to show the largest static Jahn–Teller distortion?

(A)

$[Cu(H_2O)_6]^{2+}$ (d $^9$ ) and $[Mn(H_2O)_6]^{3+}$ (d $^4$ , high‑spin)

(B)

$[Fe(H_2O)_6]^{2+}$ (d $^6$ , high‑spin) and $[Co(H_2O)_6]^{2+}$ (d $^7$ , high‑spin)

(C)

$[Ti(H_2O)_6]^{3+}$ (d $^1$ ) and $[V(H_2O)_6]^{3+}$ (d $^2$ )

(D)

$[Ni(H_2O)_6]^{2+}$ (d $^8$ ) and $[Zn(H_2O)_6]^{2+}$ (d $^{10}$ )

Q9. Which of the following best explains which of Ce, Eu and Yb are most likely to form stable divalent ions and why?

(A)

Only Ce forms a stable +2 state because Ce prefers to attain an empty $4f$ subshell.

(B)

All three (Ce, Eu, Yb) commonly form stable +2 species because $4f$ electrons are similarly bound across the series.

(C)

Eu and Yb readily form stable +2 species because Eu $^{2+}$ attains a half‑filled $4f^7$ configuration and Yb $^{2+}$ attains a filled $4f^{14}$ configuration, whereas Ce prefers +4.

(D)

Only Yb forms stable +2 because filled 5d orbitals stabilise Yb $^{2+}$ .

Q10. Compare octahedral $[Ni(H_2O)_6]^{2+}$ and tetrahedral $[NiCl_4]^{2-}$ (both Ni $^{2+}$ , d $^8$ ). Using $\Delta_t\approx\frac{4}{9}\Delta_o$ and the Laporte selection rule, which statement is correct about their visible absorption and colour intensity?

(A)

Octahedral $[Ni(H_2O)_6]^{2+}$ is more intensely coloured but absorbs at longer wavelength than tetrahedral $[NiCl_4]^{2-}$ .

(B)

Tetrahedral $[NiCl_4]^{2-}$ is more intensely coloured and absorbs at longer wavelength than octahedral $[Ni(H_2O)_6]^{2+}$ .

(C)

Octahedral $[Ni(H_2O)_6]^{2+}$ is less intensely coloured but absorbs at longer wavelength than tetrahedral $[NiCl_4]^{2-}$ .

(D)

Both complexes have comparable $\lambda_{\max}$ , but $[NiCl_4]^{2-}$ is more intensely coloured due to relaxation of the Laporte rule.

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