Coordination Compounds and the d- and f-Block — Practice

Question 1

The spin-only magnetic moment of Fe³⁺ (3d⁵) is approximately:
(a) 5.92 BM (b) 2.83 BM (c) 4.90 BM (d) 1.73 BM

Accepted Answer

(a) 5.92 BM.

Fe³⁺ = [Ar]3d⁵ has 5 unpaired electrons; μ = √(5×7) = √35 = 5.92 BM.

Question 2

Which ion is colourless?
(a) Zn²⁺ (b) Cu²⁺ (c) Fe³⁺ (d) Ni²⁺

Accepted Answer

(a) Zn²⁺.

Zn²⁺ is 3d¹⁰ (fully filled), so no d–d transition is possible and it is colourless. Cu²⁺, Fe³⁺ and Ni²⁺ have partly filled d sub-shells and are coloured.

Question 3

Which transition metal shows the maximum number of oxidation states?
(a) Mn (b) Fe (c) Cu (d) Zn

Accepted Answer

(a) Mn.

Manganese ([Ar]3d⁵4s²) can lose from two up to all seven of its 4s and 3d electrons, showing states from +2 to +7 — the widest range in the first series.

Question 4

The lanthanoid contraction is responsible for the nearly identical sizes of:
(a) Zr and Hf (b) Na and K (c) Fe and Co (d) Cl and Br

Accepted Answer

(a) Zr and Hf.

The steady size decrease across the lanthanoids offsets the expected increase, so Hf ends up almost the same size as the lighter Zr — making the pair very hard to separate.

Question 5

When acidic KMnO₄ acts as an oxidising agent, the oxidation state of Mn changes from +7 to:
(a) +2 (b) +4 (c) +3 (d) 0

Accepted Answer

(a) +2.

In acidic medium MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so Mn goes from +7 to +2, gaining 5 electrons.

Question 6

The oxidation state of cobalt in [Co(NH₃)₆]Cl₃ is:
(a) +3 (b) +2 (c) +1 (d) +6

Accepted Answer

(a) +3.

The six NH₃ are neutral ligands, so the three chloride counter-ions balance a +3 metal: cobalt is +3, with coordination number 6.

Question 7

The IUPAC name of K₄[Fe(CN)₆] is:
(a) potassium hexacyanoferrate(II) (b) potassium hexacyanoferrate(III) (c) tetrapotassium ferricyanide (d) potassium hexacyanoiron(II)

Accepted Answer

(a) potassium hexacyanoferrate(II).

It is an anionic complex, so the metal takes the -ate suffix (ferrate). Six CN⁻ (charge −6) with four K⁺ (+4) make the complex −4, so iron is +2 → hexacyanoferrate(II).

Question 8

Which ligand is ambidentate?
(a) NO₂⁻ (b) NH₃ (c) H₂O (d) Cl⁻

Accepted Answer

(a) NO₂⁻.

The nitrite ion can coordinate through nitrogen (nitro, –NO₂) or through oxygen (nitrito, –ONO), so it is ambidentate and gives rise to linkage isomerism.

Question 9

[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are examples of:
(a) ionisation isomerism (b) linkage isomerism (c) optical isomerism (d) geometrical isomerism

Accepted Answer

(a) ionisation isomerism.

The two compounds give different ions in solution (sulphate versus bromide) because the ligand and counter-ion are swapped — ionisation isomerism.

Question 10

The hybridisation of iron in the low-spin complex [Fe(CN)₆]⁴⁻ is:
(a) d²sp³ (b) sp³d² (c) sp³ (d) dsp²

Accepted Answer

(a) d²sp³.

Fe²⁺ is 3d⁶; the strong-field CN⁻ pairs all six electrons into three t2g orbitals, freeing two inner 3d orbitals for d²sp³ (inner-orbital) octahedral bonding — diamagnetic.

Question 11

The number of unpaired electrons in [Fe(CN)₆]³⁻ (Fe³⁺, strong field) is:
(a) 1 (b) 5 (c) 3 (d) 0

Accepted Answer

(a) 1.

Fe³⁺ is d⁵; the strong-field CN⁻ gives a low-spin t2g⁵ eg⁰ configuration with just one unpaired electron (whereas the weak-field [Fe(H₂O)₆]³⁺ is high spin with five).

Question 12

Which ligand produces the largest crystal-field splitting?
(a) CO (b) H₂O (c) Cl⁻ (d) F⁻

Accepted Answer

(a) CO.

Carbon monoxide is the strongest-field ligand in the spectrochemical series (I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO), producing the largest Δo.

Question 13

In an octahedral complex, the d orbitals split into:
(a) lower t2g and higher eg (b) lower eg and higher t2g (c) three equal sets (d) no splitting

Accepted Answer

(a) lower t2g and higher eg.

In an octahedral field the three t2g orbitals are stabilised (point between the ligands) and the two eg orbitals are raised (point at the ligands), separated by Δo. The tetrahedral case is inverted.

Question 14

Cis–trans isomerism is NOT shown by:
(a) tetrahedral complexes (b) square-planar MA₂B₂ (c) octahedral MA₄B₂ (d) square-planar [Pt(NH₃)₂Cl₂]

Accepted Answer

(a) tetrahedral complexes.

All four positions of a tetrahedron are equivalent, so there is no cis/trans distinction. Square-planar and octahedral complexes can show geometrical isomerism.

Question 15

Ethylenediamine (en) is a ______ ligand:
(a) bidentate (b) monodentate (c) ambidentate (d) hexadentate

Accepted Answer

(a) bidentate.

Ethylenediamine has two nitrogen donor atoms, so it binds the metal at two points, forming a stable five-membered chelate ring — it is bidentate.

Practice — Coordination Compounds and the d- and f-Block

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