Coordination CompoundsNEET MCQs with solutions

Not quite — the answer is C.

NO2− can coordinate through nitrogen (nitro mode) or through oxygen (nitrito mode) and is therefore ambidentate. NH3 has only one donor atom and is monodentate. OH− coordinates only through oxygen. en is bidentate, not ambidentate, because it uses two different donor atoms simultaneously rather than offering alternative donor sites.

Not quite — the answer is A.

A homoleptic complex contains only one type of ligand in its coordination sphere. In K3[Al(C2O4)3], all three ligands are oxalate. The coordination spheres of the remaining three compounds each contain at least two different ligand types. Counter-ions are not considered when determining homoleptic character.

Not quite — the answer is B.

[Co(NH3)5Cl]Cl2 ionises to give one complex cation [Co(NH3)5Cl]2+ and two free Cl- ions. Total ions = 3. The chloride inside the coordination sphere does not ionise.

Not quite — the answer is A.

NO2- is an ambidentate ligand that can bind through N (nitro) or O (nitrito), producing linkage isomers. Option B is ionisation isomerism, Option C is coordination isomerism, and Option D is geometrical isomerism.

Not quite — the answer is C.

Primary valence equals the oxidation state. Co is +3, so primary valence = 3. Secondary valence equals the coordination number = 6 (six NH3 donor atoms directly bonded). The three Cl- counter-ions satisfy the primary valence; the six NH3 ligands satisfy the secondary valence.

Not quite — the answer is A.

All four complexes contain Co3+ so Δo depends only on ligand-field strength. The spectrochemical series gives NH3 > H2O > F- > Cl- among these ligands. Therefore [Co(NH3)6]3+ has the largest Δo.

Not quite — the answer is A.

Ionisation isomers differ in which ion is inside the coordination sphere vs outside as counter ion — e.g. [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl. Option D is the trap — linkage isomerism involves ambidentate ligands, not ion exchange.

Not quite — the answer is B.

Ambidentate ligands (e.g. NO₂⁻, SCN⁻) can coordinate through two different donor atoms — this gives linkage isomers. Option A is the trap — ionisation isomerism involves ion exchange, not donor atom switching.

Not quite — the answer is A.

Inner orbital octahedral complexes use (n−1)d orbitals for hybridisation, giving d²sp³. Outer orbital complexes use nd orbitals and show sp³d² instead. The key distinction is which d orbitals are involved.

Not quite — the answer is B.

Co³⁺ is d⁶; NH₃ (strong field) causes pairing of electrons into three 3d orbitals, freeing two 3d orbitals for d²sp³ hybridisation. [CoF₆]³⁻ uses weak-field F⁻ and remains outer orbital (sp³d²).

Not quite — the answer is A.

I⁻ causes the minimum crystal field splitting among all common ligands — it sits at the extreme weak-field end of the spectrochemical series. Among halides, field strength decreases as F⁻ > Cl⁻ > Br⁻ > I⁻, making Br⁻ the second-weakest option here.

Not quite — the answer is D.

The correct order is F⁻ < H₂O < NH₃ < en — en (ethylenediamine) is a stronger field ligand than NH₃ due to the chelate effect and greater donor strength. CN⁻ and CO sit above en. Option A wrongly places H₂O above en.

Not quite — the answer is B.

en (ethylenediamine) has two −NH₂ donor atoms that coordinate simultaneously to a metal, forming a five-membered chelate ring. NH₃, CO, and H₂O are all monodentate — they bind through only one donor atom and cannot form rings.

Not quite — the answer is C.

EDTA (ethylenediaminetetraacetic acid) donates through two N and four O atoms — six donor sites total, making it hexadentate. Oxalate is bidentate, en is bidentate, and dien is tridentate.

Not quite — the answer is A.

Primary valency = oxidation state; it is ionic and satisfied by counter ions outside the coordination sphere. Secondary valency (not primary) is non-ionisable and directional. Option C confuses primary with secondary valency roles.

Not quite — the answer is B.

[Co(NH₃)₆]Cl₃ dissociates into [Co(NH₃)₆]³⁺ (1 ion) + 3 Cl⁻ = 4 ions total. Option C (3) is the trap — students confuse number of Cl⁻ ions with total ion count. All NH₃ ligands remain inside the coordination sphere.

Not quite — the answer is A.

CFT explains how degenerate d orbitals of the central metal ion split into sets of different energies when surrounded by ligands. p, s, and f orbitals are not involved in this splitting in transition metal complexes.

Not quite — the answer is A.

In octahedral field, ligands approach along axes. The dxy, dxz, dyz orbitals (t₂g) point between axes and experience less repulsion (lower energy); dx²-y² and dz² (eg) point directly at ligands and experience greater repulsion (higher energy).

Not quite — the answer is A.

Ni²⁺ reacts with dimethylglyoxime (dmgH) to form a characteristic bright red chelate precipitate — this is the standard spot test for Ni²⁺. EDTA forms colourless complexes. Potassium ferrocyanide and thiocyanate are used for Fe and Co detection respectively.

Not quite — the answer is B.

EDTA forms highly stable 1:1 complexes with most metal ions regardless of charge, making it ideal for complexometric titrations. It does not oxidise, reduce, or precipitate the metal — it sequesters it as a soluble chelate.

Not quite — the answer is C.

d⁵ high-spin has the configuration t₂g³eg², with each orbital singly occupied. CFSE = 3(−0.4Δo) + 2(+0.6Δo) = −1.2 + 1.2 = 0. d⁶ high-spin is the trap — it gives −0.4Δo, not zero.

Not quite — the answer is A.

d³ high-spin: t₂g³eg⁰; CFSE = 3(−0.4Δo) = −1.2Δo. d⁵ high-spin is the trap — it also has 3 electrons in t₂g but gains 2 in eg, cancelling to zero. d³ has no eg electrons so the full −1.2Δo is retained.

Not quite — the answer is A.

Secondary valency (coordination number) is directional — ligands attach at fixed angles around the metal, directly determining geometry. Primary valency is non-directional. Option D is the key trap — students confuse which valency is directional.

Not quite — the answer is D.

In [Co(NH₃)₃Cl₃], all three Cl⁻ ions are inside the coordination sphere satisfying secondary valency — none are free in solution to react with Ag⁺. Options A, B, C each have 3, 2, 1 ionisable Cl⁻ respectively.