Coordination Compounds — Class 12 MCQs with Answers

Coordination Compounds — important questions & MCQs with answers (Class 12 Chemistry)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 82 are timed and scored when you sign in.

1. Q1Easy

   Ligands are:

   A.Catalysts

   B.Acids

   C.Lewis bases that donate electron pair to central metal✓ Correct

   D.Solvents

   Answer: C. Lewis bases that donate electron pair to central metal

   Explanation: A ligand donates a lone pair to the central metal to form a coordinate bond, acting as a Lewis base — not a catalyst, acid, or mere solvent, though some solvents like H₂O do double as ligands.

2. Q2Easy

   Cis-trans isomerism is shown by complexes of type:

   A.[MA₂B₂] square planar✓ Correct

   B.[MA₄] tetrahedral

   C.[MA₆] octahedral

   D.[MA₃]

   Answer: A. [MA₂B₂] square planar

   Explanation: Cis-trans isomerism needs two non-equivalent positions for like ligands to occupy — square planar [MA₂B₂] has adjacent and opposite arrangements, but tetrahedral or [MA₆]-type complexes have only one possible arrangement, so they show none.

3. Q3Easy

   Coordination number of [Co(NH₃)₆]³⁺:

   A.6✓ Correct

   B.3

   C.4

   D.2

   Answer: A. 6

   Explanation: Coordination number counts the donor atoms directly bonded to the metal — six NH₃ molecules each donate one N lone pair to Co³⁺, giving coordination number 6, not the charge or the atom count of NH₃.

4. Q4Easy

   Linkage isomerism arises due to:

   A.Different ligands of same composition

   B.Different geometries

   C.Different counter ions

   D.Ambidentate ligand bonding through different atoms✓ Correct

   Answer: D. Ambidentate ligand bonding through different atoms

   Explanation: Linkage isomerism happens only with an ambidentate ligand — one that can attach to the metal through either of two different donor atoms, like NO₂⁻ through N or O — giving compounds of identical formula but different bonded atoms.

5. Q5Easy

   Who proposed the theory of coordination compounds explaining primary and secondary valencies?

   A.Linus Pauling

   B.Alfred Werner✓ Correct

   C.Niels Bohr

   D.J.J. Thomson

   Answer: B. Alfred Werner

   Explanation: Alfred Werner (1893) explained coordination compounds with his theory of primary valency (oxidation state) and secondary valency (coordination number), winning the 1913 Nobel Prize — not Bohr or Pauling, whose atomic models came later.

6. Q6Easy

   [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br show which isomerism?

   A.Linkage

   B.Geometrical

   C.Ionisation✓ Correct

   D.Optical

   Answer: C. Ionisation

   Explanation: [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br have the same formula but swap which ion — Br⁻ or SO₄²⁻ — sits inside the coordination sphere versus outside as the free counter ion, so they ionise differently in water: ionisation isomerism.

7. Q7Medium

   EDTA is a:

   A.Monodentate

   B.Hexadentate ligand✓ Correct

   C.Bidentate

   D.Free ion

   Answer: B. Hexadentate ligand

   Explanation: EDTA⁴⁻ binds through six donor atoms — two N and four O — wrapping the metal in ring structures as a hexadentate chelating ligand, unlike simple bidentate ligands such as ethylenediamine, which use only two N atoms.

8. Q8Medium

   Cis and trans are types of:

   A.Ionisation

   B.Optical isomerism

   C.Linkage isomerism

   D.Geometric isomerism✓ Correct

   Answer: D. Geometric isomerism

   Explanation: Cis places two identical ligands adjacent (90°) and trans places them opposite (180°) around the same metal — this purely spatial arrangement is geometric isomerism, distinct from optical isomerism, which concerns non-superimposable mirror images.

9. Q9Medium

   What is the oxidation state of cobalt in [Co(en)₂Cl₂]Cl?

   A.+1

   B.+3✓ Correct

   C.+2

   D.+4

   Answer: B. +3

   Explanation: en is neutral and the two Cl⁻ ligands inside the brackets contribute −2, while the lone Cl⁻ outside the brackets means the complex ion itself is +1 overall, so x + 0 − 2 = +1 gives cobalt = +3.

10. Q10Medium

    Number of geometrical isomers of [Co(NH₃)₄Cl₂]⁺ (octahedral) is:

    A.1

    B.4

    C.3

    D.2✓ Correct

    Answer: D. 2

    Explanation: An octahedral MA₄B₂ complex like [Co(NH₃)₄Cl₂]⁺ can have its two Cl⁻ adjacent (cis, 90°) or opposite (trans, 180°) — exactly two distinct geometrical isomers, both achiral since each retains a mirror plane.

11. Q11Medium

    The IUPAC name of [Pt(NH₃)₂Cl₂] is:

    A.Diamminedichloridoplatinate(II)

    B.Dichloridodiammineplatinum(IV)

    C.Platinum diammine dichloride

    D.Diamminedichloridoplatinum(II)✓ Correct

    Answer: D. Diamminedichloridoplatinum(II)

    Explanation: With NH₃ neutral and the complex itself neutral, Pt must be +2; ligands are cited alphabetically by name — 'ammine' before 'chlorido' — and a neutral complex keeps the plain metal name, giving diamminedichloridoplatinum(II).

12. Q12Hard

    Haemoglobin contains a coordination compound of:

    A.Copper

    B.Iron with porphyrin✓ Correct

    C.Magnesium

    D.Zinc

    Answer: B. Iron with porphyrin

    Explanation: Haemoglobin's heme group is an iron(II)–porphyrin complex, and it is the Fe²⁺ centre — not Mg²⁺ (found in chlorophyll instead) or Cu — that reversibly binds O₂ for transport.

13. Q13Hard

    Crystal Field Theory: in octahedral field, d-orbitals split into:

    A.Random

    B.All same energy

    C.Three sets

    D.t₂g (lower) and eg (higher)✓ Correct

    Answer: D. t₂g (lower) and eg (higher)

    Explanation: In an octahedral field, ligand lone pairs approach along the axes, so the axially-pointing eg orbitals (dz², dx²-y²) rise in energy while the t₂g set (dxy, dyz, dxz) drops — the gap is called Δo, not five degenerate orbitals as in the free ion.

14. Q14Hard

    Magnetic moment (μ in BM, spin only) of [Fe(H₂O)₆]³⁺ (high spin d⁵) is approximately:

    A.5.92✓ Correct

    B.2.83

    C.4.90

    D.1.73

    Answer: A. 5.92

    Explanation: H₂O is weak field, so Fe³⁺'s five d electrons stay unpaired (high spin d⁵); with n = 5 unpaired electrons, the spin-only formula μ = √(n(n+2)) gives √35 ≈ 5.92 BM, the highest magnetic moment possible for any octahedral transition metal ion.

Coordination Compounds — FAQs

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