CBSE Class 12 Chemistry HOTs Coordination Compounds Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): [Ni(CN)₄]²⁻ is square planar and diamagnetic.
Reason (R): CN⁻ is a strong field ligand that pairs up the 3d electrons of Ni²⁺.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Ni²⁺ is 3d⁸. CN⁻ pairs the electrons and frees one 3d orbital, so dsp² hybridisation gives a square planar, diamagnetic ion. The pairing is the cause of both properties, so R explains A.
In simple words: Strong field pairing gives dsp² and no unpaired electrons.

Teacher's Note:
1. Say which orbitals hybridise to support the assertion.

Question 2. Assertion (A): Zinc(II) complexes are colourless.
Reason (R): Zn²⁺ has a completely filled 3d subshell.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Colour in complexes comes from d-d transitions. In Zn²⁺ (3d¹⁰) every d orbital is full, so no electron can move to a higher d level and no visible light is absorbed. R explains A.
In simple words: No vacant d position means no d-d transition.

Teacher's Note:
1. d⁰ and d¹⁰ ions are colourless.

Question 3. Assertion (A): Tetrahedral complexes do not show geometrical isomerism.
Reason (R): In a tetrahedral arrangement all four positions are equivalent.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Each position in a tetrahedron is adjacent to the other three, so there is no cis or trans distinction. R explains A.
In simple words: Every ligand position is the same, so swapping ligands gives the same shape.

Teacher's Note:
1. Square planar complexes, not tetrahedral ones, show cis-trans isomerism.

Question 4. Assertion (A): [Co(NH₃)₆]³⁺ is a diamagnetic complex.
Reason (R): NH₃ is a weak field ligand and does not pair the d electrons of Co³⁺.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (c) Co³⁺ is d⁶. NH₃ is a strong field ligand, so it pairs all six electrons (t₂g⁶) and the complex is diamagnetic. The assertion is true, but the reason is false.
In simple words: NH₃ is strong field, not weak, and that is why the complex has no unpaired electrons.

Teacher's Note:
1. Check the position of NH₃ in the spectrochemical series.

Question 5. Assertion (A): trans-[CoCl₂(en)₂]⁺ is optically inactive.
Reason (R): The trans isomer has a plane of symmetry.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) The plane containing the two en ligands and the metal divides the trans ion into identical halves. A molecule with a plane of symmetry is superimposable on its mirror image, so it is optically inactive.
In simple words: A mirror plane inside the molecule cancels any chirality.

Teacher's Note:
1. Contrast with the cis isomer, which is chiral.

Question 6. Which of the following are outer orbital complexes? (more than one may be correct)
(a) [FeF₆]³⁻
(b) [Fe(CN)₆]³⁻
(c) [Ni(NH₃)₆]²⁺
(d) [Co(NH₃)₆]³⁺
Answer:
Ans. (a) and (c)
F⁻ is a weak field ligand and the 3d electrons of Fe³⁺ are not paired, so [FeF₆]³⁻ uses 4d orbitals (sp³d²). In Ni²⁺ (3d⁸) no two 3d orbitals can be freed, so [Ni(NH₃)₆]²⁺ is also sp³d².
CN⁻ and NH₃ pair the electrons of Fe³⁺ and Co³⁺ and free 3d orbitals, so those two are inner orbital (d²sp³).
In simple words: Weak field ligands, or ions with too many d electrons to free two 3d orbitals, give outer orbital complexes.

Teacher's Note:
1. Check the ligand strength and the d count together.

Question 7. Which of the following complexes can show geometrical isomerism? (more than one may be correct)
(a) [Pt(NH₃)₂Cl₂] (square planar)
(b) [Co(NH₃)₄Cl₂]⁺
(c) [Zn(NH₃)₂Cl₂] (tetrahedral)
(d) [Co(NH₃)₃Cl₃]
Answer:
Ans. (a), (b) and (d)
[Pt(NH₃)₂Cl₂] has cis and trans forms, [Co(NH₃)₄Cl₂]⁺ has cis and trans forms, and [Co(NH₃)₃Cl₃] has fac and mer forms.
A tetrahedral complex such as [Zn(NH₃)₂Cl₂] has only one arrangement.
In simple words: Square planar and octahedral complexes with mixed ligands can have isomers; tetrahedral ones cannot.

Teacher's Note:
1. Check the geometry first.

Question 8. [Fe(H₂O)₆]³⁺ is strongly paramagnetic (5.92 BM) whereas [Fe(CN)₆]³⁻ has a moment of only 1.73 BM, although both contain Fe³⁺. Explain.
Answer:
Fe³⁺ is d⁵.
H₂O is a weak field ligand, so Δ₀ < P and the electrons stay unpaired in t₂g³ e_g²: n = 5, μ = \( \sqrt{35} \) = 5.92 BM.
CN⁻ is a strong field ligand, so Δ₀ > P and the electrons pair as t₂g⁵ e_g⁰: n = 1, μ = \( \sqrt{3} \) = 1.73 BM.
In simple words: Same metal ion, but the ligand strength decides how many electrons pair up.

Teacher's Note:
1. Always use the ligand position in the spectrochemical series.

Question 9. Three cobalt(III) compounds A, B and C have the formulae CoCl₃.6NH₃, CoCl₃.5NH₃ and CoCl₃.4NH₃. Excess AgNO₃ precipitates 3, 2 and 1 mole of AgCl per mole of A, B and C respectively. (i) Write the formula of each, (ii) name them and (iii) arrange them in increasing order of molar conductance.
Answer:
(i) A: [Co(NH₃)₆]Cl₃, B: [Co(NH₃)₅Cl]Cl₂, C: [Co(NH₃)₄Cl₂]Cl. The ionisable Cl⁻ equals the moles of AgCl, and the coordination number of Co is 6.
(ii) A: hexaamminecobalt(III) chloride; B: pentaamminechloridocobalt(III) chloride; C: tetraamminedichloridocobalt(III) chloride.
(iii) The number of ions is 4 (A), 3 (B) and 2 (C), so the conductance increases in the order C < B < A.
In simple words: More ionisable chloride means more ions and higher conductance.

Teacher's Note:
1. C has the least ionisable chloride and so the lowest conductance.

Question 10. Calculate the spin-only magnetic moments of [Mn(CN)₆]³⁻ (strong field) and [MnCl₆]³⁻ (weak field), both containing Mn³⁺.
Answer:
Mn³⁺ is d⁴ and μ = \( \sqrt{n(n+2)} \) BM.
[Mn(CN)₆]³⁻: strong field CN⁻, low spin t₂g⁴ e_g⁰, n = 2, μ = \( \sqrt{8} \) = 2.83 BM.
[MnCl₆]³⁻: weak field Cl⁻, high spin t₂g³ e_g¹, n = 4, μ = \( \sqrt{24} \) = 4.90 BM.
In simple words: Same d⁴ ion: the strong field ligand leaves two unpaired electrons, the weak field one leaves four.

Teacher's Note:
1. The d⁴ to d⁷ ions are the ones that can be high or low spin.

Question 11. CrCl₃.6H₂O exists as three isomers: violet, grey-green and dark green. They give 3, 2 and 1 mole of AgCl respectively with excess AgNO₃. Write the formula of each and name the isomerism.
Answer:
Violet: [Cr(H₂O)₆]Cl₃ (3 Cl⁻ outside, 3 AgCl).
Grey-green: [CrCl(H₂O)₅]Cl₂.H₂O (2 Cl⁻ outside, 2 AgCl).
Dark green: [CrCl₂(H₂O)₄]Cl.2H₂O (1 Cl⁻ outside, 1 AgCl).
The isomers differ in the number of water molecules bonded to the metal, so this is solvate (hydrate) isomerism.
In simple words: Water inside the sphere replaces chloride, so fewer chlorides are free to precipitate.

Teacher's Note:
1. All three isomers have the formula CrCl₃.6H₂O.

Question 12. Lead poisoning is treated by giving the patient EDTA as its calcium salt. Explain the chemistry involved.
Answer:
EDTA⁴⁻ is a hexadentate chelating ligand. It wraps around a metal ion through two N and four O atoms and forms five chelate rings.
The Pb²⁺ ion forms a more stable chelate with EDTA than Ca²⁺ does, so Pb²⁺ displaces Ca²⁺ from the calcium-EDTA complex.
The soluble lead-EDTA complex is excreted in urine, removing lead from the body.
In simple words: EDTA grabs lead more firmly than calcium and carries it out of the body.

Teacher's Note:
1. This is an application of chelate stability.
2. EDTA is also used to estimate the hardness of water.