VKC ScienceChemistry for written exams Telegram

UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2020 · Paper I

Q4(d). Diamagnetic [Ni(CN)₄]²⁻ and paramagnetic [NiCl₄]²⁻

  • 10 marks
  • Answered
Question

The complex [Ni(CN)₄]²⁻ is diamagnetic, but [NiCl₄]²⁻ is paramagnetic and has two unpaired electrons. Explain these observations and deduce the structure of two complexes.

Answer in brief

Ni(II) is d⁸. Strong-field CN⁻ pairs the eight electrons in four d orbitals, leaving d_(x²-y²) empty for dsp² bonding: square planar and diamagnetic. Weak-field Cl⁻ cannot cause pairing; Ni uses sp³: tetrahedral, e⁴t₂⁴ with two unpaired electrons (μ=2.83 BM).

Key points

  • [Ni(CN)₄]²⁻: the strong-field CN⁻ ligands force the two unpaired 3d electrons to pair, leaving one 3d orbital empty. One 3d + 4s + two 4p give dsp² hybrids, pointing to the corners of a square: [Ni(CN)₄]²⁻: square planar,…
  • [NiCl₄]²⁻: Cl⁻ is a weak-field ligand; the 3d arrangement of Ni²⁺ (two unpaired) is kept and the outer 4s and 4p are used: sp³, tetrahedral: [NiCl₄]²⁻: tetrahedral, two unpaired electrons
  • removing the two ligands on z leaves d_(x²-y²) (pointing at the four ligands) far above the others. For d⁸ with a strong-field ligand the gap is larger than the pairing energy: all eight electrons pair in the four lower…
  • Δₜ≈49Δₒ is small, and Cl⁻ is weak, so no pairing: e⁴t₂⁴ with two unpaired electrons in t₂. Four large Cl⁻ ions are also less crowded at 109.5° than at 90°.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Valence bond picture

  1. [Ni(CN)₄]²⁻: the strong-field CN⁻ ligands force the two unpaired 3d electrons to pair, leaving one 3d orbital empty. One 3d + 4s + two 4p give dsp² hybrids, pointing to the corners of a square: [Ni(CN)₄]²⁻: square planar, diamagnetic
  2. [NiCl₄]²⁻: Cl⁻ is a weak-field ligand; the 3d arrangement of Ni²⁺ (two unpaired) is kept and the outer 4s and 4p are used: sp³, tetrahedral: [NiCl₄]²⁻: tetrahedral, two unpaired electrons

Crystal-field picture (the reason)

  1. Square planar: removing the two ligands on z leaves d_(x²-y²) (pointing at the four ligands) far above the others. For d⁸ with a strong-field ligand the gap is larger than the pairing energy: all eight electrons pair in the four lower orbitals, d_(x²-y²) stays empty. The empty orbital also gains extra stabilisation, so d⁸ with strong fields favours square planar geometry (as in [PtCl₄]²⁻).
  2. Tetrahedral: Δₜ≈49Δₒ is small, and Cl⁻ is weak, so no pairing: e⁴t₂⁴ with two unpaired electrons in t₂. Four large Cl⁻ ions are also less crowded at 109.5° than at 90°.
  3. Magnetic moment: μₛₒ=√(n(n+2))=√(2×4), μ=2.83 BM; the measured value is somewhat higher because the t₂ set has orbital angular momentum.
  4. Colour supports the structures: square planar [Ni(CN)₄]²⁻ is yellow; tetrahedral [NiCl₄]²⁻ is blue, with more intense bands (no centre of symmetry).

Application. the square-planar, diamagnetic red complex of Ni(II) with dimethylglyoxime, [Ni(dmgH)₂], is the standard gravimetric precipitate for nickel.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q4(d), 2020 Paper I: solution page 20
Solution page 20 of 45