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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 · 2016 · Paper I

Q8(a)(i). Low-spin and high-spin complexes; [CoF₆]³- and [Co(NH₃)₆]³+

  • 15 marks
  • Answered
Question

What are low-spin and high-spin complexes? Indicate spin states of CoF₆³⁻ and Co(NH₃)₆³⁺ complexes. Explain their magnetic character.

Answer in brief

When Δₒ<P the electrons spread over t_(2g) and e_g with the most parallel spins (high spin); when Δₒ>P they pair in t_(2g) (low spin). [CoF₆]³-: high spin t_(2g)⁴e_g², four unpaired electrons, paramagnetic, μₛ=4.90 BM. [Co(NH₃)₆]³+: low spin t_(2g)⁶, diamagnetic.

Key points

  • In an octahedral field the d orbitals split into t_(2g) (lower) and e_g (upper), Δₒ apart. For d¹–d³ and d⁸–d¹⁰ there is only one way to fill them.
  • For d⁴–d⁷ there is a choice. The fourth electron can pair in t_(2g), paying the pairing energy P (Coulomb repulsion and lost exchange energy), or enter e_g, paying Δₒ.
  • electrons occupy e_g before pairing: maximum number of unpaired electrons, a high-spin (outer-orbital) complex.
  • electrons pair in t_(2g) first: minimum number of unpaired electrons, a low-spin (spin-paired, inner-orbital) complex.

The full model answer

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

High spin and low spin

  1. In an octahedral field the d orbitals split into t_(2g) (lower) and e_g (upper), Δₒ apart. For d¹–d³ and d⁸–d¹⁰ there is only one way to fill them.
  2. For d⁴–d⁷ there is a choice. The fourth electron can pair in t_(2g), paying the pairing energy P (Coulomb repulsion and lost exchange energy), or enter e_g, paying Δₒ.
  3. Weak field, Δₒ<P: electrons occupy e_g before pairing: maximum number of unpaired electrons, a high-spin (outer-orbital) complex.
  4. Strong field, Δₒ>P: electrons pair in t_(2g) first: minimum number of unpaired electrons, a low-spin (spin-paired, inner-orbital) complex.
  5. Which case occurs depends on the ligand (spectrochemical series), the metal's charge and its period (4d and 5d complexes are nearly always low spin).

The two Co(III) complexes

ComplexFieldConfigurationnμₛ=√(n(n+2))CFSE
[CoF₆]³-F⁻, weak (π donor): Δₒ<Pt_(2g)⁴e_g²44.90 BM-0.4 Δₒ
[Co(NH₃)₆]³+NH₃, stronger: 22 900 > 21 000 cm⁻¹t_(2g)⁶00-2.4 Δₒ+2P
  1. Both contain Co³+: x+6(-1)=-3 and x+6(0)=+3 give x=+3, d⁶.
  2. [CoF₆]³-: F⁻ is low in the spectrochemical series, so Δₒ is below the pairing energy of Co³+ (21 000 cm⁻¹). Four unpaired electrons: paramagnetic; the measured moment is nearly 5 BM, close to √(24)=4.90 BM. It is one of the very few high-spin Co(III) complexes. VB description: outer-orbital sp³d².
  3. [Co(NH₃)₆]³+: Δₒ=22 900 cm⁻¹ exceeds P, so all six electrons pair in t_(2g): no unpaired electron, diamagnetic (measured moment close to zero). VB description: inner-orbital d²sp³.
  4. Net stabilisation of the low-spin form: -2.4(22 900)+2(21 000)=-12 960 cm⁻¹, against -0.4(22 900)=-9160 cm⁻¹ for the high-spin form, so low spin wins.

Application. magnetic susceptibility (Gouy balance) counts the unpaired electrons and so decides the spin state; the spin state also controls colour and lability: low-spin [Co(NH₃)₆]³+ is kinetically inert, while high-spin [CoF₆]³- is reactive towards water.

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Model answer to Q8(a)(i), 2016 Paper I: solution page 41
Solution page 41 of 46
Read the whole 2016 solution2016 question paper

All parts of Question 8