Coordination Compounds: Conceptual Doubts Cleared (2)

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Tags Coordination Compounds

Question

For the complex [Co(NH3)6]3+[Co(NH_3)_6]^{3+}, determine (a) the oxidation state of cobalt, (b) the coordination number, (c) the hybridisation, (d) whether it is paramagnetic or diamagnetic, and (e) the number of unpaired electrons. (Atomic number of Co = 27.) JEE Main 2024 standard.

Solution — Step by Step

NH3_3 is a neutral ligand. The complex has charge +3+3. So Co+6×0=+3\text{Co} + 6 \times 0 = +3, giving Co3+^{3+}.

There are 6 NH3_3 ligands, each binding through one nitrogen atom. Coordination number = 6.

Co has electronic configuration [Ar]3d74s2[Ar]3d^7 4s^2. Co3+^{3+} is [Ar]3d6[Ar]3d^6. NH3_3 is a strong-field ligand (causes pairing). In a strong octahedral field, d6d^6 pairs up: all 6 electrons occupy the lower t2gt_{2g} orbitals.

This frees up two inner 3d3d orbitals plus 4s4s and three 4p4p, giving d2sp3d^2sp^3 inner-orbital hybridisation (octahedral geometry).

With all dd electrons paired, the complex is diamagnetic. Unpaired electrons = 0.

Final answers: (a) +3, (b) 6, (c) d2sp3d^2sp^3, (d) diamagnetic, (e) 0 unpaired electrons.

Why This Works

The strength of NH3_3 as a ligand (it’s a moderate-to-strong field ligand on the spectrochemical series) determines whether dd electrons pair up. For d6d^6 Co3+^{3+} with a strong-field ligand, pairing is energetically favourable: all 6 electrons sit in the lower-energy t2gt_{2g} set, leaving the upper ege_g empty.

This pairing also makes inner dd orbitals available for hybridisation with 4s4s and 4p4p, hence “inner-orbital” d2sp3d^2sp^3. Weak-field ligands (like F−^-, H2_2O for some metals) leave electrons unpaired and force “outer-orbital” sp3d2sp^3d^2 hybridisation using 4d4d orbitals.

Alternative Method

Crystal field splitting energy comparison:

  • If Δo>P\Delta_o > P (pairing energy): low-spin (all paired). NH3_3 on Co3+^{3+} qualifies.
  • If Δo<P\Delta_o < P: high-spin (max unpaired). H2_2O on Co3+^{3+} borderline.

This same conclusion arrives without thinking about hybridisation explicitly.

Students often forget that NH3_3 is not always a strong-field ligand — it depends on the metal. For Co3+^{3+}, it is strong; for some other metals, it’s borderline. Always check the spectrochemical position relative to the metal in question.

NEET asks magnetic moment in Bohr magnetons: μ=n(n+2)\mu = \sqrt{n(n+2)} BM, where nn is the number of unpaired electrons. With n=0n = 0, μ=0\mu = 0 — diamagnetic. With n=4n = 4, μ≈4.9\mu \approx 4.9 BM — paramagnetic.

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