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🧪 Chemistry  ·  Coordination Compounds  ·  NEET & JEE

Which of the following has the most CFSE stabilisation in an octahedral field?

Answer: d 3 (t 2g 3 , CFSE = -1.2 Delta_o) and d 6 low spin (t 2g 6 , CFSE = -2.4 Delta_o).

  • A d<sup>3</sup> (t<sub>2g</sub><sup>3</sup>, CFSE = -1.2 Delta_o) and d<sup>6</sup> low spin (t<sub>2g</sub><sup>6</sup>, CFSE = -2.4 Delta_o)
  • B d<sup>0</sup>, which by definition has zero d electrons and therefore zero CFSE
  • C d<sup>10</sup>, which has all orbitals fully occupied giving zero net CFSE
  • D d<sup>5</sup> high spin, which has one electron in every orbital giving zero net CFSE

Correct answer: A. d<sup>3</sup> (t<sub>2g</sub><sup>3</sup>, CFSE = -1.2 Delta_o) and d<sup>6</sup> low spin (t<sub>2g</sub><sup>6</sup>, CFSE = -2.4 Delta_o)

Explanation: d<sup>6</sup> low spin (t<sub>2g</sub><sup>6</sup> e<sub>g</sub><sup>0</sup>): CFSE = 6(-0.4) = -2.4 Delta_o, the maximum CFSE for any dn in octahedral field. d<sup>3</sup> gives -1.2 Delta_o (all in t<sub>2g</sub>).

Octahedralcoordination no. 6e.g. [Co(NH3)6]3+Tetrahedralcoordination no. 4e.g. [Ni(CO)4]Square Planarcoordination no. 4e.g. [Pt(NH3)2Cl2]

The three common coordination geometries: octahedral (6 ligands), tetrahedral (4 ligands), and square planar (4 ligands in one plane).

Concept context

Study of compounds where a central metal atom is bonded to surrounding ligands. Covers nomenclature, types of isomerism, bonding theories (VBT, CFT), and applications in medicine, photography, and industry.

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