The CFSE for octahedral [CoCl_6]^4- is 18,000 cm–1. The CFSE for tetr…

The correct answer is Option (4) → 8,000 cm^–1

In an octahedral complex the six ligands point straight at two of the d orbitals, which split the d orbitals by Δ_o. In a tetrahedral complex there are only four ligands, and none of them points directly at any d orbital. The splitting is therefore inverted and smaller. For the same metal, the same ligands and the same metal–ligand distance:

    \( \displaystyle \Delta_t = \frac{4}{9}\,\Delta_o \)

Both complexes here contain \( \ce{Co^{2+}} \) and \( \ce{Cl^{-}} \) ligands; the oxidation state of cobalt is +2 in each:

    \( \displaystyle [\ce{CoCl6}]^{4-}: x + 6(-1) = -4 \;\Rightarrow\; x = +2 \qquad [\ce{CoCl4}]^{2-}: x + 4(-1) = -2 \;\Rightarrow\; x = +2 \)

The question uses the term CFSE for this splitting energy. Substituting:

    \( \displaystyle \Delta_t = \frac{4}{9} \times 18000\ \mathrm{cm^{-1}} = 8000\ \mathrm{cm^{-1}} \)

So the value for tetrahedral \( \ce{[CoCl4]^{2-}} \) is 8,000 cm⁻¹.

Shortcut: 4/9 is less than one half, so Δ_t must be below 18000/2 = 9000 cm⁻¹. Only Option (4) is.

NCERT (2026-27 edition):
Crystal Field Splitting in Tetrahedral Coordination Entities