Read the passage carefully and answer the Questions.
In the formation of coordination complexes, if the inner d orbital (n-1d) is used in hybridisation, the complex, is called an inner orbital or low spin or spin paired complex. And if it uses outer d orbital (nd) in hybridisation (like \( sp^3d^2 \)), it is called outer orbital or high spin or spin free complex. The degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions in the complex. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. For coordination complexes, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula, \( μ = \sqrt{n (n+2)} \) where \( n \) is the number of unpaired electrons and \( μ \) is the magnetic moment in units of Bohr magneton (BM). The coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine.
The crystal field splitting energy for tetrahedral \( [CoCl_4]^{2-} \) is \( 8000\, cm^{-1} \), the crystal field splitting energy for octahedral \( [CoCl_6]^{4-} \) will be .........
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(18000\, cm^{-1}\)
In a tetrahedral complex only four ligands surround the metal ion, and none of them points directly at any d orbital. The d orbitals therefore split less than in an octahedral complex, and the order of the two sets is inverted. For the same metal ion, the same ligands and the same metal–ligand distances, the two splitting energies are related by:
\( \displaystyle \Delta_t = \frac{4}{9}\,\Delta_o \)
Both complexes contain cobalt in the same oxidation state with the same ligand, \( \ce{Cl-} \):
\( \ce{[CoCl4]^2-} \): \( x + 4(-1) = -2 \Rightarrow x = +2 \)
\( \ce{[CoCl6]^4-} \): \( x + 6(-1) = -4 \Rightarrow x = +2 \)
Here \( \Delta_t = 8000\ \mathrm{cm^{-1}} \). So:
\( \displaystyle \Delta_o = \frac{9}{4}\,\Delta_t \)
\( \displaystyle \Delta_o = \frac{9}{4} \times 8000\ \mathrm{cm^{-1}} = 18000\ \mathrm{cm^{-1}} \)
So the value for octahedral \( \ce{[CoCl6]^4-} \) is 18000 cm−1.
Shortcut: The octahedral splitting must be larger than the tetrahedral one, which leaves only Options (1) and (4). The factor is 9/4 = 2.25, not 2, so the value is 18000 cm−1.
NCERT (2026-27 edition):
Crystal Field Splitting in Tetrahedral Coordination Entities – “In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting. For the same metal, the same ligands and metal-ligand distances, it can be shown that Δt = (4/9) Δo.”