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 CFSE for tetrahedral $[CoCl_4]^{2-}$ is $8000\, cm^{-1}$, the CFSE for octahedral $[CoCl_6]^{4-}$ will be .........

$18000\, cm^{-1}$

The correct answer is Option (4) → $18000\, cm^{-1}$

In crystal field theory,the relationship between tetrahedral crystal field splitting (Δₜ) and octahedral crystal field splitting (Δ₀) is

Δₜ = (4/9) Δ₀

Given: Δₜ = 8000 cm⁻¹

So, Δ₀ = (9/4) × Δₜ

 Δ₀ = (9/4) × 8000

Δ₀ = 9 × 2000 Δ₀ = 18000 cm⁻¹