Question

The CFSE for octahedral $[CoC{l}_6{]}^{4-}$ is 18000 $c{m}^{-1}$. The CFSE for tetrahedral $[CoC{l}_4{]}^{2-}$ will be

Answer 1

CFSE for octahedral and tetrahedral complexes are closely related to each other by formula, ${△}_r=\frac{4}{9}{△}_0.$
where, ${△}_0=CFSE$ for octahedral complex, ${△}_t=CFSE$ for tetrahedral complex
According to question, ${△}_0=18000c{m}^{-1}\therefore {△}_t=\frac{4}{9}{△}_0=\frac{4}{9}\times 18000c{m}^{-1}=4\times 2000c{m}^{-1}=8000c{m}^{-1}$
Hence, correct choice is (c).