Question

The CFSE for $[CoC{l}_6{]}^{4-}$ complex is 18000 $c{m}^{-1}$ the $\Delta$ for $[CoC{l}_4{]}^{2-}$ will be

• A. $18000c{m}^{-1}$
• B. $16000c{m}^{-1}$
• C. $8000c{m}^{-1}$
• D. $2000c{m}^{-1}$

Answer 1

The correct option is C $8000c{m}^{-1}$

Given CFSE for $[CoC{l}_6{]}^{4-}=18000c{m}^{-1}$
$[CoC{l}_4{]}^{2-}\to$ Tetrahedral complex
Hence, ${\Delta }_t=?$
$\because {\Delta }_t=\frac{4}{9}{\Delta }_0\therefore {\Delta }_t\text{for}[CoC{l}_4{]}^{2-}=\frac{4}{9}\times 18000c{m}^{-1}=8000c{m}^{-1}$