Question

Calculate the Crystal Field Stabilization Energy (CFSE) of $[Co{F}_6{]}^{3-}$.

• A. $-0.6{\Delta }_t$
• B. $0.6{\Delta }_t$
• C. $-0.4{\Delta }_o$
• D. $0.4{\Delta }_o$

Answer 1

The correct option is C $-0.4{\Delta }_o$

$\text{Hybridisation of Co in ground state}:3d^{7}4s^2$
$\text{Hybridisation of}C{o}^{3+}\text{in excited state}:3d^{6}4s^0$. Since it is high spin complex
Hence, $C{o}^{3+}:t_{2g}^4{e}_g^2$
$\text{CFSE = Number of electrons in}{t}_{2g}\times (-0.4){\Delta }_o+\text{Number of electrons in}{e}_g\times \left(0.6\right){\Delta }_o$
$\Rightarrow \text{CFSE =}4\times (-0.4){\Delta }_o+2\times \left(0.6\right){\Delta }_o$
$\Rightarrow \text{CFSE =}-1.6{\Delta }_o+1.2{\Delta }_o=-0.4{\Delta }_o$