Question

The crystal field stabilisation energy of $\left[Co\right(N{H}_3{)}_6]C{l}_3$ is :

• A. $-7.2{\Delta }_0$
• B. $-0.4{\Delta }_0$
• C. $-2.4{\Delta }_0$
• D. $-3.6{\Delta }_0$

Answer 1

The correct option is C $-2.4{\Delta }_0$

The cobalt ion in $\left[Co\right(N{H}_3{)}_6]C{l}_3$ undergoes $d{s}^2{p}^3$ hybridization to form inner orbital or low spin or spin paired complex.

6 electrons are present in $t_{2g}$ level and 0 electrons are present in $e_g$ level.

The crystal field stabilization energy of $\left[Co\right(N{H}_3{)}_6]C{l}_3$ is $CFSE=6\times (-0.4{\Delta }_0)+0\times \left(0.6{\Delta }_0\right)=-2.4{\Delta }_0$