Question

Low spin complex of $d^6-$ cation in an octahedral field will have the following crystal field stabilization energy
$({\Delta }_0=$ crystal field splitting energy in an octahedral field, $P=$ Electron pairing energy)

• A. $\frac{-12}{5}{\Delta }_0+P$
• B. $\frac{-12}{5}{\Delta }_0+3P$
• C. $\frac{-2}{5}{\Delta }_0+2P$
• D. $\frac{-2}{5}{\Delta }_0+P$

Answer 1

The correct option is B $\frac{-12}{5}{\Delta }_0+3P$

We know that,
$\text{Crystal field stabilization energy}=(-0.4x+0.6y){\Delta }_0+zP$
Where $x=$ number of electrons occupying $t_{2g}$ orbital
$y=$ number of electrons occupying $e_g$ orbital
$z=$ number of pairs of electrons
For low spin $d^6$ complex electronic configuration $=t_{2g}^6{e}_g^0$
$\therefore x=6,y=0,z=3$
$\text{Hence crystal field stabilization energy}(-0.4\times 6+0\times 0.6){\Delta }_0+3P$
$=\frac{-12}{5}{\Delta }_0+3P$