Question

Low spin complex of $d^6$-cation in an octahedral field will have the following 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

Answer: (B)

$C.F.S.E.=\left(-0.4x+0.6y\right){\Delta }_0+zP$
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
$=t_{2g}^6{e}_g^0$ or $t_{2g}^{2,2,2}{e}_g^0$
$\therefore x=6,y=0,z=3$
$C.F.S.E=\left(-0.4\times 6+0\times 0.6\right){\Delta }_0+3P$
$=\frac{-12}{5}{\Delta }_0+3P$