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
$= t^{6}_{2g} e^{0}_{g}$ or $t^{2,2,2}_{2g} e^{0}_{g}$
$\therefore x =6, y =0, z =3$
$C.F.S.E =\left(-0.4 \times 6 +0 \times0.6\right)\Delta_{0} +3P$
$ = \frac{-12}{5} \Delta_{0} +3P$