Chemistry - Co-ordination Compounds Question with Solution | TestHub

(1) ${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}{\mathrm{Co}}^{3+}\Rightarrow 3{\mathrm{d}}^{6}4{\mathrm{s}}^0$

$=[-0.4\times 6+0.6\times (0)]{\Delta }_0=-2.4{\Delta }_0$
(2)
$\begin{array}{rl}& {\left[\mathrm{Mn}(\mathrm{SCN}{)}_6\right]}^4\\ & {\mathrm{Mn}}^{2+}\Rightarrow 3{\mathrm{d}}^{5}4{\mathrm{s}}^0\end{array}$

$\begin{array}{rl}& \mu =\sqrt{35}\text{ B.M. }=5.96\text{ B.M. }\\ & \mathrm{CFSE}=(-0.4\times 3+0.6\times 2){\Delta }_0\end{array}$
So ${\Delta }_0=0$
(3)
$\begin{array}{rl}& {\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{-4}{\mathrm{Fe}}^{2+}\Rightarrow 3{\mathrm{d}}^{6}4{\mathrm{s}}^0\end{array}$

$\mathrm{CFSE}=-2.4{\Delta }_0$
(4) ${\left[{\mathrm{FeF}}_6\right]}^{4-}$
${\mathrm{Fe}}^{2+}\Rightarrow 3{\mathrm{d}}^{6}4{\mathrm{s}}^0$

$\begin{array}{rl}& \mu =\sqrt{24}\text{ B.M. }=4.89\text{ B.M. }\\ & \text{ CFSE }=(-0.4\times 4+0.6\times 2){\Delta }_0=-1.2{\Delta }_0\end{array}$