Chemistry - Coordination Chemistry Question with Solution | TestHub

${\mathrm{\Delta }}_0=\left(-0.4\times {\mathrm{nt}}_{2\mathrm{g}}\right)+\left(0.6\times {\mathrm{ne}}_{\mathrm{g}}\right)$

where, $\mathrm{n}=$number of electrons in ${\mathrm{nt}}_{2\mathrm{g}}$ and ${\mathrm{ne}}_{\mathrm{g}}$.

$\mathrm{CFSE}$ of given $\mathrm{\alpha }$ -electrons configuration are as follows

a. For high spin, ${\mathrm{d}}^6$;

${\mathrm{\Delta }}_0\left(\mathrm{CFSE}\right)=-0.4\times 4+0.6\times 2=-0.4$

b. For low spin, ${\mathrm{d}}^4$;

${\mathrm{\Delta }}_0\left(\mathrm{CFSE}\right)=-0.4\times 4+0.6\times 0=1.6$

c. For low spin ${\mathrm{d}}^5$

${\mathrm{\Delta }}_0\left(\mathrm{CFSE}\right)=-0.4\times 5+0.6\times 0=-2.0$

d. For high spin, ${\mathrm{d}}^7$

${\mathrm{\Delta }}_0\left(\mathrm{CFSE}\right)=-0.4\times 5+0.6\times 2=-0.8$

Neglecting $-\mathrm{ve}$ sign (as it indicates that energy is released).

Hence, maximum ${\mathrm{\Delta }}_0$ $\left(\mathrm{CFSE}\right)$ value is shown by low spin ${\mathrm{d}}^5$ configuration.