Chemistry - Coordination Chemistry Question with Solution | TestHub

For $[\mathrm{Cr}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$, Cr is in the +2 oxidation state.

The electronic configuration of Cr is $[\mathrm{Ar}]3d^{5}4s^1$.

For ${\mathrm{Cr}}^{2+}$, the electronic configuration is $[\mathrm{Ar}]3d^4$.

The ligand ${\mathrm{H}}_2\mathrm{O}$ is a weak field ligand, so it forms a high spin complex. However, we need to calculate for both high spin and low spin states.

Given: Mean pairing energy $P=23500{\mathrm{cm}}^{-1}$

Magnitude of ${\Delta }_0=13900{\mathrm{cm}}^{-1}$

For a $d^4$ octahedral complex:

High Spin State (x):

Electrons fill $t_{2g}$ and $e_g$ orbitals according to Hund's rule.

Configuration: $t_{2g}^3{e}_g^1$

CFSE (x) $=[3\times (-0.4{\Delta }_0)+1\times (0.6{\Delta }_0)]+0P$

CFSE (x) $=[-1.2{\Delta }_0+0.6{\Delta }_0]=-0.6{\Delta }_0$

CFSE (x) $=-0.6\times 13900{\mathrm{cm}}^{-1}=-8340{\mathrm{cm}}^{-1}$

Low Spin State (y):

Electrons pair up in $t_{2g}$ orbitals before filling $e_g$ orbitals.

Configuration: $t_{2g}^4{e}_g^0$

CFSE (y) $=[4\times (-0.4{\Delta }_0)+0\times (0.6{\Delta }_0)]+1P$ (1 pairing energy is added)

CFSE (y) $=[-1.6{\Delta }_0]+P$

CFSE (y) $=-1.6\times 13900{\mathrm{cm}}^{-1}+23500{\mathrm{cm}}^{-1}$

CFSE (y) $=-22240{\mathrm{cm}}^{-1}+23500{\mathrm{cm}}^{-1}=1260{\mathrm{cm}}^{-1}$

Calculate $\left(\frac{\mathrm{y}-\mathrm{x}}{100}\right)$:

$\left(\frac{1260-(-8340)}{100}\right)=\left(\frac{1260+8340}{100}\right)=\left(\frac{9600}{100}\right)=96$