Q. 46 Using crystal field theory, draw energy level diagram, wrile electronic

configuration of the central metal atom/ion and determine the magnetic

moment value in the following.

(a)  ${\left[Co{F}_{6}\right]}^{3-},{\left[Co{\left({H}_{2}0\right)}_{6}\right]}^{2+},{\left[Co{\left(CN\right)}_{6}\right]}^{3-}$

(b) $Fe{{F}_{6}}^{3-}{\left[Fe{\left({H}_{2}0\right)}_{6}\right]}^{2+},{\left[Fe{\left(CN\right)}_{6}\right]}^{4-}$

${\left[Co{F}_{6}\right]}^{3-}$

F is a weak field ligand. Configuration of Co${}^{3+}=3{d}^{6}\left(or{t}_{2g}^{4}{e}_{g}^{2}$

Number of unpaired electrons (n) =4

Magnetic moment ($\mu$)=$\sqrt{n\left(n+2\right)}=\sqrt{4\left(4+2\right)}=\sqrt{24}=4.9BM$
${\left[Co\left({H}_{2}{O}_{6}\right]}^{2+}$


${H}_{2}O$ is a weak field ligand. Configuration of Co There is no unpaired electon, so it is diamagnetic

$\mu =0$

(b) ${\left[Fe{F}_{6}\right]}^{3-}$ $\mu =\sqrt{5\left(5+2\right)}$
$=\sqrt{35}=5.92BM$
${\left[Fe\left({H}_{2}{O}_{6}\right]}^{2+}$ $\mu =\sqrt{4\left(4+2\right)}$
$=\sqrt{24}=$
${\left[Fe{\left(CN\right)}_{6}\right]}^{4-}$ Since , CN is a strong filed ligand, all the electrons get paired.

Because there is no unpaired electron, so it is diamagnetic in nature.