Chemistry - Coordination Chemistry Question with Solution | TestHub

$\left(\mathrm{I}\right) {\left[\mathrm{Cr}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$

${\mathrm{Cr}}^2=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{4}4{\mathrm{s}}^0$

It is ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridised low spin complex as ${\mathrm{CN}}^{-}$ is a strong field ligand.

$\left(\mathrm{II}\right) {\left[{\mathrm{RuCl}}_6\right]}^{2-}$

Low spin complex as Ru belongs to 4d series, as elements belonging to 4d and 5d series forms low spin complexes even with weak field ligand.

${\mathrm{Ru}}^{+4}=\left[\mathrm{Kr}\right]4{\mathrm{d}}^{4}5{\mathrm{s}}^0$

${\mathrm{t}}_{2\mathrm{g}}$ set contains $4$ electrons.

$\left(\mathrm{III}\right) {\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$

${\mathrm{Cr}}^{+2}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{4}4{\mathrm{s}}^0$

Its configuration is ${\mathrm{t}}_{2\mathrm{g}}^3{\mathrm{e}}_{\mathrm{g}}^1$

It has $4$ unpaired ${\mathrm{e}}^{-}$ as ${\mathrm{H}}_2\mathrm{O}$ is weak field ligand.

So, its Magnetic moment, $\mu =4.9 \mathrm{B}.\mathrm{M}$

$\left(\mathrm{IV}\right) {\left[\mathrm{Fe}\left({\mathrm{H}}_2\mathrm{O}\right)6\right]}^{2+}$

${\mathrm{Fe}}^{2+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4 {\mathrm{s}}^0$

$={\mathrm{t}}_{2\mathrm{g}}^4{\mathrm{e}}_{\mathrm{g}}^2$

It has $4$ unpaired ${\mathrm{e}}^{-}$, its Magnetic moment, $\mathrm{\mu }=4.9 \mathrm{B}.\mathrm{M}.$