Chemistry - Coordination Chemistry Question with Solution | TestHub

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(i) ${\left[{\mathrm{FeF}}_6\right]}^{3-}{\mathrm{Fe}}^{3+}\Rightarrow \left[\mathrm{Ar}\right]3 {\mathrm{d}}^5, {\mathrm{F}}^{-}$ is WFL

$\mathrm{n}=5, \mu =\sqrt{35} \mathrm{B}.\mathrm{M}$

(ii)${\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}\Rightarrow {\mathrm{Co}}^{3+}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^6,{\mathrm{NH}}_3$ is SFL

$\mathrm{n}=0, \mathrm{\mu }=0 \mathrm{B}.\mathrm{M}$

(iii) ${\left[{\mathrm{NiCl}}_4\right]}^{2-}{\mathrm{Ni}}^{2+}\Rightarrow \left[\mathrm{Ar}\right]3 {\mathrm{d}}^8,{\mathrm{Cl}}^{-}\to \mathrm{WFL}$

$n=2, \mu =\sqrt{8} \mathrm{B}.\mathrm{M}$

(iv) ${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}{\mathrm{Cu}}^{2+}\Rightarrow \left[\mathrm{Ar}\right]3 {\mathrm{d}}^9,{\mathrm{NH}}_3\to \mathrm{SFL}$

$n=1, \mu =\sqrt{3} \mathrm{B}.\mathrm{M}$

Thus correct order of spin only magnetic moment is

$\left(\mathrm{i}\right)>\left(\mathrm{iii}\right)>\left(\mathrm{iv}\right)>\left(\mathrm{ii}\right)$