Chemistry - Coordination Chemistry Question with Solution | TestHub

(1) ${\left[{\mathrm{FeF}}_6\right]}^{3-}: {\mathrm{Fe}}^{3+}=3{\mathrm{d}}^5 ({\Delta }_{\mathrm{O}}<\mathrm{P})$

Number of unpaired ${\mathrm{e}}^{-}=5$

$\mathrm{\mu } = \sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}$

$\therefore \mu =\sqrt{35} \mathrm{BM}$

(2) ${\left[{\mathrm{CoF}}_6\right]}^{3-}: {\mathrm{Co}}^{3+}=3{\mathrm{d}}^6\left({\Delta }_{\mathrm{O}}<\mathrm{P}\right)$

Number of unpaired ${\mathrm{e}}^{-}=4$

$\mathrm{\mu } = \sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}$

$\therefore \mu =\sqrt{24} \mathrm{BM}$

(3) ${\left[\mathrm{Co}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}^{3-}: {\mathrm{Co}}^{3+}=3{\mathrm{d}}^6\left({\Delta }_{\mathrm{O}}>\mathrm{P}\right)$

Number of unpaired ${\mathrm{e}}^{-}=0$

$\mathrm{\mu } = \sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}$

$\therefore \mu =0 \mathrm{BM}$

So, the correct order is ${\left[{\mathrm{FeF}}_6\right]}^{3-}>{\left[{\mathrm{CoF}}_6\right]}^{3-}>{\left[\mathrm{Co}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]}^{3-}$