Chemistry - Coordination Chemistry Question with Solution | TestHub

(P) $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_3{\left({\mathrm{NO}}_2\right)}_3\right]\to$

${\mathrm{Ma}}_3{\mathrm{b}}_3$ Type ⟶ Show G.I. but Not O.I.

→ Ambidentate ligand ⟶ show linkage isomerism

$\frac{{\mathrm{Co}}^{+3}}{3\mathrm{d}+\text{ SFL }}\to \frac{3{\mathrm{d}}^{6}4{\mathrm{s}}^{0}4{\mathrm{p}}^{0}4{\mathrm{d}}^0}{{\Delta }_0>\text{ P.E. }}:{\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0$ : Diamagnetic

$\mathrm{P}\to 1,2,4$

(Q) $\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_4(\mathrm{CN}{)}_2\right]$

${\mathrm{Ma}}_4{\mathrm{b}}_2$ Type ⟶ Show G.I. but Not O.I.

${\mathrm{CN}}^{\ominus }⟶$ Ambidentate ligand → show linkage isomerism $\frac{{\mathrm{Ni}}^{+2}}{3\mathrm{d}+\text{ SFL }}\to \frac{3{\mathrm{d}}^{8}4{\mathrm{s}}^{0}4{\mathrm{p}}^{0}4{\mathrm{d}}^0}{\underline{{\Delta }_0>\text{ P.E. }}}:t_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^2:\frac{2\text{ unpaired electron }}{\mu =\sqrt{\mathrm{n}(\mathrm{n}+2)}}=\sqrt{2(2+2)}=2.84$ B.M. $\mathrm{Q}\to 1,2,5$

(R) $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2{\left({\mathrm{NO}}_2\right)}_2\right]$

${\mathrm{Ma}}_2{\mathrm{b}}_2{\mathrm{c}}_2$ Type → Show G.I. as well O.I.

${\mathrm{NO}}_2^{\ominus }\to$ Ambidentate ligand → show linkage isomerism

$\frac{{\mathrm{Pt}}^{+4}}{5\mathrm{d}+\text{ metal }}:\frac{5{\mathrm{d}}^{6}6{\mathrm{s}}^{0}6{\mathrm{p}}^{0}6{\mathrm{d}}^0}{{\Delta }_0>\text{ P.E. }}:t_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0:$ Diamagnetic

$\mathrm{R}\to 1,2,3,4$

(R) $\left[\mathrm{Zn}{\left({\mathrm{NH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{O}\right)}_2\right]$ $\frac{\left[\mathrm{Zn}(\mathrm{gly}{)}_2\right]}{M(\mathrm{AB}{)}_2}:\frac{{\mathrm{Zn}}^{+2}}{\frac{\text{ Tetrahedral complex }}{\underline{\left[\mathrm{M}(\mathrm{AB}{)}_2\right]}}}:3{\mathrm{d}}^{10}4{\mathrm{s}}^0\to$ Diamagnetic

(can not show geometrical but can show optical isomerism)

$\mathrm{S}\to 3,4$