Among (a) - (d), the complexes that can display geometrical isomerism are: (a) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{3}\right] \mathrm{Cl}^{+}$ (b) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}_{5}\right]^{-}$ (c) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}\left(\mathrm{NO}_{2}\right)\right]$ (d) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right]^{2+}$

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Coordination Compounds

Among (a) - (d), the complexes that can display geometrical isomerism are:
(a)

[Pt (NH_{3})_{3}] Cl^{+}

(b)

[Pt (NH_{3}) Cl_{5}]^{-}

(c)

[Pt (NH_{3})_{2} Cl (NO_{2})]

(d)

[Pt (NH_{3})_{4} ClBr]^{2 +}

Answer: (B)

Geometrical isomerism needs ligands that can be set either next to or opposite one another. The complex

[Pt (NH X_{3}) X_{2} Cl (NO X_{2})]

is of the type

M A_{2} BC

and can place its two different ligands cis or trans, and

[Pt (NH X_{3}) X_{4} ClBr] X^{2 +}

of the type

M A_{4} BC

can likewise set its

Cl

and

Br

adjacent or across. The other two lack a ligand arrangement that allows a second placement, so cis-trans isomerism appears only in

[Pt (NH X_{3}) X_{2} Cl (NO X_{2})]

and

[Pt (NH X_{3}) X_{4} ClBr] X^{2 +}

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option