JEE Advanced 2023 · Paper 2, Chemistry question 5

$\ce{[Pt(NH3)2Br2]}$ is square planar $\ce{MA2B2}$, so it shows geometrical (cis-trans) isomerism.

(A) $\ce{[Pt(en)(SCN)2]}$: the bidentate en must span cis positions, so the two SCN are always cis. There is no geometrical isomer; only linkage isomerism (SCN/NCS) is possible. ✗

(B) $\ce{[Zn(NH3)2Cl2]}$ is tetrahedral and has no geometrical isomers. ✗

(C) $\ce{[Pt(NH3)2Cl4]}$ is octahedral $\ce{MA2B4}$, so it has cis and trans forms. ✓

(D) $\ce{[Cr(en)2(H2O)(SO4)]+}$ is octahedral $\ce{M(AA)2BC}$, so it has cis and trans forms (the cis form is also optically active). ✓