JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Coordination Compounds

Analyzing each complex:

**P. $[\mathrm{Cr}({\mathrm{NH}}_3{)}_4{\mathrm{Cl}}_2]\mathrm{Cl}$**: ${\mathrm{Cr}}^{3+}$ has electronic configuration $3d^3$, so it is paramagnetic. The complex has formula ${\mathrm{Ma}}_4{\mathrm{b}}_2$, which exhibits cis-trans isomerism. → P matches 3

**Q. $[\mathrm{Ti}({\mathrm{H}}_2\mathrm{O}{)}_5\mathrm{Cl}]({\mathrm{NO}}_3{)}_2$**: ${\mathrm{Ti}}^{3+}$ has configuration $3d^1$, so it is paramagnetic. The complex can exhibit ionisation isomerism (${\mathrm{Cl}}^{-}$ and ${\mathrm{NO}}_3^{-}$ can exchange positions inside and outside the coordination sphere). → Q matches 1

**R. $[\mathrm{Pt}(\mathrm{en})({\mathrm{NH}}_3)\mathrm{Cl}]{\mathrm{NO}}_3$**: ${\mathrm{Pt}}^{2+}$ has configuration $5d^8$, and in square planar geometry, all electrons are paired, so it is diamagnetic. The complex can exhibit ionisation isomerism (${\mathrm{Cl}}^{-}$ and ${\mathrm{NO}}_3^{-}$ exchange). → R matches 4

**S. $[\mathrm{Co}({\mathrm{NH}}_3{)}_4({\mathrm{NO}}_3{)}_2]{\mathrm{NO}}_3$**: ${\mathrm{Co}}^{3+}$ has configuration $3d^6$, and in octahedral strong field (with ${\mathrm{NH}}_3$), all electrons are paired, so it is diamagnetic. The complex exhibits cis-trans isomerism. → S matches 2

Therefore, the correct matching is P-3, Q-1, R-4, S-2, which corresponds to option (B).