JEE Main 2007 — Coordination Compounds Question with Solution

${}_{27}{\mathrm{Co}}^{2+}$-$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{7}4{\mathrm{s}}^0$
As ${\mathrm{Cl}}^{-}$is weak field ligand so no pairing up. Hence it is $s{p}^3$ hybridized giving tetrahedral geometry. ${\mathrm{Fe}}^{2+}$-$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^{6}4{\mathrm{s}}^0$
Due to ${\mathrm{Cl}}^{-}$, back pairing is not observed so it will be ${\mathrm{sp}}^3$ hybridized giving tetrahedral geometry. ${\mathrm{Ni}}^{2+}-1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3p^{6}3{\mathrm{d}}^{8}4{\mathrm{s}}^0$
Because weak ligand, back pairing is not observed so it will be ${\mathrm{sp}}^3$ i.e. tetrahedral geometry. All the complexes of ${\mathrm{Pt}}^{2+}$ are square planar including those with weak field ligand such as halide ions thus (D) is correct.