On treatment of ${[Pt {({NH}_{3})}_{4}]}^{2 +}$ with concentrated HCl, two compounds (I) and (II) having the same formula, $[Pt {({NH}_{3})}_{2} {Cl}_{2}]$ are obtained, (I) can be converted into (II) by boiling with dilute HCl. A solution of (I) reacts with oxalic acid to form $[Pt {({NH}_{3})}_{2} (C_{2} O_{4})]$ whereas (II) does not react. Point out the correct statement of the following.

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(I) cis, (II) trans; both tetrahedral

(I) cis, (II) trans; both square planar

(l) trans, (II) cis; both tetrahedral

(I) trans, (II) cis; both square planar

answer is B.

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Detailed Solution

Chelate ligands i.e. oxalato being too small to span the transpositions and thus prefer to occupy the cis-position. So (I) is cis and (II) is trans. Platinum is in +2 oxidation state with 5d⁸ valence shell electron configuration. This configuration has higher CFSE and thus favours square planar geometry.

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