When `[Ni(NH_(3))_(4)]^(2+)` is treated with conc `HCI`, two compounds having the formula `Ni(NH_(3))(2)CI_(2)` are formed. A solution of I reacts with oxalic acid to form `Ni(NH_(3))_(2)(C_(2)O_(4))`. II does not react with oxalic acid. Deduce the configuration of I and II and the geometry of Ni (II) complexes .

Options

A. (a) (I) cis, (II) trans, both tetrahedral
B. (b) (I) cis, (II) trans, both square planar
C. (c) (I) trans, (II) cis, both tetrahedral
D. (d) (I) trans, (II) cris, both square planar

Answer

B

Explanation

Chelate ligands i.e. oxalato being too small to span the trans-positions and thus prefer to occupy the cis-position. So (I) is cis and (II) is trans. Platinum is in `+2` oxidation state with `5d^8` valence shell electron configuration. This configuration has higher CFSE and thus favours square planar geometry.