(a) \( Ni(CO)_4 \), (b) \( [Ni(CN_4)]^{2-} \) and (c) \( [Pt(NH_3)_2Cl_2] \) are respectively ______ complexes.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (a) Tetrahedral, (b) square planar and (c) square planar
The shape of a four-coordinate complex follows from its hybridisation. If the metal ion keeps an empty inner d orbital (often after a strong field ligand pairs the electrons), it uses dsp2 hybridisation (square planar); otherwise it uses sp3 (tetrahedral).
(a) \( \ce{Ni(CO)4} \): Ni is 0. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); the strong field CO pushes the 4s electrons into 3d, giving 3d10. No 3d orbital is free, so 4s and 4p give sp3: tetrahedral.
(b) \( \ce{[Ni(CN)4]^2-} \): x + 4(−1) = −2, so Ni is +2, 3d8 (the two 4s electrons lost). \( \ce{CN^{-}} \) pairs the eight electrons into four 3d orbitals, leaving one empty: dsp2, square planar.
(c) \( \ce{[Pt(NH3)2Cl2]} \): x + 0 + 2(−1) = 0, so Pt is +2. Pt (Z = 78) is \( \mathrm{[Xe]\,4f^{14}\,5d^{9}\,6s^{1}} \); losing the 6s electron and one 5d electron gives 5d8. Heavier d8 ions form dsp2, square planar complexes; that is why this complex has cis and trans isomers (cis-platin), which a tetrahedral complex of this type cannot have.
Shortcut: \( \ce{[Ni(CN)4]^2-} \) is square planar, which leaves Options (2) and (4); the platinum complex shows cis–trans isomerism, so it is square planar too: Option (4).
Note: It seems that there is a mistake in the original NTA paper, which prints [Ni(CN4)]2− when it should read [Ni(CN)4]2−.