A) \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]
B) \[{{[PtC{{l}_{4}}]}^{2-}}\]
C) \[{{[Pd{{(CN)}_{4}}]}^{2-}}\]
D) \[{{[Zn{{(CN)}_{4}}]}^{2-}}\]
Correct Answer: D
Solution :
(1)\[N{{i}^{+2}}\] \[3{{d}^{8}}\] \[ds{{p}^{2}}\] Planar \[CN\to SFL\]pairing\[\to \]Sq. planar splitting (2)\[P{{t}^{+2}}\] \[5{{d}^{8}}\]\[ds{{p}^{2}}\] \[Cl\] \[P{{t}^{+2}}\]Forms a sq. Planar complex (3)\[{{[Pd{{(CN)}_{4}}]}^{2-}}\] \[ds{{p}^{2}}\] \[P{{d}^{+2}}\] \[4{{d}^{8}}\]\[P{{d}^{+2}}\]forms a sq. planar complex (4)\[Z{{n}^{+2}}\] \[3{{d}^{10}}\]\[s{{p}^{3}}\]tetrahedral hence, non-planar
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