$C o^{2 +} (a q .) + S C N^{-} (a q .) \to C o m p l e x (X)$
$N i^{2 +} (a q .) + d i m e t h y l g l y o x i m e N H_{4} O H - ---- \to C o m p l e x (Y)$
The coordination number of cobalt and nickel in both the complexes (X) and (Y) is four. The geometry of complexes (X) and (Y) are respectively :

tetrahedral and square planar

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both tetrahedral

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square planar and tetrahedral

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both square planar

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Solution

The correct option is A tetrahedral and square planar
The complexes X and Y are $[C o (S C N)_{4}]^{2 -}$ and $[N i (d m g)_{2}]$ . The central metal ions in these complexes undergo $s p^{3}$ and $d s p^{2}$ hybridization respectively. Hence they have tetrahedral and square planar geometries.

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Similar questions

C o^{2 +} (a q .) + S C N^{-} (a q .) \to C o m p l e x (X)

N i^{2 +} (a q .) + d i m e t h y l g l y o x i m e N H_{4} O H - ---- \to C o m p l e x (Y)

C o^{2 +} (a q .) + S C N^{-} (a q) \to

(X)

N i^{2 +} (a q) +

N H_{4} O H - ---- \to

(Y)

N i^{2 +}, P t^{2 +}

Z n^{2 +},

The geometries of the ammonia complexes of $N i^{2 +}, P t^{2 +} a n d Z n^{2 +},$ respcetively, are

[N i (C O)_{4}]

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