Geometric shapes of the complexes (p), (q) and (r) formed by the following reactions respectively are:
$C u^{2 +} (a q) + K C N (a q) (e x c e s s) \to c o m p l e x (p)$
$C o^{2 +} (a q) + K N O_{2} (s) + H^{+} (a q) \to c o m p l e x (q)$
$Z n^{2 +} (a q) + N a O H (a q) (e x c e s s) \to c o m p l e x (r)$ .

Tetrahedral, octahedral and square planar

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Tetrahedral, octahedral and tetrahedral

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Square planar, octahedral and tetrahedral

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Octahedral, octahedral and tetrahedral

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Solution

The correct option is B Tetrahedral, octahedral and tetrahedral
(A) $C u^{2 +} (a q) + 4 C N^{-} (a q) (e x c e s s) \to [C u (C N)_{4}]^{3 -} (a q) (p)$
(B) $C o^{2 +} (a q) + 3 K^{+} (a q) + 7 N O_{2}^{-} (s) + 2 H^{+} (a q) \to K_{3} [C o (N O_{2})_{6}] ↓ (q) + N O ↑ + H_{2} O$
(C) $Z n^{2 +} (a q) + 4 O H^{-} (a q) (e x c e s s) \to [Z n (O H)_{4}]^{2 -} (r)$
The coordination number of $C u^{2 +}$ and $Z n^{2 +}$ ions with $3 d^{10}$ electron configuration is four. They undergo $s p^{3}$ hybridization and form tetrahedral compounds. The coordination number of $C o^{2 +}$ with $3 d^{6}$ electronic configuration is six. It undergoes $d^{2} s p^{3}$ hybridization and forms octahedral complex.

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