Question

Geometric shapes of the complexes (p), (q) and (r) formed by the following reactions respectively are:
$C{u}^{2+}\left(aq\right)+KCN\left(aq\right)\left(excess\right)\to complex\left(p\right)$
$C{o}^{2+}\left(aq\right)+KN{O}_2\left(s\right)+H^{+}\left(aq\right)\to complex\left(q\right)$
$Z{n}^{2+}\left(aq\right)+NaOH\left(aq\right)\left(excess\right)\to complex\left(r\right)$.

• A. Tetrahedral, octahedral and square planar
• B. Tetrahedral, octahedral and tetrahedral
• C. Square planar, octahedral and tetrahedral
• D. Octahedral, octahedral and tetrahedral

Answer 1

The correct option is B Tetrahedral, octahedral and tetrahedral

(A) $C{u}^{2+}\left(aq\right)+4C{N}^{-}\left(aq\right)\left(excess\right)\to \left[Cu\right(CN{)}_4{]}^{3-}\left(aq\right)\left(p\right)$
(B) $C{o}^{2+}\left(aq\right)+3K^{+}\left(aq\right)+7N{O}_2^{-}\left(s\right)+2H^{+}\left(aq\right)\to K_3\left[Co\right(N{O}_2{)}_6]\downarrow (q)+NO\uparrow +H_{2}O$
(C) $Z{n}^{2+}\left(aq\right)+4O{H}^{-}\left(aq\right)\left(excess\right)\to \left[Zn\right(OH{)}_4{]}^{2-}\left(r\right)$

The coordination number of $C{u}^{2+}$ and $Z{n}^{2+}$ ions with $3d^{10}$ electron configuration is four. They undergo $s{p}^3$ hybridization and form tetrahedral compounds. The coordination number of $C{o}^{2+}$ with $3d^6$ electronic configuration is six. It undergoes $d^{2}s{p}^3$ hybridization and forms octahedral complex.