Question

An aqueous solution of metal ion $M_1$ reacts separately with reagents $Q$ and $R$ in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion $M_2$ always forms tetrahedral complexes with these reagents. Aqueous solution of $M_2$ on reaction with reagent $S$ gives white precipitate which dissolves in excess of $S$. The reactions are summarised in the scheme given below
Scheme

$M_1,Q$ and $R$, respectively are
(1) $Z{n}^{2+},KCN$ and $HCl$
(2) $N{i}^{2+},HCl$ and $KCN$
(3) $C{d}^{2+},KCN$ and $HCl$
(4) $C{o}^{2+},HCl$ and $KCN$

• A. $Z{n}^{2+}$, KCN and HCl
• B. $N{i}^{2+}HClandKCN$
• C. $C{d}^{2+}$, KCN and HCl
• D. $C{o}^{2+}$,HCl and KCN

Answer 1

Answer: (B)

PLAN This problem can be solved by using concept of chemical
reactions of transition metal ions colour and structure of
transition metal compounds.
Here, among given four option $N{i}^{2+}$ and $Z{n}^{2+}$ has ability to
form tetrahedral as well as square planar complex depending
upon types of reagent used.
$N{i}^{2+}$ on reaction with $KCN$ forms square planar complex
$[Ni(CN{)}_4]2^{2-}$ due to strong field strength of CN.
$N{i}^{2+}+KCN⟶\underset{\text{Square planar}}{[Ni(CN{)}_4{]}^{2-}}$
While on reaction with HCl, $N{i}^{2+}$ forms stable tetrahedral
complex $[Ni(Cl{)}_4{]}^{2-}$.
$Z{n}^{2+}$ on the other hand, on reaction with KCN as well as HCl
produces tetrahedral complex because of its $d^{10}$ electronic
configuration.
Complete reaction sequence can be shown as
$\underset{\text{Tetrahedral}}{[NiC{l}_4{]}^{2-}}\underset{\text{excess}}{\overset{HCl(Q)}{\leftarrow }}\underset{(M_1)}{N{i}^{2+}}\underset{\text{excess}}{\overset{KCN(R)}{\to }}\underset{\text{Square planar}}{[Ni(CN{)}_4{]}^{2-}}$
$\underset{\text{Tetrahedral}}{[ZnC{l}_4{]}^{2-}}\underset{\text{excess}}{\overset{HCl(Q)}{\leftarrow }}\underset{(M_2)}{Z{n}^{2+}}\underset{\text{excess}}{\overset{KCN(R)}{\to }}\underset{\text{Tetrahedral}}{[Zn(CN{)}_4{]}^{2-}}$
$\downarrow KOH(x)$
$\underset{\text{White ppt }}{Zn(OH{)}_2}\underset{\text{excess}}{\overset{KOH}{\to }}\underset{\text{Soluble}}{[Zn(OH{)}_4{]}^{2-}}$