Q.
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) $Ni ^{2+}, HCl$ and $KCN$
(3) $Cd ^{2+}, KCN$ and $HCl$
(4) $Co ^{2+}, HCl$ and $KCN$
Organic Chemistry – Some Basic Principles and Techniques
Solution:
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 $Ni^{2+}$ and $Zn^{2+}$ has ability to
form tetrahedral as well as square planar complex depending
upon types of reagent used.
$Ni^{2+}$ on reaction with $KCN$ forms square planar complex
$[Ni(CN)_4]2^{2-}$ due to strong field strength of CN.
$Ni^{2+} + KCN \longrightarrow \underset {\text{Square planar}}{[Ni(CN)_4]^{2-}}$
While on reaction with HCl, $Ni^{2+}$ forms stable tetrahedral
complex $[Ni(Cl)_4]^{2-}$.
$Zn^{2+}$ 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
$\underset {\text{Tetrahedral}}{[NiCl_4]^{2-}} \xleftarrow [\text{excess}] {HCl(Q)} \, \underset{(M_1)}{Ni^{2+}} \xrightarrow [\text{excess}] {KCN (R)} \, \underset {\text{Square planar}} {[Ni (CN)_4]^{2-}}$
$\underset {\text{Tetrahedral}}{[ZnCl_4]^{2-}} \xleftarrow [\text{excess}] {HCl(Q)} \, \underset {(M_2)}{Zn^{2+}} \xrightarrow [\text{excess}] {KCN (R)} \, \underset {\text{Tetrahedral}} {[Zn (CN)_4]^{2-}}$
$\big \downarrow KOH (x)$
$\underset{\text{White ppt }}{Zn(OH)_2} \xrightarrow [\text{excess}] {KOH} \, \underset {\text{Soluble}} {[Zn (OH)_4]^{2-}}$
