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  4. An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarised in the scheme given below Scheme <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/chemistry/45dffda5b5b2ac71a58604d56a636bbe-.png /> Reagent S is (1) K 4[ Fe ( CN 6]. (2) Na 2 HPO 4 (3) K 2 CrO 4 (4) KOH

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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
image
Reagent $S$ is
(1) $K _{4}\left[ Fe \left( CN _{6}\right]\right.$
(2) $Na _{2} HPO _{4}$
(3) $K _{2} CrO _{4}$
(4) $KOH$

Organic Chemistry – Some Basic Principles and Techniques

Solution:

$Zn^{2+}$ on treatment with excess of KOH produces $[Zn(OH)_4]^{2-}$