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 Reagent S is (1) K 4[ Fe ( CN 6]. (2) Na 2 HPO 4 (3) K 2 CrO 4 (4) KOH

Question

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 (A) K4 [Fe (CN)6] (B) Na2 HPO4 (C) K2 CrO4 (D) KOH

Tardigrade Admin · Accepted Answer

Zn2+ on treatment with excess of KOH produces [Zn(OH)4]2-

$K_{4} [F e (CN)_{6}]$

$N a_{2} H P O_{4}$

$K_{2} C r O_{4}$

KOH

$Z n^{2 +}$ on treatment with excess of KOH produces $[Z n (O H)_{4}]^{2 -}$

K4​[Fe(CN)6​]

Na2​HPO4​

K2​CrO4​

KOH

Zn2+ on treatment with excess of KOH produces [Zn(OH)4​]2−

$K_{4} [F e (CN)_{6}]$

$N a_{2} H P O_{4}$

$K_{2} C r O_{4}$

$Z n^{2 +}$ on treatment with excess of KOH produces $[Z n (O H)_{4}]^{2 -}$