Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $MC{l}_2.6H_{2}O\left(X\right)$ and $N{H}_{4}Cl$ gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1 :3 electrolyte. The reaction of X with excess $HCl$ at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for the complex Y. Among the following options, which statement(s) is/are correct?

• A. The hybridization of the central metal ion in Y is $d^{2}s{p}^3$
• B. Z is a tetrahedral complex
• C. Addition of silver nitrate to Y gives only two equivalents of silver chloride
• D. When X and Z are in equilibrium at $0^{\circ }C$, the colour of the solution is pink

Answer 1

Answer: (A), (B), (D)

When X and Z are in equilibrium at $0^{\circ }C$, the colour of the solution is pink

$\underset{(\mu =3.87B.M.)}{\underset{Octahedral\left(pink\right)}{\left[\underset{\left(X\right)}{Co(H_{2}O{)}_6}\right]}}C{l}_2\underset{air}{\overset{Aq.N{H}_3+N{H}_{4}Cl}{\to }}\underset{(\mu =0)}{\underset{\left(Y\right)}{\left[Co\right(N{H}_3{)}_6]C{l}_3}}$

1:3 elctrolyte means $3C{l}^{-}andone\left[Co\right(N{H}_3{)}_6{]}^{3+}$

Y is inner-orbital complex having $d^{2}s{p}^3$ hybridization and thus, all the electrons are paired in the complex.

$[\underset{\left(X\right)}{Co(H_{2}O{)}_6}{]}^{2+}+HCl(excess)\stackrel{0^{\circ }C}{\rightleftharpoons }\underset{(\mu =3.87B.M.)}{\underset{tetrahedral\left(blue\right)}{[CoC{l}_4{]}^{2-}}},\Delta H=+ve$