Question

Addition of excess aqueous ammonia to a pink coloured aqueous solutions of $MC{l}_2.6H_{2}O\left(X\right)\text{and}N{H}_{4}Cl$ 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 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 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°C, the colour of the solution is pink

Answer 1

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

When X and Z are in equilibrium at 0°C, the colour of the solution is pink


(A). Hybridisation of (Y) is $d^{2}s{p}^3\text{as}N{H}_3$ is strong field ligand.

(B). $[CoC{l}_4{]}^{2-}$ have $s{p}^3$ hybridisation as $C{l}^{-}$ is weak field ligand.

(C). $\left[Co\right(N{H}_3{)}_6]C{l}_3+3AgN{O}_3(aq.)\to 3AgCl$

(D).
$\begin{array}{c}[CoC{l}_4{]}^{2-}\\ \text{(Blue)}\end{array}+6H_{2}O\rightleftharpoons \begin{array}{c}\left[Co\right(H_{2}O{)}_6{]}^{2+}\\ \text{Pink}\end{array}+4C{l}^{-}$

$\Delta H=(-ve)$ (exothermic). When ice is added to the solution the equilibrium shifts right side hence pink colour will remain predominant.
So, correct answer is (A, B & D)