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 `I behaves as I :3 electrolyte. The reaction of X with excess HCI 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 statements 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 I arc in equilibrium at 0C. the colour of the solution is pink

Answer 1

Answer: (A)

The hybridization of the central metal ion in Y is $d^{2}s{p}^3$

Since, it is given that spin only magnetic moment for $Y=0$, this implies that $Y$ is a low spin complex. Also, $Y$ is an octohedral complex. Hence, the hybridization of central metal ion in $Y$ is $d^{2}s{p}^3$.