Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $MC{l}_2\cdot 6H_{2}O(X)$ 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 colured complex $Z$. The calculated spin only magnetic moment of $X$ and $Z$ is $3.87B.M$., whereas it is zero for complex $Y$. Among the following options, which statement(s) is(are) correct ?

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

Answer 1

Answer: (B), (C), (D)

$X=CoC{l}_2\cdot 6H_{2}O$ i.e. $\left[Co{\left(H_{2}O\right)}_6\right]C{l}_2\Rightarrow s{p}^3{d}^2$
$Y=\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3\Rightarrow d^{2}s{p}^3$
$Z=\left[Co{\left(H_{2}O\right)}_{3}Cl\right]Cl\cdot 2H_{2}O\Rightarrow s{p}^3$(Tetrahedral)
Addition of silver nitrate to $Y$ give three equivalents of silver chloride.