Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $MCl _{2} \cdot 6 H _{2} O (X)$ and $NH _{4} Cl$ 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 $3.87\, B.M$., 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 $d^2sp^3$
• 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

The Correct Option is D

$X = CoCl _{2} \cdot 6 H _{2} O $ i.e. $\left[ Co \left( H _{2} O \right)_{6}\right] Cl _{2} \,\,\,\, \Rightarrow sp ^{3} d ^{2} $
$Y =\left[ Co \left( NH _{3}\right)_{6}\right] Cl _{3} \,\,\,\, \Rightarrow d ^{2} sp ^{3} $
$Z =\left[ Co \left( H _{2} O \right)_{3} Cl \right] Cl \cdot 2 H _{2} O \,\,\,\,\Rightarrow sp ^{3} $(Tetrahedral)
Addition of silver nitrate to $Y$ give three equivalents of silver chloride.