Coordination Compounds 2 Question 50
52. Addition of excess aqueous ammonia to a pink coloured aqueous solution of $M Cl _2 \cdot 6 H _2 O(X)$ 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., whersas it is zero for complex $Y$.
Among the following options, which statement(s) is (are) correct?
(2017 Adv.)
(a) The hybridisation of the central metal ion in $Y$ is $d^{2} s p^{3}$
(b) Addition of silver nitrate to $Y$ given only two equivalents of silver chloride
(c) When $X$ and $Y$ are in equilibrium at $0^{\circ} C$, the colour of the solution is pink
(d) $Z$ is a tetrahedral complex
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Answer:
Correct Answer: 52. (a, b, d)
Solution:
(a) Since $NH _3$ is moderately strong ligand, hybridisation of cobalt in $Y$ is $d^{2} s p^{3}$.
(b) Cobalt is $s p^{3}$-hybridised in $\left[CoCl _4\right]^{2-}$.
(c) $\left[Co\left(NH _3\right) _6\right] Cl _3+3 AgNO _3($ aq $) \longrightarrow 3 AgCl \downarrow$
(d) $\underset{\text { Blue }}{[CoCl _4]^{2-}}+6 H _2 O \rightleftharpoons \underset{\text { Pink }}{[Co(H _2 O) _6]^{2+}}+4 Cl^{-} ; \Delta H<0$
BETA
