Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\mathrm{MCl}_{2} \cdot 6

Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\mathrm{MCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{X})$ and $\mathrm{NH}_{4} \mathrm{Cl}$ 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?\

Professor PC Thomas Classes · Answer

(a,b,c) : Magnetic moment,\ $\mu=\sqrt{n(n+2)}$ B.M.\ where, $n=$ No. of unpaired electrons\ For $X$ and $Z$ :\ $\sqrt{n(n+2)}=3.87$ B.M.\ $n^{2}+2 n-15=0 \Rightarrow n^{2}+5 n-3 n-15=0$\ $	herefore \quad n=3$\ For complex $Y$ ( $1: 3$ electrolyte) :\ $\sqrt{n(n+2)}=0 \Rightarrow n=0$\ $\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ or $\left.\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}ight)_{6}ight)ight] \mathrm{Cl}_{2}(\mathrm{X})$ is pink coloured compound.\ Adding excess of HCl at room temperature, changes $(X)$ into $\left[\mathrm{CoCl}_{4}ight]^{2-}$ $(Z)$ and on adding excess of $\mathrm{NH}_{3}$ and $\mathrm{NH}_{4} \mathrm{Cl}$ in the presence of air forms 
\ Thus, it is a $1: 3$ electrolyte. The hybridisation of Co in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}ight)_{6}ight] \mathrm{Cl}_{3}$ is $d^{2} s p^{3}$ (octahedral).\ Adding of $\mathrm{AgNO}_{3}$ to $\left[\mathrm{Co}\left(\mathrm{NH}_{3}ight)_{6}ight] \mathrm{Cl}_{3}(Y)$ gives three equivalents of AgCl .\ Complex $Z \quad\left[\mathrm{CoCl}_{4}ight]^{2-}$ has $s p^{3}$ hybridisation. Thus, it is a tetrahedral complex.\ \ When ice is added to the solution $\left(0^{\circ} \mathrm{C}ight)$, the equilibrium shifts towards right hence, pink colour will remain predominant.\