JEE Advanced 2017 Paper 1

$\mathrm{[M(H_2O)_6]Cl_2} + 6\mathrm{NH_3} \xrightarrow[\mathrm{O_2}]{\mathrm{NH_4Cl}} \mathrm{[M(NH_3)_6]Cl_3} + 6\mathrm{H_2O}$  $\underset{\text{Pink}}{(X)}$             (Y)$*\mathrm{[M(NH_3)_6]Cl_3} \rightarrow \mathrm{[M(NH_3)_6]^{3+}} + 3\mathrm{Cl^{-}}$[1:3 electrolyte]$\mathrm{[M(H_2O)_6]Cl_2} + 2\mathrm{HCl} \xrightarrow[\text{Temp}]{\text{Room}} \mathrm{[MCl_4]^{2-}} + 4\mathrm{H_2O} + 2\mathrm{H_3O^{+}}$  $\underset{\text{Pink}}{(X)}$    Excess      $\underset{\text{Blue}}{(Z)}$If M is Co then $X : \mathrm{[Co(H_2O)_6]Cl_2}$$Y : \mathrm{[Co(NH_3)_6]Cl_3}$$Z : \mathrm{[CoCl_4]^{2-}}$⇒ $\mu_x = 3.87$ due to presence of three unpaired electron in X$\mathrm{Co^{2+}};3d^7$$\mathrm{H_2O}$ is weak field ligand hence no pairingHybridization state is $sp^3d^2$⇒ $\mu_z = 3.87$BM due to presence of three unpaired electron in Z.$\mathrm{Co^{2+}};3d^7$$\mathrm{Cl^{-}}$ is weak a field ligand and coordination number is four hence hybridization state is $sp^3$$\Rightarrow \mathrm{[M(NH_3)_6]Cl_3} + 3\mathrm{AgNO_3} \rightarrow \mathrm{[M(NH_3)_6]^{3+}} + 3\mathrm{NO_3^{-}} + 3\mathrm{AgCl(s)}$$\Rightarrow \mathrm{[Co(H_2O)_6]^{2+}}4\mathrm{Cl^{-}} \rightarrow \mathrm{[CoCl_4]^{2+}} + 6\mathrm{H_2O}; \Delta H > 0$    Pink      Blue(As octahedral complex changes to tetrahedral complex with $\mathrm{H_2O}$ ligands replaced by $\mathrm{Cl^{-}}$)At 0°C, equilibrium shifts in reverse direction hence colour of the solution is pink due to $\mathrm{[Co(H_2O)_6]^{2+}}$