JEE Advanced 2017 Paper 1

[M(H2O)6]Cl2+6NH3

NH4Cl

––––––––>

O2

[M(NH3)6]Cl3+6H2O
  

(X)

Pink

             (Y)
*[M(NH3)6]Cl3→[M(NH3)6]3++3Cl−[1:3 electrolyte]
[M(H2O)6]Cl2+2HCl

Room

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Temp

[MCl4]2−+4H2O+2H3O+
  

(X)

Pink

    Excess      

(Z)

Blue

If M is Co then X:[Co(H2O)6]Cl2
Y:[Co(NH3)6]Cl3
Z:[CoCl4]2−
⇒ µx=3.87 due to presence of three unpaired electron in X
Co2+;3d7
H2O is weak field ligand hence no pairing
Hybridization state is sp3d2
⇒ µz=3.87BM due to presence of three unpaired electron in Z.
Co2+;3d7
Cl− is weak a field ligand and coordination number is four hence hybridization state is sp3
⇒[M(NH3)6]Cl3+3AgNO3→[M(NH3)6]3++3NO3−+3AgCl(s)
⇒[Co(H2O)6]2+4Cl−→[CoCl4]2++6H2O;ΔH>0
    Pink      Blue
(As octahedral complex changes to tetrahedral complex with H2O ligands replaced by Cl−)
At 0°C, equilibrium shifts in reverse direction hence colour of the solution is pink due to [Co(H2O)6]2+