JEE Advanced 2013 Paper 1

P = [FeF‌6]‌3− ox. no. of Fe = +3, configuration: - 3d‌5 4S‌D
As F‌− is weak ligand, pairing does not take place,
so it has 5 unpaired electron
Q = [V(H‌2O)‌6]‌2+ , ox. no. of V = +2, configuration: - 3d‌3 4S‌D
It has 3 unpaired electrons
R = [Fe(H‌2O)‌6]‌2+ , ox. no. of Fe = +2, configuration: - 3d‌6 4S‌D
As H‌2O is weak ligand, pairing does not take place, so it has 4 unpaired electron
⇒ order of spin only magnetic moment ⇒ Q < R < P