Hydrogen and its compoundHard
Question
Amongst the following, the lowest degree of paramagnetism per kole of the compound at 298 K will be shown by :
Options
A.MnSO4.4H2O
B.CuSO4.5H2O
C.FeSO4.4H2O
D.NiSO4.6H2O
Solution
Mn2+ : [Ar]3d5, 4s0 n = 5
Cu2+ : [Ar]3d9, 4s0 n = 1
Fe2+ : [Ar]3d6, 4s0 n = 4
Ni2+ : [Ar]3d8, 4s0 n = 2
∵ Cu2+ has least number of unpaired electrons so it show minimum paramagnetic character.
Cu2+ : [Ar]3d9, 4s0 n = 1
Fe2+ : [Ar]3d6, 4s0 n = 4
Ni2+ : [Ar]3d8, 4s0 n = 2
∵ Cu2+ has least number of unpaired electrons so it show minimum paramagnetic character.
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