I don't think anyone here will do all of these for you. Here are a couple with explanation.
MgSO4 is an ionic compound consisting of Mg^2+ + [SO4]^2-
12Mg neutral atom is 1s2 2s2 2p6 3s2 = 12 electrons
12Mg^2+ is 1s2 2s2 2p6 = 10 so there are no unpaired electrons for the 12Mg^2+ ion.
CuSO4.5H2O is Cu^2+ + [SO4]^2- + 5H2O
29Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1 and
29Cu^2+ is 1s2 2s2 2p6 3s2 3p6 3d9 4s0
So 4 H2O molecules pair with Cu^2+ and the fifth H2O molecule forms H bonds with the sulfate ions. O from two sulfate ions form coordinate bonds with Cu ion also to form an octahedral d2sp3 complex. So that leaves the lone electron in the 3d shell as the only unpaired electron for the Cu(II) ion.
Find the number of unpaired electrons on the central metal atom of following:
MgSO4
NiSO4.6H2O
CoCl2.6H2O
KCr(SO4)2.12H2O
CuSO4.5H2O
(NH4)2Cr2O7
FeSO4.7H2O
MnCl2.4H2O
FeCl3.6H2O
K3Mn(CN)6
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