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Asked by Prashant DIGHE | 29 Mar, 2020, 09:39: PM
Expert Answer
As we know greater is the number of unpaired electrons, larger is the paramagnetism.
A. [Co(ox)2(OH)2]-
O.S. of Co :
x +(-2)×2+(-1)×2=-1
x - 6 = -1
x = +5
Co+5 = [Ar] 3d4
It has 4 unpaired electrons
Similarly,
B. Ti(NH3)6]3+ =Ti3+ = [Ar] 3d1 only one unpaired electron.
C. [V(gly)2(OH)2(NH3)2]+ = V+5 =[Ar] 3d0 no unpaired electrons.
D [Fe(en)(bby)(NH3)2]2+ = Fe2+= [Ar]3d6
But as bby and NH3 are strong field electrons hence pairing occurs resulting no unpaired electrons.
Hence from above information it is clar that complex [Co(ox)2(OH)2]- has highest number of unpaird electrons hence it show highest paramagnetism.
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