Please solve This Question sirCoordinaton compound question Once asked in aipmt also
Tanvir Alam , 6 Years ago
Grade 12th pass
AIPMT> Please solve This Question sir Coordinato...
1 Answerstarun rooniy
Last Activity: 6 Years ago
Option (A)
Greater the number of unpaired electrons, larger is the paramagnetism.
[V(gly)2 (OH)2(NH3)2]+ V23 = [Ar] 4s2, 3d3 Oxidation state of V in [V(gly)2 (OH)2(NH3)2]+ is
x + (-l)x2 + (-l)x2 + (0)x2 = +1
x =+5
Vs+ =[Ar]3d° (No unpaired electron)
[Fe(en)(bpy)(NH3)2)2+ Fe26 =[Ar]4s2, 3d6
Oxidation state of Fe in [Fe(en) (bpy) (NH3)2]2+ is
x + (0) + (0) + (0) x 2 = +2
x=+2
Fe2+ = [Ar]3d6
But en, bpy and NH3 all are strong field ligands, so pairing occurs, thus no unpaired electrons. [Co(OX)2(OH)2]-
CO27 = [Ar]4s2, 3d7
Oxidation state of Co in[CO(OX)2(OH)2]- is
x + (- 2) x 2+ (-1) x 2 = -1
x-6=-1
X =+ 5
CO5+ = [Ar]3d4 (4 unpaired electrons)
[Ti(NH3)6]3+
Ti22 =[Ar]4s2, 3d2
Oxidation state of Ti in [Ti(NH3)6]3+ is +3, thus it contains 1 unpaired electron.
Hence [Co(OX)2(OH)2]- has highest paramagnetic behaviour.
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