The one that can exhibit highest paramagnetic behaviour among the following is: gly=glycinato; bpy=2, 2'−bipyridine
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a
Pdgly2
b
TiNH363+
c
CoOx2OH2−
d
FeenBipyNH322+
answer is C.
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Detailed Solution
Co(OX)2(OH)2−OX is C2O42−So the charge on Co is +5Co+5 has 3d4 configuration with Δ0>p;d4 has t2g4arrangement is two unpaired e−s (3) In TiNH363+;Ti3+has d configuration, Number of unpaired electrons=1