The one that can exhibit highest paramagnetic behaviour among the following is: gly=glycinato; bpy=2, 2'−bipyridine

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Pdgly2

TiNH363+

CoOx2OH2−

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

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