Which of the following complexes has the highest paramagnetic behaviour? (where gly = glycine, en = ethylenediamine, ox = oxalate ion and bpy = bipyridyl moieties) (Atomic numbers: Ti = 22, V = 23, Fe = 26, Co = 27)
A
[V(gly)₂(OH)₂(NH₃)₂]⁺
B
[Fe(en)(bpy)(NH₃)₂]²⁺
C
[Co(ox)₂(OH)₂]⁻
D
[Ti(NH₃)₆]³⁺
Step-by-Step Solution
Paramagnetic behaviour depends on the number of unpaired electrons (n); higher n leads to a higher magnetic moment (μ=n(n+2) B.M.). We analyze the oxidation state, coordination number, and ligand field strength for each central metal ion .
[Fe(en)(bpy)(NH3)2]2+:Fe is in the +2 state (3d6). Ligands en,bpy,NH3 are strong field nitrogen donors . They cause pairing of electrons. Configuration becomes t2g6eg0. Unpaired electrons = 0 (Diamagnetic).
[V(gly)2(OH)2(NH3)2]+: Assuming standard anionic ligands (gly−, OH−), the oxidation state is x+2(−1)+2(−1)+0=+1⇒x=+5. V5+ (3d0) has no d-electrons. Unpaired electrons = 0 (Diamagnetic).
[Ti(NH3)6]3+:Ti is in the +3 state (3d1). With one d-electron, pairing is not relevant. Unpaired electrons = 1.
[Co(ox)2(OH)2]−: Assuming Co is in a typical oxidation state (likely +3 given the context of weak field dominance in this comparison, though the charge balance in the specific typo provided varies). Ligands ox2− (oxalate) and OH− are oxygen donors and act as weak field ligands . For Co3+ (3d6), weak field ligands result in a high spin complex (t2g4eg2). Unpaired electrons = 4.
Since the Cobalt complex has the highest number of unpaired electrons (4), it exhibits the highest paramagnetic behaviour .
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