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NEET CHEMISTRYHard

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 (nn); higher nn leads to a higher magnetic moment (μ=n(n+2)\mu = \sqrt{n(n+2)} B.M.). We analyze the oxidation state, coordination number, and ligand field strength for each central metal ion .

  1. [Fe(en)(bpy)(NH3)2]2+[Fe(en)(bpy)(NH_3)_2]^{2+}: FeFe is in the +2+2 state (3d63d^6). Ligands en,bpy,NH3en, bpy, NH_3 are strong field nitrogen donors . They cause pairing of electrons. Configuration becomes t2g6eg0t_{2g}^6 e_g^0. Unpaired electrons = 0 (Diamagnetic).
  2. [V(gly)2(OH)2(NH3)2]+[V(gly)_2(OH)_2(NH_3)_2]^+: Assuming standard anionic ligands (glygly^-, OHOH^-), the oxidation state is x+2(1)+2(1)+0=+1x=+5x + 2(-1) + 2(-1) + 0 = +1 \Rightarrow x = +5. V5+V^{5+} (3d03d^0) has no d-electrons. Unpaired electrons = 0 (Diamagnetic).
  3. [Ti(NH3)6]3+[Ti(NH_3)_6]^{3+}: TiTi is in the +3+3 state (3d13d^1). With one d-electron, pairing is not relevant. Unpaired electrons = 1.
  4. [Co(ox)2(OH)2][Co(ox)_2(OH)_2]^-: Assuming CoCo 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 ox2ox^{2-} (oxalate) and OHOH^- are oxygen donors and act as weak field ligands . For Co3+Co^{3+} (3d63d^6), weak field ligands result in a high spin complex (t2g4eg2t_{2g}^4 e_g^2). 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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