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

🧠 Most Unpaired Electrons = Most Paramagnetic

Compute μ (or just unpaired e⁻ count) for each complex and pick the largest.

🗺️ Evaluate Each

| Complex | Metal/d-config | Geometry/spin | Unpaired | |---|---|---|---| | $[\mathrm{Pd(gly)_2}]$ | Pd(II) d⁸ | Square planar (always for Pd²⁺) | 0 | | $[\mathrm{Fe(en)(bpy)(NH_3)_2}]^{2+}$ | Fe(II) d⁶ | Octahedral, mixed-field; en + bpy moderately strong, NH₃ moderate — HS d⁶ on borderline; usually HS for Fe(II) with no CN/CO | 4 | | $[\mathrm{Co(ox)_2(OH)_2}]^-$ ($\Delta_o > P$) | Co(III) d⁶ | Oct LS (given $\Delta_o > P$) | 0 | | $[\mathrm{Ti(NH_3)_6}]^{3+}$ | Ti(III) d¹ | Octahedral | 1 |

So Fe(II) HS d⁶ has 4 unpaired → highest paramagnetism.

🗺️ Why Fe(II) HS, Not LS?

en and bpy are moderately strong field, but the question's choice (and JEE convention) treats Fe(II) with this mixed environment as HS. With en+bpy alone, Fe(II) leans toward LS, but adding 2 NH₃ (which is intermediate-field) keeps it HS. Empirically, $[\mathrm{Fe(en)_2(NH_3)_2}]^{2+}$ analogues are HS.

🗺️ Compute μ for Fe(II) HS

$\mu = \sqrt{4 \times 6} = \sqrt{24} = 4.90 \text{ BM}$

⚡ Quick d⁶ Spin Checker

| d⁶ | Oct HS | Oct LS | Td (HS) | |---|---|---|---| | Unpaired | 4 | 0 | 4 | | μ (BM) | 4.90 | 0 | 4.90 |

⚠️ Pd(II) Is d⁸ Sq Pl, Always Diamagnetic

Like Pt(II), Pd(II) prefers square planar with d⁸ filling — all paired → $\mu = 0$. Never get this wrong.

$\boxed{\text{Answer: (2) } [\mathrm{Fe(en)(bpy)(NH_3)_2}]^{2+}}$