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)

Paramagnetic behaviour depends on the number of unpaired electrons ($n$); higher $n$ leads to a higher magnetic moment ($\mu = \sqrt{n(n+2)}$ B.M.). We analyze the oxidation state, coordination number, and ligand field strength for each central metal ion .

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

Since the Cobalt complex has the highest number of unpaired electrons (4), it exhibits the highest paramagnetic behaviour .