where $gly =$ glycine, $en =$ ethylenediamine and $bpy =$ bipyridyl moities. (At. nos. $Ti = 22, V = 23,$$ Fe = 26, Co = 27$)
$\mathrm{Ti}^{3+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{1}$
no. of unpaired electron in $d$ orbital is one. Oxidation State of $V$ in complex $\left[\mathrm{V}(\mathrm{gly})_{2}(\mathrm{OH})_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$
$=x+2 \times 0+2 \times(-1)+2 \times 0=+1$
$\therefore x=+3$
$\mathrm{V}^{3+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2}$
no. of unpaired electron in $d$ orbital is two.
Oxidation State of $Fe$ in the given complex is $+2$
$\therefore \quad \mathrm{Fe}^{2+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$
no. of unpaired electron in $d$ orbital is four.
Oxidation State of $\mathrm{Co}$ in the given complex $\left[\mathrm{Co}(ox)_{2}(\mathrm{OH})_{2}\right]^{-}$
$=x+2 \times(-2)+2 \times(-1)=-1=x-4-2=-1$
$\therefore x=+5$
(not possible, common ox. no. of $\mathrm{Co}=+2,+3,+4$ ) $\mathrm{Co}^{5+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{4}$
It should be an inner orbital complex ($ d'sp$ hybridisation) containing only one unpaired electron. So the complex having highest paramagnetism would be the complex of iron containing four unpaired electrons.
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${{O}_{(g)}}+{{e}^{-}}=O_{(g)}^{-}\Delta {{H}^{o}}=-142\ kJ\,mo{{l}^{-1}}$
$O_{(g)}^{-}+{{e}^{-}}=O_{(g)}^{2-}\Delta {{H}^{o}}=844\ kJ\,mo{{l}^{-1}}$
(Round off to the NearestInteger)
[Use:Atomicmasses:$Na:23.0\,u \mathrm{O}: 16.0 \,\mathrm{u} \quad \mathrm{H}: 1.0 \,\mathrm{u}$, Density of $\mathrm{H}_{2} \mathrm{O}: 1.0 \,\mathrm{~g} \,\mathrm{~cm}^{-3}$ ]