Which of the following complexes exhibits the | vedclass

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(C) To determine the highest paramagnetic behaviour,we calculate the number of unpaired electrons in each complex:$1$. $[Ti(NH_3)_6]^{3+}$: $Ti$ is in

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$[V(gly)_2(OH)_2(NH_3)_2]^+$

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(C) To determine the highest paramagnetic behaviour,we calculate the number of unpaired electrons in each complex:$1$. $[Ti(NH_3)_6]^{3+}$: $Ti$ is in $+3$ oxidation state $(3d^1)$. Number of unpaired electrons $(n)$ = $1$.$2$. $[V(gly)_2(OH)_2(NH_3)_2]^+$: $V$ is in $+3$ oxidation state $(3d^2)$. Number of unpaired electrons $(n)$ = $2$.$3$. $[Fe(en)(bpy)(NH_3)_2]^{2+}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. Since $en$ and $bpy$ are strong field ligands,they cause pairing of electrons. However,even with pairing,$Fe^{2+}$ in an octahedral field typically results in $t_{2g}^6 e_g^0$ (low spin,$n=0$) or $t_{2g}^4 e_g^2$ (high spin,$n=4$). Given the ligands,it is a low-spin complex $(n=0)$.$4$. $[Co(ox)_2(OH)_2]^-$: $Co$ is in $+5$ oxidation state $(3d^4)$. This is highly unstable and unlikely. Re-evaluating the options based on standard coordination chemistry: $[Fe(en)(bpy)(NH_3)_2]^{2+}$ is a low-spin $d^6$ complex $(n=0)$. $[V(gly)_2(OH)_2(NH_3)_2]^+$ has $n=2$. $[Ti(NH_3)_6]^{3+}$ has $n=1$. The question likely contains a typo in the options or complex formulas. Based on typical competitive exam patterns for this specific question,the complex $[Fe(en)(bpy)(NH_3)_2]^{2+}$ is often intended to be high-spin or the comparison is between standard $d$-electron counts. Given the options,$[V(gly)_2(OH)_2(NH_3)_2]^+$ has the highest number of unpaired electrons $(n=2)$.

Which of the following complexes exhibits the highest paramagnetic behaviour? (where $gly = \text{glycine}$,$en = \text{ethylenediamine}$,and $bpy = \text{bipyridyl}$ moieties. At. nos. $Ti = 22, V = 23, Fe = 26, Co = 27$)

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