The one that can exhibit the highest paramagn | vedclass

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(B) To determine the highest paramagnetic behaviour,we calculate the number of unpaired electrons $(n)$ for each complex:$A) [Pd(gly)_2]$: $Pd^{2+}$ i

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$[Ti(NH_3)_6]^{3+}$

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(B) To determine the highest paramagnetic behaviour,we calculate the number of unpaired electrons $(n)$ for each complex:$A) [Pd(gly)_2]$: $Pd^{2+}$ is a $4d^8$ metal ion. $Pd$ is a $4d$ series element,so it always forms low-spin square planar complexes. $n = 0$.$B) [Ti(NH_3)_6]^{3+}$: $Ti^{3+}$ is a $3d^1$ ion. $n = 1$.$C) [Co(OX)_2(OH)_2]^{-}$: $Co^{3+}$ is a $3d^6$ ion. Given $\Delta_0 > P$,it forms a low-spin complex. $t_{2g}^6 e_g^0$. $n = 0$.$D) [Fe(en)(bpy)(NH_3)_2]^{2+}$: $Fe^{2+}$ is a $3d^6$ ion. $en$ and $bpy$ are strong field ligands. This forms a low-spin complex. $t_{2g}^6 e_g^0$. $n = 0$.Wait,re-evaluating the question options. If we consider $[Fe(en)(bpy)(NH_3)_2]^{2+}$,$Fe^{2+}$ is $3d^6$. With strong field ligands,it is $t_{2g}^6$,$n=0$. Let's re-check $[Ti(NH_3)_6]^{3+}$. $Ti^{3+}$ is $3d^1$,$n=1$. Since $1 > 0$,$[Ti(NH_3)_6]^{3+}$ has the highest number of unpaired electrons among the choices provided.

The one that can exhibit the highest paramagnetic behaviour among the following is:-
$gly = \text{glycinato}$; $bpy = 2,2'-\text{bipyridine}$

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The one that can exhibit the highest paramagnetic behaviour among the following is:-$gly = \text{glycinato}$; $bpy = 2,2'-\text{bipyridine}$