The one that can exhibit highest paramagnetic behaiour among the following is ($\ce{gly}$ is glycinato; $\ce{bpy}$ is 2,2ʹ-bipyridine):
(a) $\ce{[Pd(gly)2]}$
(b) $\ce{[Fe(en)(bpy)(NH3)2]^2+}$
(c) $\ce{[Co(ox)2(OH)2]-}~(\Delta_0 > P)$
(d) $\ce{[Ti(NH3)6]^3+}$
In option (c) $\ce{[Co(ox)2(OH)2]-}~(\Delta_0 > P)$ cobalt has valence shell configuration $\mathrm d^4.$ Since the pairing energy $P$ is less than $\Delta_0$, there should be pairing. So, the number of unpaired electrons here should be $2.$
According to me, iron in (b) $\ce{[Fe(en)(bpy)(NH3)2]^2+}$ also has valence shell configuration $\mathrm d^6.$ Since all the ligands have nitrogens as donor sites, there shouldn't be any pairing.
I was told when the element is of $\mathrm{3d}$ series and in $+2$ oxidation state, ligands having nitrogen as donor sites are considered as weak field ligands and only ligands like carbonyl and cyano are considered as strong field ligands, and thus the number of unpaired electrons should be $4$ in this case, resulting in the highest paramagnetic behavior for (b).
However, the answer says the opposite. Is the aforementioned fact wrong?