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?

  • 2
    $\begingroup$ Now, you wrote, "since all the ligands have nitrogens as donor sites, there shouldn't be any pairing." Actually, bpy is usually considered a π-acceptor ligand, which makes it more likely that the complex is low-spin. However, it's IMO quite unfair to expect a student to guess whether a given Fe(II)-bpy complex is high- or low-spin, because they're often "on the edge", i.e. need to be experimentally determined. $\endgroup$ Nov 6, 2021 at 15:46
  • $\begingroup$ @orthocresol Okay, my teacher asked me to learn it and thus this confusion $\endgroup$
    – hansika
    Nov 6, 2021 at 18:16

1 Answer 1


When taken as a 'fact', it is wrong that iron(II) complexes hexacoordinated by nitrogen atoms will always be high-spin (no pairing). This is exemplified by the complexes of 1-propyl-1H-tetrazol-κN4 (ptz): both the high-spin and low-spin complexes of $\ce{[Fe(ptz)6]^2+}$ are known, isolated and characterised; I wrote about this in a previous rather long answer. In the complexes, six nitrogens are donating to an iron(II) centre.

It is also obvious that the question and its solution hinges on whether the specific iron complex is high or low spin, as both the palladium and titanium complexes can be easily ruled out. High-spin iron(II) complexes have four unpaired electrons as you correctly noticed while low-spin iron(II) complexes have two.

Unfortunately, as the correct solution goes against your teacher's rule of thumb and as examples for both spin types exist, I cannot really help you here. At best, you might be able to look at the bpy ligand, notice that it has a multitude of π-interaction capabilities and take that into account – although I myself wouldn't know whether the π interactions would be enough to change the spin state.

I hope you aren't presented with such a question in your exam.


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