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A transition metal with 7 d electrons forms a complex ion with cyanide. I know that cyanide is a high field ligand, which means the electrons will pair before moving to the next energy level, but wouldn't there still be one unpaired electron? And wouldn't that make it paramagnetic? I was told that my prediction was wrong and that it was in fact diamagnetic, but I don't understand why.

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  • $\begingroup$ With an octahedral or tetrahedral field, you cannot have a ground-state electron configuration containing 7 electrons and all of them paired; therefore, it cannot be diamagnetic. Is it possible that the person to whom you were speaking meant your calculation of the valence electrons when s/he said your "prediction was wrong"? Is there a specific compound in question? $\endgroup$ – bobthechemist May 4 '15 at 0:59
  • $\begingroup$ It's a fake transition metal with 7 d electrons that forms a 3+ charge. The comment my evaluator left just said that my prediction was paramagnetic and that a correct prediction is not present. Was I incorrect in saying that there would be one unpaired electron? I'm not really sure what i did wrong $\endgroup$ – Sarah May 4 '15 at 2:14
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    $\begingroup$ I think the question is whether there are 7d electrons in the atom or in the ion. For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic. $\endgroup$ – Geoff Hutchison May 4 '15 at 2:53
  • $\begingroup$ It sounds like @GeoffHutchison has the right interpretation: a d7 metal, after forming a +3 charge, would have 4 electrons remaining and could be diamagnetic. If it is a hypothetical free ion, however, then the aufbau principle would still predict a paramagnetic species with four unpaired electrons (and one unfilled d-orbital). Something is misleading about your evaluator's question. $\endgroup$ – bobthechemist May 4 '15 at 14:05
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As I noted in my comment above, I think the question is whether there are 7d electrons in the atom or in the ion.

For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic.

This is all, of course, a guess given what little you posted.

Let's say I had a similar question in my class. Students might then assume this $\ce{M^{3+}}$ is octahedral. (Personally, I'd make it a bit more obvious in the question.)

Given that the ligand is high-field, you get an initial MO diagram like this:

d4-high spin

But... we're not done. This electron configuration is subject to a weak Jahn-Teller distortion. So you'd end up with an MO diagram like this:

enter image description here

That, of course, is diamagnetic.

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