Although both $\ce{Cr+}$ and $\ce{Mn^2+}$ both ions have $d^5$ configuration, still only $\ce{Mn^2+}$ is well-known while $\ce{Cr+}$ is not common. What is so special about Cr that it does not form $\ce{Cr+}$?
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$\begingroup$ Strongly related (possibly a dupe?): Cr(II) and Mn(III) - their oxidizing and reducing properties? $\endgroup$– orthocresolCommented Mar 9, 2017 at 18:30
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2$\begingroup$ @orthocresol Even if the answer is basically identical, the question is different. So: not a dupe. $\endgroup$– hBy2PyCommented Mar 9, 2017 at 19:53
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$\begingroup$ @hBy2Py I know that and to be honest, I did not think the questions were very different at all. :/ To me, this is a subset of the other. I did opt to leave it open, though. $\endgroup$– orthocresolCommented Mar 9, 2017 at 19:54
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$\begingroup$ @orthocresol $\ce{Mn^{2+}/Cr^{+}}$ have odd numbers of electrons; $\ce{Mn^{3+}/Cr^{2+}}$ have even numbers. That fact alone I would think would make for appreciably different behavior. $\endgroup$– hBy2PyCommented Mar 9, 2017 at 20:04
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1$\begingroup$ @hBy2Py The way I see it, the crux of the linked question is not about Cr(II), it is about "why can't Cr(II) get reduced to Cr(I) to attain a stable half-filled orbitals", or essentially "why can't I make Cr(I) even though it is d5". I guess you are seeing it more as "why can't I reduce Cr(II)", and fair enough, but I think we have to agree to disagree on this one. $\endgroup$– orthocresolCommented Mar 9, 2017 at 20:15
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1 Answer
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Even with the stabilization of the $d^5$ electron configuration, chromium is quite electropositive and the (second) ionization energy for removing an additional electron from $\ce{Cr^+}$ is rather low. It's outweighed by the extra electrostatic attraction or solvation energy available by having two charges on the chromium, therefore you are ordinarily going to go on to $\ce{Cr^{2+}}$.
Chromium can adopt a +1 oxidation state in certain complexes, such as $\ce{Cr(CN)_5NO^{3-}}$. See https://en.wikipedia.org/wiki/Chromium.
answered Mar 10, 2017 at 0:51