We know that half-filled and fully filled orbitals are highly stable. In ground state $\ce{Cr}$ has a $\ce{3d^5\! 4s^1}$ configuration. Therefore, the elctronic configuration of $\ce{Cr+}$ should have $\ce{3d^5\! 4s^0}$ state. The half-filled $\ce{3d}$ orbital should make the ion stable.

However, we don't see many $\ce{Cr(I)}$ compounds. There are one or two rare examples, but other than them, most compounds of $\ce{Cr}$ are $\ce{Cr(II)}$, $\ce{Cr(III)}$ and so on. Why does this happen? Why $\ce{Cr+}$ is unstable?

[If you find a reference, add it to your answer. As I am a student, it will help me immensely. But if you don't have a reference, don't hesitate to post an answer. Getting an answer is the priority.]

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    $\begingroup$ Cr(II) and Mn(III) - their oxidizing and reducing properties? conceivably a duplicate. $\endgroup$ – orthocresol Jan 27 '18 at 18:02
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    $\begingroup$ I'm afraid it might be one of these cases where there is no intuitive explanation and we're left with "because when we go through all the quantum mechanic computation we get such and such ionization and hydration energies". $\endgroup$ – Jan Rzymkowski Jan 27 '18 at 19:55
  • $\begingroup$ @JanRzymkowski , Yes, that might be true. But in most cases, there is often a simple and intuitive explanation. I put the question on the assumption that there is an explanation. I still think there can be a good explanation without the need to go into quantum mechanics. $\endgroup$ – Shoubhik Raj Maiti Jan 28 '18 at 8:59

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