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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 electronic 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?

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    $\begingroup$ Cr(II) and Mn(III) - their oxidizing and reducing properties? conceivably a duplicate. $\endgroup$
    – orthocresol
    Commented Jan 27, 2018 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
    Commented Jan 27, 2018 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. $\endgroup$
    – S R Maiti
    Commented Jan 28, 2018 at 8:59
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    $\begingroup$ i suspect its the high hydration enthalpy of Cr2+ and 3+ because stability of oxidation states is in aqueous medium $\endgroup$
    – Abhinav
    Commented Aug 9 at 5:42

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The instability of the Cr(I) oxidation state (Cr⁺) compared to Cr(II) or Cr(III) can be explained by considering several factors beyond just the stability of half-filled orbitals.

Electrochemical Considerations Ionization Energy: The first ionization energy of chromium removes the 4s electron, resulting in a Cr⁺ ion with a 3d⁵ configuration. While this 3d⁵ configuration is relatively stable due to the half-filled d-orbitals, the energy required to remove the next electron (to form Cr²⁺) is not significantly higher. As a result, the Cr²⁺ state is readily achievable.

Redox Potential: The standard reduction potential for Cr⁺ to Cr is negative, meaning that Cr⁺ is more likely to reduce back to metallic chromium rather than staying in the Cr⁺ state. This makes the Cr⁺ state less stable in aqueous solutions.

Bonding and Chemical Behavior in terms of chemical behavior, Cr⁺ is highly reducing and can easily be oxidized to Cr²⁺ or Cr³⁺, which are more stable and form stronger bonds with ligands.

inter-electronic Repulsion and Exchange Energy In Cr⁺, the electron configuration is 3d⁵. While this configuration is stabilized by exchange energy due to the presence of unpaired electrons, the stabilization is not sufficient to outweigh the other factors that favor higher oxidation states. In contrast, Cr²⁺ and Cr³⁺ have less inter-electronic repulsion and can achieve more stable configurations in the presence of ligands.

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  • $\begingroup$ Is this AI generated? $\endgroup$
    – Nilay Ghosh
    Commented Aug 10 at 14:49
  • $\begingroup$ well I took a reference from a website and took all the information and made chat GPT compress all the information $\endgroup$
    – riteshwar
    Commented Aug 10 at 14:59
  • $\begingroup$ took from both websites and chat gpt $\endgroup$
    – riteshwar
    Commented Aug 10 at 15:00
  • $\begingroup$ See: chemistry.meta.stackexchange.com/questions/5277/… $\endgroup$
    – Nilay Ghosh
    Commented Aug 10 at 17:09

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