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Melanie Shebel
Melanie Shebel
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We know that half-filled and fully filled-filled orbitals are highly stable. In ground state $\ce{Cr}$ has a $\ce{3d^5\! 4s^1}$ configuration. Therefore, the elctronicelectronic 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?

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?

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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Nilay Ghosh
Nilay Ghosh
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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 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.]

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.]

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?

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S R Maiti
S R Maiti
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Why is Cr+ unstable?

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.]