# What is the electronic configuration of Cr (+1) ion

I believe it should be $\ce{d^5}$ , but i found out that it is $\ce{d^4s^1}$. Is it so? I have not found any explanation for the following. So why the strange configuration?

• Neither of those is the electronic configuration of $\ce{Cr+}$. You might be asking for what the configuration ends in. – M.A.R. ಠ_ಠ Jan 29 '16 at 9:21
• @Ϻ.Λ.Ʀ. It's perfectly fine to talk of a $\mathrm{d^4s^1}$ configuration. Obviously it's not the absolutely technically correct description but the unwritten parts are implied. If we had to write [Ar] every time we wanted to discuss the configuration of a TM it would get very boring. – orthocresol Jan 29 '16 at 11:47

I checked my notes and it seems that the electronic configuration of $\ce{Cr+}$ should be $[\ce{Ar}]\mathrm{3d^5}$, the reference being a rather old book: Phillips & Williams' Inorganic Chemistry (1966). I tried to look in slightly newer textbooks, but I couldn't find any information on the monovalent ions.
I would not claim to be 100% certain on this, but it can certainly be rationalised by the increase in $Z_\mathrm{eff}$ in the $\ce{Cr+}$ ion as compared to neutral $\ce{Cr}$ (one less electron means less shielding and greater effective nuclear charge). That would serve to stabilise the 3d orbital more than it stabilises the 4s orbital, and therefore favour the $\mathrm{3d^5}$ configuration over the $\mathrm{3d^44s^1}$ configuration.