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?
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2$\begingroup$ Neither of those is the electronic configuration of $\ce{Cr+}$. You might be asking for what the configuration ends in. $\endgroup$– M.A.R.Commented Jan 29, 2016 at 9:21
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4$\begingroup$ @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. $\endgroup$– orthocresolCommented Jan 29, 2016 at 11:47
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1 Answer
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I assume we are talking about the free ions in the gas phase.
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.
In general, the electronic configurations of transition metals and their ions are a balance of delicate factors. It's not easy to rationalise the electronic configurations fully and there are several factors that will affect the configuration adopted, e.g. 1) nuclear charge and relative energy of 3d and 4s orbitals, 2) relative e-e repulsions in 3d and 4s orbitals, 3) exchange energy.
Some interesting discussion on the topic can be found here:
answered Jan 29, 2016 at 14:18
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1$\begingroup$ you can also check here physics.nist.gov/PhysRefData/ASD/levels_form.html 3d^5 is lower by 1.48 eV $\endgroup$– DavePhDCommented Jan 29, 2016 at 14:45