Question

How many valence electrons does `"Cr"` have?

Answer 1

The valence electrons of chromium include its `4s` and `3d` electrons, because they are close enough in energy that more than one electron can be used to bond.

Its electron configuration as an atom is `[Ar]3d^5 4s^1`, so it has `6` valence electrons.

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There most certainly cannot be `4` or `2` (that's just impossible... you can't disregard some `3d` electrons and not the rest and still account for all valence electrons).

For chromium, it is enough of a stabilization to maximize its total spin state by having all unpaired electrons in a `3d^5 4s^1` configuration instead of `3d^4 4s^2`:

`ul(uarr color(white)(darr))`
`-"7.46 eV"`

`ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))`
`" "" "" "" "" "-"10.75 eV"`

If other websites tell you that the electron configuration is otherwise, they're not correct, because all textbooks I've ever read in 4 years of college tell me `3d^5 4s^1`. Energy levels were acquired from this appendix.

There is clear evidence that chromium's orbitals need to be able to hold `12` electrons to make the bonds in this compound:

(You can count six bonds and one interaction per chromium atom.)

So it needs to use `6` orbitals to bond in this compound. Valence orbitals are for holding valence electrons, and valence electrons are used for bonding.

If chromium can bond like this, then it must always have the capacity to bond like this, so it wouldn't make sense if it could use all `6` valence electrons in one compound and not be able to in another.

Therefore, its valence electrons must include those in the `3d` orbitals, giving us `1+5 = bb(6)` total.