Question

What is the electron configuration and oxidation state of vanadium in `"VBr"_3`?

Answer 1

The anion in `"VBr"_3` is `"Br"^(-)`, or bromide, contributing a total of a `"3-` charge to the neutral compound `"VBr"_3`.

To make this compound neutral, we must cancel out the `3-` charge with a `3+` charge. Naturally, `3 - 3 = 0`. So, the cation is `"V"^(3+)`.

Vanadium is atomic number `23`, so its electron configuration is `[Ar] 3d^3 4s^2`.

Thus, `"V"^(3+)` has an electron configuration of `color(blue)([Ar]3d^2)`, having removed two `4s` electrons and THEN one `3d` electron, due to the `4s` being higher in energy than the `3d`.

Thus, vanadium has TWO valence electrons.