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

1 Answer
Truong-Son N.
Apr 29, 2016

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.

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