Why are V^(V+) salts generally colourless?

1 Answer
Jul 7, 2017

Well, V^(V+) has no d"-electrons......."

Explanation:

Atomic vanadium, Z=23, has an electronic configuration of [Ar] 3d^3 4s^2. The colours of transition metal ions are, to a first approximation, related to electronic transitions of the d-"electrons".......

Given that V^(V+) has NEITHER "d-electrons" nor "s-electrons", there are no valence electrons to give rise to a absorption in the visible region.

On the other hand, V^(+III), (and V^(+II), and V^(+III), V^(+IV)) are conceived to have 3d electrons, and their electronic transitions gives rise to colour......

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The illustration displays solution of V(IV+), V(IV+), V(III+), and V(II+)..............

In fact vanadium has a very large redox manifold, and with many accessible oxidation states displays a rainbow of colours.........