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

Jul 7, 2017

Well, ${V}^{V +}$ has no $d \text{-electrons.......}$

#### Explanation:

Atomic vanadium, $Z = 23$, has an electronic configuration of $\left[A r\right] 3 {d}^{3} 4 {s}^{2}$. The colours of transition metal ions are, to a first approximation, related to electronic transitions of the $d - \text{electrons}$.......

Given that ${V}^{V +}$ has NEITHER $\text{d-electrons}$ nor $\text{s-electrons}$, there are no valence electrons to give rise to a absorption in the visible region.

On the other hand, ${V}^{+ I I I}$, (and ${V}^{+ I I}$, and ${V}^{+ I I I}$, ${V}^{+ I V}$) are conceived to have $3 d$ electrons, and their electronic transitions gives rise to colour......

The illustration displays solution of $V \left(I V +\right)$, $V \left(I V +\right)$, $V \left(I I I +\right)$, and $V \left(I I +\right)$..............

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