# The formula for calculating the Highest Oxidation State for a Transition metal can be given as follows: Number of Unpaired d-electrons + Two s-orbital electrons. This is true for all the transition metals except two. Which of the following two?

$\begin{array}{1 1} \text{Copper and Chromium} \\ \text{Copper and Scandium } \\ \text{Zinc and Scandium} \\ \text{Chromium and Zinc} \end{array}$

Answer: Copper and Chromium do not follow the general formula for transition metal oxidation states.
This is because copper has 9 d-electrons, which would produce 4 paired d-electrons and 1 unpaired d-electron.
Since copper is just 1 electron short of having a completely full d-orbital, it steals an electron from the s-orbital, allowing it to have 10 d-electrons.
Similarly, chromium has 4 d-electrons, only 1 short of having a half-filled d-orbital, so it steals an electron from the s-orbital, allowing chromium to have 5 d-electrons.
edited Jul 29, 2014