Question

The oxidation state of $Cr$ in $Cr{O}_5$ is

• A. +4
• B. +2
• C. -2
• D. +6

Answer 1

The correct option is D

+6

Oxidation number of Cr in CrO₅ (Blue perchromate)

If we assume the oxidation state of $Cr$ in $Cr{O}_5$ to be n, then by the usual method we get n - 10 = 0 $⟹$ n = +10

This cannot be true as maximum O.N. of $Cr$ cannot be more than +6 as the $Cr$ atom has only five electrons in 3d orbitals and one electron in 4s orbital. This erroneous, exceptional value is due to the fact that we are not taking peroxide linkages into account. In fact, four oxygen atoms in $Cr{O}_5$ seem to be connected via peroxide linkages. The chemical structure of $Cr{O}_5$ is

As you can see, there is one 'normal' double bond. For the Oxygen atom connected via a double bond, let us assign the usual O.N = -2. For the peroxide linked Oxygen atoms, O.N = -1 each.
Therefore, the evaluation of O.N. of Cr should be made as follows

n + 1$\times$(-2) + 4$\times$(-1)=0

Or x - 2 - 4 = 0 $⟹$ x= + 6.