Question

What is the electron configuration of Mo^3+?

Answer 1

`"[Kr]" 4"d"^3`

or

`1"s"^2 2"s"^2 2"p"^6 3"s"^2 3"p"^6 3"d"^10 4"s"^2 4"p"^6 4"d"^3`

Explanation:

The electronic configuration of ground state Mo is

`1"s"^2 2"s"^2 2"p"^6 3"s"^2 3"p"^6 3"d"^10 4"s"^2 4"p"^6 4"d"^5 5"s"^1`

or in condensed form,

`"[Kr]" 4"d"^5 5"s"^1`

Normally, when atoms get ionized, they lose the electron that has the highest energy level.

However, even though the `5"s"` orbital is lower in energy than the `4"d"` orbital, the electrons in the `4"d"` orbitals shield the electron in the `5"s"` orbitals from the nucleus' attraction. This means that it is easier for the electron in the `5"s"` orbital to leave.

So, the `5"s"` electron get ionized first. After the `5"s"` electron leave, the next two electrons to be ionized comes from the `4"d"` orbital. Therefore, the electronic configuration of `"Mo"^{3+}` is

`"[Kr]" 4"d"^3`