# What is the electron configuration of Mo^3+?

Feb 11, 2016

${\text{[Kr]" 4"d}}^{3}$

or

$1 {\text{s"^2 2"s"^2 2"p"^6 3"s"^2 3"p"^6 3"d"^10 4"s"^2 4"p"^6 4"d}}^{3}$

#### Explanation:

The electronic configuration of ground state Mo is

$1 {\text{s"^2 2"s"^2 2"p"^6 3"s"^2 3"p"^6 3"d"^10 4"s"^2 4"p"^6 4"d"^5 5"s}}^{1}$

or in condensed form,

${\text{[Kr]" 4"d"^5 5"s}}^{1}$

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

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

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

${\text{[Kr]" 4"d}}^{3}$