What is the electron configuration of Mo^3+?

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What is the electron configuration of Mo^3+?

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Answer 1 · 2016-02-11T11:27:07

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

or

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

Explanation:

The electronic configuration of ground state Mo is

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

or in condensed form,

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

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

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

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

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

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What is the electron configuration of Mo^3+?

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