What is the electron configuration of Mo^3+?

1 Answer
Feb 11, 2016

Answer:

#"[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#