Which of the following is the correct electron configuration for the atom?

Question

If the last electron added to an atom can be described with quantum number as n = 3 and l = 2, which of the following is the correct electron configuration for the atom?

-[Ar]4s²3d²

-[Ar]5s²

-[Ne]3s²3p⁴

-[Ne]4s3d⁵

Answer 1

The principal quantum number n describes the size and energy of an orbital, whereas the angular momentum quantum number l describes its shape, so whether it is an s, p, d or f orbital.

l is always equal to n-1. If l=0 the orbital is s, equal to 1 is p and equal to 2 is d.

Based on this, we can immediately rule out B and C since they do not have any d electrons. Between A and D, the major difference is the noble gas used for the shorthand configuration. If you look at the periodic table you will see that the noble before the 4s period is Ar and not Ne, hence A is the right answer. To further confirm, the 3d² orbital has n=3 and l=2.