# Question ceb8f

Dec 17, 2015

Trick question!

#### Explanation:

The trick here is to realize that although this specific electron configuration is possible in theory, it doesn't actually take place in practice.

I'll show you how to find the element that could have this electron configuration first.

As you know, a neutral atom has equal numbers of protons in its nucleus and of electrons surrounding its nucleus. This means that you can find the element's atomic number, which of course tells you the number of protons it has in its nucleus, by counting how many electrons are included in its electron configuration.

In this case, the number of electrons used will be

${\overbrace{2}}^{\textcolor{red}{{1}^{\text{st" "energy shell")) + overbrace(2 + 6)^(color(blue)(2^"nd" "energy shell")) + overbrace(2 + 6 + 4)^(color(green)(3^"rd" "energy shell")) + overbrace(2)^(color(orange)(4^"th" "energy shell}}}} = 24$

A quick look in the periodic table will reveal that the element with the atomic number equal to $24$ is chromium, $\text{Cr}$.

Now, the reason I said that this is a trick question is because chromium's actual electron configuration is

$\text{Cr: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5} 4 {s}^{1}$

or, using the noble gas shorthand notation

"Cr: " ["Ar"] 3d^5 4s^1#

Now, why is this the case with chromium? This excellent answer covers exactly why that happens

socratic.org/questions/why-is-the-electron-configuration-of-chromium

You can also check out this answer