# What is the ground-state electron configuration of the element Kr?

Jun 19, 2017

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{10} 4 {p}^{6}$
OR
$\left[\text{Kr}\right]$
OR
$\left[\text{Ar}\right] 4 {s}^{2} 3 {d}^{10} 4 {p}^{6}$

#### Explanation:

First: Determine the number of electrons. Krypton has a total of 36 electrons

Know how many electrons each orbital can hold and their order.
(Refer to the following pictures as notes)

Third: Write out the electron configuration

For Krypton and most of the elements there are more the just one way (usually two) to write the electron configuration. One way is to write out the entire electron configuration by going through each orbital or we can use a shorthand notation using the noble gases as a starting point. I will go through both methods:

First Method: (Long way)

We know that Krypton has $36$ electrons.

We know that the $s , p , d$ orbital can hold a max of $2 , 6 , 10$ electrons respectively. Using this as a guide and going through the periodic table we find the electron configuration to be:

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

Second Method (Shorthand)

The key to using this method is to identify the noble gas closest to the desired element that is at a lower energy (Has a lower atomic number if I'm loosely speaking). In essence, the shorthand notation tells us the configuration by using a noble gas element as our starting point instead of starting all the way at the $1 s$ orbital. This method becomes really useful for elements that have a lot of electrons like Krypton because it becomes a hassle to write out such a long electron configuration.

Coincidently, Krypton itself is a noble gas so we could write the electron configuration as $\left[\text{Kr}\right]$. But we could also write out the configuration beginning using $\text{Ar}$ since it is the closest noble gas with the lower energy. From there we write out the remaining configuration like we did in the first method. The end result would then be: *$\left[\text{Ar}\right] 4 {s}^{2} 3 {d}^{10} 4 {p}^{6}$

*Notice how the configuration $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$ is the electron configuration for $\text{Ar}$ so by starting with $\text{Ar}$ we indicate that it is the configuration of everything that came before. In essence, $\text{Ar}$ is just a representation for $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$ without us having to write it all out.

All in all, the three given answers are correct ways of figuring out the ground-state electron configuration of Krypton .