# What is the atomic number of an element whose atom has the following electron configuration: 1s2 2s2 2p6 3s2 3p6?

Nov 12, 2015

$Z = 18$

#### Explanation:

The electron configuration for a neutral atom must account for all the electrons that surround that atom's nucleus.

Moreover, a neutral atom will always the same number of protons in its nucleus as it does number of electrons surrounding its nucleus.

Hence, you can say that for a neutral atom, the atomic number, which tells you the number of protons in the nucleus, will also tell you how many electrons said atom has.

So, the electron configuration for this atom is

$\text{X: } \textcolor{b l u e}{1} {s}^{2} \textcolor{red}{2} {s}^{2} \textcolor{red}{2} {p}^{6} \textcolor{g r e e n}{3} {s}^{2} \textcolor{g r e e n}{3} {p}^{6}$

Notice that this atom has electrons spread out on three energy levels

• ${\textcolor{b l u e}{1}}^{\text{st" " energy level}} \to$ two electrons
• ${\textcolor{red}{2}}^{\text{nd" " energy level}} \to$ eight electrons
• ${\textcolor{g r e e n}{3}}^{\text{rd" " energy level}} \to$ eight electrons

The total number of electrons that can be found in this atom is

$2 + 8 + 8 = {\text{18 e}}^{-}$

Therefore, the atomic number of the atom will be

$\text{atomic number} = Z = 18$

A quick look in the periodic table will show that the element in question is argon, $\text{Ar}$.