Question

Which element has the noble gas notation, $\left[Xe\right]6s^2$?

Answer 1

To work backwards and find the element with a given configuration, we first need to find the atomic number of the noble gas mentioned in the notation.

$Xe$ is a noble gas with atomic number $54$. This means that the notation of $\left[Xe\right]$ represents a count of$54$ electrons.

The superscript on a subshell represents the number of electrons present in that subshell. So, $6s^2$ implies that our element has two electrons in the $6s$ subshell.

Therefore, the total number of electrons in the element $=54+2=56$

Number of electrons in an atom $=$ Number of protons $=$ Atomic number of the element

Therefore, the atomic number of the element with $56$ electrons is $56$, and the element with an atomic number of $56$ in the periodic table is barium.

Hence, barium $\left(Ba\right)$ has the noble gas notation of $\left[Xe\right]6s^2$.