# How do you calculate number of valence electrons in an arsenic atom?

Apr 22, 2018

Ground state, neutral arsenic atoms have five valence electrons.

Refer to the explanation.

#### Explanation:

For the representative elements, groups 1,2, and 13-18 (IA-VIIIA), the number of valence electrons in their atoms can be determined from their group number. For groups 1 and 2, the group number is the number of valence electrons. For groups 13-18, the number of valence electrons is 10 minus the group number. For the A groups, the group number is the number of valence electrons.

Arsenic is in group 15, so its atoms have 5 valence electrons.

We can also determine the number of valence electrons of the representative elements by determining their ground state electron configurations. The highest energy (outermost) s and p sublevels, also called the valence shell, contain the valence electrons.

Arsenic has atomic number 33. This means arsenic atoms have 33 protons and, if neutral, 33 electrons. The electron configuration for a ground state, neutral arsenic atom is:

$\text{1s"^2"2s"^2"2p"^6"3s"^2"3p"^6"3d"^10"4s"^2"4p"^3}$ or ["Ar"]"3d"^10"4s"^2"4p"^3".

So we can see that the electron configuration for arsenic has 5 electrons in the valence shell of the 4th energy level.

Apr 22, 2018

And arsenic lies in the old $\text{Group V}$ or the new $\text{Group 15}$...and even if these numbers are NOT written on the copy of the Periodic Table you have in your possession, you can very soon write them on your copy, for instance on the copy you have been provided in an exam....
And so arsenic lies is the old $\text{Group V}$ and new $\text{Group 15}$ of the Periodic Table, and the neutral $\text{element}$ thus, like its congener, phosphorus, has $V = 5$, or $15 - 10 = 5 \cdot \text{valence electrons}$... Why do I go thru this rigmarole? Well, it automatically tells me a formula for a arsenic compound of $A {s}_{2} {O}_{5}$ is kosher, and so is $A {s}_{2} {O}_{3}$...i.e. $\text{arsenic(V) and arsenic(III)}$...the arsenic atom has UP to five valence electrons to contribute, and oxidation states of $A s \left(+ V\right)$ and $A s \left(+ I I I\right)$ should be reasonable.