# Question #26b0e

Apr 7, 2015

The first thing you need to do when writing electron configurations is to determine exactly how many electrons your element or ion has.

Tin is located in period 5, group 14 of the periodic table, and has an atomic number equal to 50. That means that a neutral tin atom will have 50 electrons surrounding its nucleus.

Add your electrons starting from the lowest energy orbital, the 1s-orbital, and move on up; remember to fill all orbitals located on the same energy level before moving up to a higher energy level.

You'd get

$1 {s}^{\textcolor{red}{2}}$ $\to$ 2 electrons used, 48 to go;
$2 {s}^{\textcolor{red}{2}}$ $\to$ another 2 electrons used, 46 to go;
$2 {p}^{\textcolor{red}{6}}$ $\to$ another 6 electrons used, 40 to go;
$3 {s}^{\textcolor{red}{2}}$ $\to$ another 2 electrons used, 38 to go;
$3 {p}^{\textcolor{red}{6}}$ $\to$ another 6 electrons used, 32 to go;
$4 {s}^{\textcolor{red}{2}}$ $\to$ another 2 electrons used, 30 to go;

Moving on up

$3 {d}^{\textcolor{red}{10}}$ $\to$ another 10 electrons used, 20 to go;
$4 {p}^{\textcolor{red}{6}}$ $\to$ another 6 electrons used, 14 to go;
$5 {s}^{\textcolor{red}{2}}$ $\to$ another 2 electrons used, 12 to go;
$4 {d}^{\textcolor{red}{10}}$ $\to$ another 10 electrons used, 2 to go;
$5 {p}^{\textcolor{red}{2}}$ $\to$ the last 2 electrons used, 0 to go.

And there you have it - the electron configuration for tin is

$\text{Sn} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{10} 4 {p}^{6} 5 {s}^{2} 4 {d}^{10} 5 {p}^{2}$

As practice, you can try and name the element that has the electron configuration shown in the diagram I've used - notice that its configuration ends at $4 s$, so count all the electrons used to get to that orbital and look up a matching atomic number (it's a neutral atom, so this method works).