# Question #51981

Sep 15, 2017

$n = 3 , l = 1 , {m}_{l} = - 1 , {m}_{s} = + \frac{1}{2}$

#### Explanation:

For starters, grab a Periodic Table and look for silicon, $\text{Si}$. You should find it in period 3, group 14.

Now, notice that silicon has an atomic number equal to $14$. This means that a neutral silicon atom has $14$ protons inside its nucleus and $14$ electrons surrounding its nucleus.

In your case, you need to find the set of quantum numbers that describe the $\text{13th}$ electron in a neutral silicon atom.

To do that, use the fact that the Periodic Table can be organized in blocks.

In your case, the last $2$ electrons added to a neutral silicon atom are located in the $p$ block.

This means that these electrons will have the angular momentum quantum number, $l$, equal to $1$.

$l = 1 \to$ describes an electron located in the $p$ block

The period in which silicon is located will give you the value of the principal quantum number, $n$. In your case, silicon is located in period $3$, so the principal quantum number will be equal to $3$.

$n = 3 \to$ describes an electron located on the third energy level

Now, in order to find the magnetic quantum number, ${m}_{l}$, which gives you the orientation of the orbital in which the electron is located, you need to use the fact that the $p$ subshell can hold a total of three $p$ orbitals.

$l = 1 \implies {m}_{l} = \left\{\begin{matrix}- 1 \to \text{the p"_xcolor(white)(.)"orbital" \\ color(white)(+)0 -> "the p"_zcolor(white)(.)"orbital" \\ +1 -> "the p"_ycolor(white)(.)"orbital}\end{matrix}\right.$

In your case, the $\text{13th}$ electron will be added to the first empty $3 p$ orbital, which, by convention, will have ${m}_{l} = - 1$.

${m}_{l} = - 1 \to$ describes and electron located in a ${p}_{x}$ orbital

Finally, the spin quantum number, ${m}_{s}$, which tells you the spin of the electron, is equal to

${m}_{s} = \left\{\begin{matrix}+ \frac{1}{2} \to \text{the electron is added to an empty orbital" \\ -1/2 -> "the electron is added to a half-filled orbital}\end{matrix}\right.$

Since the $\text{13th}$ is added to the first empty $3 {p}_{x}$ orbital, you can say that you have

${m}_{s} = + \frac{1}{2} \to$ describes and electron that has spin-up

Therefore, you can say that the quantum number set that describes the $\text{13th}$ electron in a neutral silicon atom is

$n = 3 , l = 1 , {m}_{l} = - 1 , {m}_{s} = + \frac{1}{2}$

This electron is located on the third energy level, in the $3 p$ subshell, in the $3 {p}_{x}$ orbital, and has spin-up