# What is the correct set of quantum numbers (n, l, ml, ms) for the highest energy electron in the ground state of tin, Sn?

Nov 17, 2015

Here's what I got.

#### Explanation:

Your starting point here will be tin's electron configuration.

Tin, $\text{Sn}$, is located in period 5, group 14 of the periodic table and has an atomic number equal to $50$. This tells you that a neutral tin atom will have a total of $50$ electrons surrounding its nucleus.

So, the electron configuration for tin looks like this - I'll use the noble gas shorthand notation

"Sn: " ["Kr"] 4d^10 color(red)(5)s^2 color(red)(5)p^2

Now, you're looking for the electron that's highest in energy. Notice that you have two electrons that match this profile, both located in the 5p-subshell.

As you know, four quantum numbers are used to describe the location and spin of an electron in an atom.

So, let's start with the first one, the principal quantum number, $n$, which tells you the energy level on which an electron resides.

In your case, the highest energy level is equal to $\textcolor{red}{5}$, which means that you have

$n = \textcolor{red}{5}$

Next, the angular momentum quantum number, $l$, which tells you the subshell in which you can find the electron. The angular momentum quantum number can take values that range from $0$ to $n - 1$. You know that you have

• $l = 0 \to$ the s-subshell
• $l = 1 \to$ the p-subshell
• $l = 2 \to$ the d-subshell
• $l = 3 \to$ the f-subshell

and so on. In your case, you're looking for the value of $l$ that describes the p-subshell, which means that you will pick

$l = 1$

Next, the magnetic quantum number, ${m}_{l}$, which tells you the exact orbital in which you can find the electron. Now, you need to be aware of Hund's rules for filling degenerate orbitals.

More specifically, you need to know that all degenerate orbitals must be half-filled before any of them can take in a second electron. Furthermore, all half-filled degenerate orbitals contain electrons of the same spin.

The p-subshell contains a total of three orbitals, given by the values of ${m}_{l}$

• ${m}_{l} = - 1 \to$ the $5 {p}_{x}$ orbital
• ${m}_{l} = \textcolor{w h i t e}{-} 0 \to$ the $5 {p}_{y}$ orbital
• ${m}_{l} = \textcolor{w h i t e}{-} 1 \to$ the $5 {p}_{z}$ orbital

Since tin's 5p-subshell contains two electrons, it follows that these electrons will occupy distinct 5p-orbitals. More specifically, you will have

${m}_{l} = - 1 \to$ one electron in the $5 {p}_{x}$ orbital
${m}_{l} = \textcolor{w h i t e}{-} 0 \to$ one electron in the $5 {p}_{y}$ orbital

Finally, the spin quantum number, ${m}_{s}$, which can only take two possible values, $\pm \frac{1}{2}$, is restricted by Hund's rules to have the same value for both electrons, ${m}_{s} = + \frac{1}{2}$.

Therefore, the quantum number sets for the two highest-energy electrons found in a tin atom are

$n = 5 , l = 1 , {m}_{l} = - 1 , {m}_{s} = + \frac{1}{2}$

This electron is located on the fifth energy level, in the p-subshell, in the $5 {p}_{x}$ orbital, and has spin-up.

$n = 5 , l = 1 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

This electron is located on the fifth energy level, in the p-subshell, in the $5 {p}_{y}$ orbital, and has spin-up.