# What is a possible set of four quantum numbers (n,l ,ml ,ms ) for the highest-energy electron in Ga?

Nov 3, 2015

Here's what I got.

#### Explanation:

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

Gallium, $\text{Ga}$, is located in period 4, group 13 of the periodic table and has an atomic number equal to $31$. This means that a neutral gallium atom will have a total of $31$ electrons surrounding its nucleus.

The electron configuration of gallium looks like this - I'll use the noble gas shorthand notation

"Ga: " ["Ar"]3d^10 4s^2 4p^1

Now, you're interested in finding the possible sets of quantum numbers that describe the highest-energy electron that belongs to a gallium atom.

As you know, the quantum numbers are defined

So, the highest-energy electron found in gallium is located in a 4p-orbital, which means that right from the start you know that the value of its principal quantum number, $n$, will be $4$.

Now for the angular momentum quantum number, $l$, which describes the subshell in which the electron resides.

Notice that the fourth energy level has total of $4$ subshells, each corresponding to a different value of $l$

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

SInce your electron is located in the p-subshell, it follows that its $l$ value wil be $1$.

The magnetic quantum number, ${m}_{l}$, tells you exactly in which orbital you can expect to find the electron.

For the p-subshell, $l = 1$, the magnetic quantum number can take the values

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

Since the p-subshell only contains one electron, you can place it in the first available p-orbital, which is ${p}_{x}$, for which ${m}_{l} = - 1$.

Finally, the spin quantum number, ${m}_{s}$, can only take one of two possible values

• ${m}_{s} = - \frac{1}{2} \to$ a spin-down electron
• ${m}_{s} = + \frac{1}{2} \to$ a spin-up electron

Since the orbital only contains one electron, it follows that it could be either spin-up or spin-down, so you get two possible sets of quantum numbers

n=4 -> l=1 -> m_; = -1 -> m_2 = -1/2

A spin-down electron located in the $4 {p}_{x}$ orbital

n=4 -> l=1 -> m_; = -1 -> m_2 = +1/2

A spin-up electron located in the $4 {p}_{x}$ orbital