# How many electrons in magnesium share the quantum numbers l=0 and m_l = 1 ?

Apr 24, 2016

Zero.

#### Explanation:

As you know, four quantum numbers are used to describe the position and spin of an electron in an atom. Your goal here is to identify how many electrons located in an atom of magnesium, $\text{Mg}$, share the quantum numbers

$l = 0 \text{ }$ and $\text{ } {m}_{l} = 1$

Start by writing the complete electron configuration for a neutral atom of magnesium. Magnesium is located in period 3, group 2 of the periodic table and has an atomic number equal to $12$.

This means that a neutral magnesium atom will have a total of $12$ electrons surrounding its nucleus. Its electron configuration would look like this

$\text{Mg: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2}$

Now, the angular momentum quantum number, $l$, tells you the subshell in which the electron resides. The possible values of $l$, which depend on the value of $n$, the principal quantum number, correspond to

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

In your case, the value $l = 0$ corresponds to the s-subshell.

Now, the magnetic quantum number, ${m}_{l}$, tells you the exact orbital in which the electron resides. As you can see, ${m}_{l}$ depends on the value of $l$

${m}_{l} = - l , - \left(l - 1\right) , \ldots , - 1 , 0 , 1 , \ldots , \left(l - 1\right) , l$

The only possible value for ${m}_{l}$ that would correspond to an electron located in an s-subshell, regardless of energy level, is

${m}_{l} = 0 \to$ corresponds to the s-orbital

You can thus say that no electrons share the quantum numbers $l = 0$ and ${m}_{l} = 1$, in a magnesium atom, or in any atom for that matter.

You will never find an atom in which an electron has the quantum number $l = 0$ and ${m}_{l} = 1$ because the only possible value for ${m}_{l}$ when $l = 0$ is ${m}_{l} = 0$.

$\textcolor{w h i t e}{a}$
ALTERNATIVE INTERPRETATION

Just in case you had to determine how many electrons share the quantum number $l = 0$ and how many share the quantum number ${m}_{l} = 1$, I'll write out a quick explanation here.

If you're looking for electrons that have $l = 0$, you're essentially looking for electrons located in the s-subshell. As you can see by looking at magnesium's electron configuration, the atom has three s-subshells filled with electrons

• $1 {s}^{2} \to$ two electrons located in the s-subshell of the first energy level
• $2 {s}^{2} \to$ two electrons located in the s-subshell of the second energy level
• $3 {s}^{2} \to$ two electrons located in the s-subshell of the third energy level

Therefore, a total of six electrons have $l = 0$ in a neutral magnesium atom.

Now, in a magnesium atom, ${m}_{l} = 1$ corresponds exclusively to electrons located in one of the three orbitals located in the p-subshell.

More specifically, you have

• ${m}_{l} = - 1 \to$ 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

A magnesium atom has the p-subshell of the second energy level completely filled with electrons, which means that the ${p}_{z}$ orbital holds two electrons.

Therefore, a total of two electrons have ${m}_{l} = 1$ in a neutral magnesium atom.