# What are the quantum numbers of the five electrons of Boron?

Nov 11, 2015

Here's what I got.

#### Explanation:

Boron, $\text{B}$, is located in period 2, group 13 of the periodic table, and has an atomic number equal to $5$.

This means that a neutral boron atom will have a total of $5$ electrons surrounding its nucleus.

Now, your tool of choice here will be boron's electron configuration, which looks like this

$\text{B: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{1}$

Since you have five electrons, you will need five sets of quantum numbers.

So, to make things interesting, let's start removing electrons one by one from the boron atom and describing them as we go.

The first electron will come from the orbital that's highest in energy. In this case, this electron will come from a 2p-orbital.

Now, the energy level is given by the principal quantum number, $n$, which in this case is equal to $2$. The subshell in which the electron can be found is given by the angular momentum quantum number, $l$.

Notice that $l$ can take values from $0$ to $n - 1$. For an electron located on the second energy level, there are only two possible subshells in which to reside

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

The actual orbital in which the electron can be found is given by the magnetic quantum number, ${m}_{l}$.

For the p-subshell, you have a total number of three possible orbitals

• ${m}_{l} = - 1 \to$ the ${p}_{x}$ orbital
• ${m}_{l} = 0 \to$ the ${p}_{y}$ orbital
• ${m}_{l} = + 1 \to$ the ${p}_{z}$ orbital

Finally, the spin of the electron, which is given by the spin quantum number, can either by spin-up, ${m}_{s} = + \frac{1}{2}$, or spin-down, ${m}_{s} = - \frac{1}{2}$.

So, for the first electron, a valid set of quantum numbers will be

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

Now for the second and third electrons, which will come from a 2s-orbital. This time, you will have

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

and

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

Now you're down to the last two electrons, which reside on the first energy level, in the 1s-orbital

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

and

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