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

##### 1 Answer

Here's what I got.

#### Explanation:

*Boron*,

This means that a *neutral* boron atom will have a total of

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

#"B: " 1s^2 2s^2 2p^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*, **subshell** in which the electron can be found is given by the *angular momentum quantum number*,

Notice that

#l=0 -># the 2s-subshell#l=1 -># the 2p-subshell

The **actual orbital** in which the electron can be found is given by the *magnetic quantum number*,

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

#m_l = -1 -># the#p_x# orbital#m_l = 0 -># the#p_y# orbital#m_l = +1 -># the#p_z# orbital

Finally, the spin of the electron, which is given by the *spin quantum number*, can either by spin-up,

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

#n = 2, l=1, m_l = -1, m_s = +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 = -1/2#

and

#n=2, l=0, m_l=0, m_s = +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 = -1/2#

and

#n=1, l=0, m_l = 0, m_s = +1/2#