# Question #2d40e

Jan 17, 2017

${\text{6 e}}^{-}$

#### Explanation:

You are asked to find the number of electrons located in a sodium atom that can have the spin quantum number equal to $+ \frac{1}{2}$.

As you know, the spin quantum number, ${m}_{s}$, or simply $s$, denotes the spin of an electron inside an orbital and can take two possible values

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

Now, an orbital can hold a maximum of $2$ electrons of opposite spins, as stated by the Pauli Exclusion Principle. In other words, a fully-occupied orbital must hold an electron that has ${m}_{s} = + \frac{1}{2}$ and an electron that has ${m}_{s} = - \frac{1}{2}$.

This means that a starting point here would be to write the electron configuration for a neutral sodium atom and look for orbitals that are completely filled, since you know that

$\textcolor{red}{\underline{\textcolor{b l a c k}{{\text{1 fully-occupied orbital" -> "1 electron with m}}_{s} = + \frac{1}{2}}}}$

The electron configuration for a neutral sodium atom looks like this -- for neutral atoms, the atomic number, which gives you the number pf protons located in the nucleus, also gives you the number of electrons that surround the nucleus

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

Here you have $Z = 11$, which means that the electron configuration must account for a total of $11$ electrons

The electron configuration for a neutral sodium atom features

• $1 {s}^{2} \to$ the 1s orbital is completely filled
• $2 {s}^{2} \to$ the 2s orbital is completely filled
• $2 {p}^{6} = \left\{\begin{matrix}2 {p}_{x}^{2} \\ 2 {p}_{y}^{2} \\ 2 {p}_{z}^{2}\end{matrix}\right. \to$ all three 2p orbitals are completely filled
• $3 {s}^{1} \to$ the 3s orbital is only partially filled

So, you know that you have

${\text{5 fully-occupied orbitals" -> "5 electrons that have m}}_{s} = + \frac{1}{2}$

Now, the $3 s$ orbital is only partially filled, meaning that it contains $1$ electron. By convention, we designate ${m}_{s} = + \frac{1}{2}$ for electrons that occupy an empty orbital, so the number of electrons that meet the required criterion becomes

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{5 e"^(-) + "1 e"^(-) = "6 e}}^{-}}}}$

This is how the electron diagram for a neutral sodium atom would look like.

Here the electrons that have ${m}_{s} = + \frac{1}{2}$ are shown using up arrows, $\uparrow$, and the electrons that have ${m}_{s} = - \frac{1}{2}$ are shown using down arrows, $\downarrow$.

As you can see, if we follow the aforementioned convention, we find that $6$ electrons have spin-up in a neutral sodium atom.