# One of the outermost electrons in a strontium atom in the ground state can be described by what sets of four quantum numbers?

Nov 21, 2015

Here's what I got.

#### Explanation:

Your starting point for this problem will be the electron configuration for a neutral strontium atom in its ground state, which looks like this

$\text{Sr: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} \textcolor{red}{5} {s}^{2}$

As you can see, strontium's outermost electrons are located on the fifth energy level, in the 5s-orbital.

In order to completely describe these two electrons, you need to list the values of the four quantum numbers used to characterize their location and spin.

So, you know that the two electrons are located on the fifth energy level, which means that the principal quantum number, $n$, will be equal to $\textcolor{red}{5}$.

The angular momentum quantum number, $l$, which describes the subshell in which the electrons reside, will take the value specific for the s-subshell, i.e. $l$ will be equal to $0$.

For the s-subshell, the magnetic quantum number, ${m}_{l}$, can only take one possible value, and that is ${m}_{l} = 0$. Here ${m}_{l} = 0$ describes the s-orbital found in the s-subshell.

Finally, the spin quantum number, ${m}_{s}$, can take two possible values, ${m}_{2} = \pm \frac{1}{2}$.

Since two electrons occupy the 5s-orbital, one will have ${m}_{s} = \frac{1}{2}$, which describes spin-up, and the other ${m}_{s} = - \frac{1}{2}$, which describes spin-down.

Therefore, the quantum sets that describe the two outermost electrons electrons in a strontium atom will be

$n = 5 , l = 0 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

and

$n = 5 , l = 0 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$