# What are the possible quantum numbers for the last (outermost) electron in Ca ? (Z=20)

## What are the possible quantum numbers for the last (outermost) electron in Ca ? (Z=20)

Dec 19, 2017

$n = 4 , l = 0 , {m}_{l} = 0 , {m}_{s} + - \frac{1}{2}$

#### Explanation:

Start by looking for calcium in the Periodic Table. You'll find it in period 4, group 2.

Now, you know that the period in which an element is located gives you the principal quantum number, $n$, of its outermost electrons.

In this case, you have

$n = 4$

As you know, the Periodic Table can be organized in terms of blocks. The block in which an element is located tells you the energy subshell in which its outermost electrons are located.

Calcium is located in the $s$ block, which means that its outermost electrons are located in the $s$ subshell. The angular momentum quantum number, $l$, that describes the energy subshell in which an electron is located, can take the possible values

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

and so on. In your case, you have

$l = 0$

The magnetic quantum number, ${m}_{l}$, tells you the orientation of the orbital in which the electron is located. For the $s$ subshell, you have a single orbital, the $s$ orbital.

$l = 0 \implies {m}_{l} = 0$

Finally, the spin quantum number, ${m}_{s}$, which tells you the spin of the electron, can take the possible values

${m}_{s} = \left\{+ \frac{1}{2} , - \frac{1}{2}\right\}$

By convention, we assign a positive spin to an electron that occupies an empty orbital and a negative spin to an electron that occupies a full orbital.

In this case, the fact that calcium is located in group $2$ tells you that it has $2$ valence electrons. The first valence electron is added to the $4 s$ orbital when the orbital is empty, so we assign it a positive spin, or ${m}_{2} = + \frac{1}{2}$.

This implies that the second valence electron that is added to the $4 s$ orbital will have a negative spin, ${m}_{s} = - \frac{1}{2}$.

This means that you have

$n = 4 , l = 0 , {m}_{l} = 0 , {m}_{s} + - \frac{1}{2}$

The quantum number set describes an electron located in the fourth energy shell, in the $s$ subshell, in the $4 s$ orbital, that has spin-down.