Question 4460e

Nov 6, 2016

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

Explanation:

I assume that you want to write out a quantum number set that describes the highest-energy electron located in a potassium atom, i.e. the element's valence electron.

For starters, keep in mind that we're looking for four quantum numbers, three used to describe the electron's position in the atom and one used to describe its spin.

Now, let's figure out the values of the four quantum numbers by using the Periodic Table of Elements. More specifically, let's use the fact that the Periodic Table is organized in blocks.

Potassium, $\text{K}$, is located in period 4, group 1, which means that you can find it in the s block, shown here in $\textcolor{\mathrm{da} r k g r e e n}{\text{darkgreen}}$.

You can thus say that potassium's highest-energy electron will have

1)" " n = 4 -> the value of the principal quantum number is given by the period in which the element is located.

In this case, potassium is located in period 4, so its valence electron will have $n = 4$.

2)" "l = 0 -> the value of the angular momentum quantum number is given by the block in which the element is located.

Elements located in the s block will have their highest-energy electron located in the s subshell. As you know, the angular momentum quantum number can take the values

• $l = 0 \to$ the s subshell
• $l = 1 \to$ the p subshell
• $l = 2 \to$ the d subshell
• $l = 3 \to$ the f subshell

Therefore, you can say that potassium's valence electron will have $l = 0$.

3)" "m_l = 0 -> the magnetic quantum number can be determined using the group in which the element is located.

In this case, the s subshell can only hold one orbital, the s-orbital. This means that the value of ${m}_{l}$, which is determined by the value of $l$, can only be ${m}_{l} = 0$.

4)# $\text{ } {m}_{s} = + \frac{1}{2} \to$ the spin quantum number is assigned by keeping in mind that each orbital can only hold two electrons, one having spin-up and one having spin-down.

In this case, the single electron located in the s-orbital is considered to have spin-up.

Therefore, you can say that the quantum number set that describes the highest-energy electron in an atom of potassium will be

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

This set describes an electron located in the fourth energy shell, in the s subshell, in the 4s orbital, that has spin-up.