# Question 9de81

Apr 8, 2017

Both (1) $\text{S}$ and (5) $\text{K}$ have an electron with those quantum numbers in the ground state.

#### Explanation:

The quantum numbers

$n = 3$ is the principal quantum number.

The "3" tells you that the electron is in the third energy level.

$l = 1$ is the secondary quantum number.

$l$ can have any integer value from $0$ to $n \text{-1}$. Since $n = 3$, the allowed values of $l$ are $0 , 1 , 2$.

The value of $l$ tells you the shape of the orbital.

• $l = 0$ corresponds to an $\text{s}$ orbital.
• $l = 1$ corresponds to a $\text{p}$ orbital.
• $l = 2$ corresponds to a $\text{d}$ orbital.

The value $n = 3 , l = 1$ tells us that this is a "3p electron.

${m}_{l}$ is the magnetic quantum number.

${m}_{l}$ tells us the orientation of the orbital in a magnetic field.

${m}_{l}$ can tale any integer value from $+ l$ to $\text{-"l}$. The three allowed values of $- 1 , 0 , \text{and} + 1$ correspond to the three $\text{3p}$ orbitals.

$\text{s}$ is the spin quantum number. It tells us the relative direction of the spin of the electron on its axis.

$\text{s}$ can have only the values $\text{+½}$ or $\text{-½}$.

Name the element

We are looking for an element that has a $\text{3s}$ electron in its configuration.

From the Periodic Table. that could be any element with an atomic number greater than 12.

Two of the elements in your list have a $\text{3p}$ electron:

$\text{S}$ has the electron configuration ${\text{1s"^2 "2s"^2 "2p"^6 "3s"^2 "3p}}^{4}$.

$\text{K}$ has the electron configuration $\text{1s"^2 "2s"^2 "2p"^6 "3s"^2 "3p" ^6 "4s}$.

Both sulfur and potassium contain a "3p"# electron in the ground state.