QJEEM24Ch0482: JEE Main 2024 Chemistry - Class 11 JEEM24Ch0482 Solution: The four quantum numbers for the electron in the outer most…

The four quantum numbers for the outermost electron of potassium can be determined by first understanding the electronic configuration of potassium (atomic number 19). The electron configuration is as follows:

$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1$

The outermost electron resides in the 4s orbital.

The four quantum numbers are as follows:

Principal quantum number ( $n$ ): This indicates the energy level or shell number of the electron and for the 4s orbital, $n = 4$ .

Azimuthal quantum number ( $l$ ): This indicates the subshell or shape of the orbital. For an $s$ orbital, $l = 0$ .

Magnetic quantum number ( $m$ ): This indicates the orientation of the orbital in space. Since there is only one orientation for $s$ orbitals, $m = 0$ .

Spin quantum number ( $s$ ): This can be either $+\frac{1}{2}$ or $-\frac{1}{2}$ . Typically, the first electron in an orbital has a spin of $+\frac{1}{2}$ .

Thus, the four quantum numbers for the outermost electron of potassium are:

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

This matches with Option B:

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