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# What are the possible quantum number for the last (outermost) electron in the quantum number in Calcium (Z $=$ $20$)

Last updated date: 23rd Apr 2024
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Hint: First of all, know that we define four different quantum numbers. Each quantum number can be defined by the position or energy of the subshell. Like principal quantum numbers can be defined by the position of that element in the periodic table i.e. in which period or group.

So, as we know that Quantum numbers are basically – Principal quantum number, which is denoted by ‘n’, Angular momentum quantum number, which is denoted by ‘l’, Magnetic quantum number, denoted by ‘m’ and spin quantum number, that is denoted by the letter ‘s’.
So, now since we have basic knowledge about the quantum numbers let’s get started on finding the possible quantum number for the last (outermost) electron in the quantum number in Calcium
So, as we know that Calcium has an atomic number and its position in the periodic table is - ${4^{th}}$ period and the ${2^{nd}}$ group.
Since, the period number tells us about the ‘Principal quantum number’,
So, our $n = 4$
i.e. Principal quantum number is $4$
Now coming next to Angular momentum quantum number, as we know the periodic table is divided into blocks and the element lies in the ‘s-block’. And this gives us the value of the Angular momentum quantum number. i.e. it specifies subshell.
Therefore , $l = 0$, the s subshell
Now next we will calculate the Magnetic quantum number, this tells us about the orientation of the orbital in which the electron is revolving.
Since, for the ‘s subshell,’ we have single orbital,
So $m = 0$
Now about the spin quantum number , it has values $s = + \dfrac{1}{2}{\text{ and - }}\dfrac{1}{2}$
Therefore, our possible quantum number for the element calcium is:
$n = 4$ , $l = 0$, $m = 0$, $s = + \dfrac{1}{2}{\text{ and - }}\dfrac{1}{2}$

Note:
The spin quantum number tells us about the spin of the electron. By the convention or default we generally assign $+ \dfrac{1}{2}{\text{ }}$ value to the electron that occupies the empty orbital and then we assign ${\text{ - }}\dfrac{1}{2}$ value to the electron that occupies the half filled orbital.