Question

A neutral atom of an element has $2K,8L{\text{ and }}5M$ electrons. What is the maximum number of electrons having the same spin?
(a) $5$
(b) $8$
(c) $9$
(d) $3$

Answer 1

The orientation of an electron in an orbital is decided by its spin axis. An electron can spin in any two directions, either clockwise (up) or anticlockwise (down). If up spin is taken as $+ \dfrac{1}{2}$, then down will be $- \dfrac{1}{2}$.

Complete step by step answer:
Spin of an electron is given by a fourth quantum number, denoted by ${m_s}{\text{ or }}s$. By Pauli’s exclusion principle, an orbital can have only two electrons, and those two must have opposite spin. That’s why the value of one spin is taken as $+ \dfrac{1}{2}$, and value for other is taken as $- \dfrac{1}{2}$. ${\text{K,L,M}}$ are the shells of an atom with $K$ being the first shell, and so on. By the formula of $2{n^2}$, where $n$ is the shell number, $K$ can have maximum $2{\left( 1 \right)^2} = 2$ electrons, $M$ can have maximum $2{\left( 2 \right)^2} = 8$ electrons, and $L$ can have maximum $2{\left( 3 \right)^2} = 18$ electrons. As Given in the question, an element has $2K,8L{\text{ and }}5M$ electrons means, this element will have total $2 + 8 + 5 = 15$ electrons, and since the number of electrons in a neutral atom is equal the atomic number, so, this element is Phosphorus with atomic number ${\text{15}}$. To calculate the spin of its electrons, its electronic configuration should be observed, which is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^3}$. Filling out the electrons in the respective shells according the electronic configuration:

$\uparrow \downarrow$		$\uparrow \downarrow$		$\uparrow \downarrow$	$\uparrow \downarrow$	$\uparrow \downarrow$		$\uparrow \downarrow$		$\uparrow$	$\uparrow$	$\uparrow$

$1{s^2}$ $2{s^2}$ $2{p^6}$ $3{s^2}$ $3{p^3}$
From the electronic configuration, it can be seen that $9$ electrons have the same up spin.

So, the correct answer is Option C .

Note:
It should be carefully observed that there are ${\text{6}}$ electrons also which have the same down spin but it was not there in the option, so it was not considered. Spin of electrons is also related to spin angular momentum, which is given by $S = \sqrt {s\left( {s + 1} \right)} \dfrac{h}{{2\pi }}{\text{, where }}h{\text{ is Planck's constant and }}s = \pm \dfrac{1}{2}$.