Question

In an atom, which has $2K,8L,18M$ and $2N$ electrons in the ground state. The total number of electrons having magnetic quantum number $m = 0$ is
A. $6$
B. $10$
C. $7$
D. $14$

Answer 1

We are given an electronic shell and thus we will find the atomic number of the element by finding the total number of electrons present in the element. Therefore we will write the electronic configuration of elements and thus find the atomic orbitals. Thus we can find the total number of electrons that have a magnetic quantum number equal to zero.

Complete Answer:
Since we are given with number of shells present in the element which are given as $2K,8L,18M$ and $2N$ and thus we can find the total number of electrons present in the element as:
$\Rightarrow 2 + 8 + 18 + 2 = 30$
Thus it is zinc having atomic number $30$ and its electronic configuration can be written as:
$\Rightarrow 1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},4{s^2},3{d^{10}}$
Now we know that each orbital contains one m orbital as it is a spin quantum number. Spin quantum number depicts the rotation of electrons inside the orbitals. Therefore the total number of orbitals in the above electronic configuration is seven. Since we know that each orbitals accounts for two electrons and thus seven orbitals accounts for:
$\Rightarrow 7 \times 2 = 14$
Therefore the number of electrons which have the quantum number $m = 0$ are fourteen electrons.

Therefore we can say that the correct option is option D.

Note:
It must be noted that the given configuration represents the number of electrons present in the given shell. It must be noted that each subshell contains an orbital which contains electrons having $m = 0$. The number of electrons available in each orbital having spin quantum number equal to zero is two. The electronic configuration is written according to the Auf-Bau principle.