Question

In an atom, the total number of electrons having quantum numbers $n = 4$, $\left| {{m_1}} \right| = 1$ and ${m_s} = - \dfrac{1}{2}$ are:

Answer 1

Hint: To answer this question, you must have knowledge of the quantum numbers and the range of their values. From the given value of principal quantum number i.e, $n = 4$, find the values of azimuthal quantum number ($l$). For each value of $l$, there are $(2l + 1)$ values of magnetic quantum number (${m_l}$) possible.

Complete step by step answer:
Electrons are present in atomic orbitals and atomic orbitals are precisely distinguished by the terms known as quantum numbers. There are four quantum numbers as described below:

- Principle quantum number ‘$n$’: It defines the shell and also determines the size and energy of the orbital. It is a positive integer with values of $n$= 1, 2, 3...... .

- Azimuthal Quantum number ‘$l$’: It identifies the subshell and determines the shape of the orbitals. For a given value of $n$, values of $l$ ranges from 0 to $(n - 1)$.

- Magnetic orbital quantum number ‘${m_l}$’: It gives information about the orientation of the orbital. For a given value of $l$, ${m_l}$ has $(2l + 1)$ values possible.

- Spin quantum number ‘${m_s}$’: It refers to the orientation of the spin of the electron present in an atomic orbital. ${m_s}$ can have two values: ${m_s} = + \dfrac{1}{2}$ and ${m_s} = - \dfrac{1}{2}$.
Now, we are given $n = 4$, $\left| {{m_1}} \right| = 1$ and ${m_s} = - \dfrac{1}{2}$.
For $n = 4$, $l$=$0{\text{ }}to{\text{ }}(n - 1) = 0{\text{ }}to{\text{ }}(4 - 1)$
Therefore, possible values of $l$ are 0, 1, 2 and 3.

Now, we know for each value of $l$, there are $(2l + 1)$ values of ${m_l}$, given by:
${m_l} = - l,{\text{ }} - (l - 1),{\text{ }} - (l - 2)...{\text{ }}0,1...{\text{ }}(l - 2),{\text{ }}(l - 1),{\text{ }}l{\text{ }}$
Therefore, for $l = 0,{\text{ }}{m_l} = 0$
For $l = 1,{\text{ }}{m_l} = - 1,{\text{ }}0,{\text{ }} + 1$
For $l = 2,{\text{ }}{m_l} = - 2,{\text{ }} - 1,{\text{ }}0,\; + 1,\; + 2$
For $l = 3,\;{m_l} = - 3,{\text{ }} - 2,{\text{ }} - 1,{\text{ }}0,{\text{ }} + 1,{\text{ }} + 2,{\text{ }} + 3$
Given, $\left| {{m_1}} \right| = 1$, which means ${m_l} = \pm 1$.
Hence form the above calculation, there are total 6 electrons which have ${m_l} = \pm 1$ and ${m_s} = - \dfrac{1}{2}$
Therefore, in an atom, the total number of electrons having quantum numbers $n = 4$, $\left| {{m_1}} \right| = 1$ and ${m_s} = - \dfrac{1}{2}$ are six.

Note: For each value of $l$, there is a corresponding subshell assigned and the total number of orbitals in each subshell are as follows:

Value of $l$	0	1	2	3
Subshell notation	$s$	$p$	$d$	$f$
Number of orbitals	1	3	5	7