Question

What is the maximum number of electrons in an atom that can have the following set of quantum numbers? $n = 3,{m_l} = 2,{m_s} = + \dfrac{1}{2}$ ?

Answer 1

Hint : A shell with energy level $n$ will have a maximum of $2{n^2}$ electrons. And they will reside in $n - 1$ subshells each with $2n - 1$ orbitals. An orbital can have a maximum of two electrons each with different spin quantum numbers.

Complete Step By Step Answer:
We know that quantum numbers are a series of numbers that define the position and energy of an electron in an atom. Principal quantum number, azimuthal quantum number, magnetic quantum number, and spin quantum number are the four types of quantum numbers. And for an electron in an atom, it is uniquely described by these four numbers. That is, no two different electrons can have the same set of quantum numbers.
The principal quantum number represents the energy level or shell in which the electron is present. And the shell corresponding to the quantum number $n$ can occupy a maximum of $2{n^2}$ electrons. So, in the case of $n = 3$ , there are $2{(3)^2} = 18$ electrons in that shell. In other words, a maximum of $18$ electrons can take the value $n = 3$ . Now nothing about the angular momentum quantum number $l$ - which specifies the subshell in which the electron resides - is mentioned directly in the question. But we already know that the electrons in a shell with energy level $n$ can take the values $1,2,...,n - 1$ for their $l$ . That is, here the maximum value $l$ can take is 2. And we are concerned with the orbitals $3s,3p{\text{ and }}3d$ . The magnetic quantum number ${m_l}$ can take the values \left\\{ { - l, - (l - 1),..., - 1,0,1,...,l - 1,l} \right\\} and corresponds to each orientation of orbitals in each subshell. Now, since it is given that ${m_l} = 2$ , it represents the electrons in the $3d$ subshell and specifically in the orbital corresponding to ${m_l} = 2$ . So, there are a maximum of two electrons which can take the values $n = 3{\text{ and }}{m_l} = 2$ since each orbital can have only two electrons. Further the spin quantum number is also given, ${m_s} = + \dfrac{1}{2}$ . From Pauli’s exclusion principle, we know that the two electrons in the same orbital will have different spins. So, one will take the value ${m_s} = + \dfrac{1}{2}$ and the other will take ${m_s} = - \dfrac{1}{2}$ . Hence, we find the maximum number of electrons that can have the set $n = 3,{m_l} = 2,{m_s} = + \dfrac{1}{2}$ as a quantum number to be one.

Note :
Remember that in this question since $n = 3$ and ${m_l} = 2$ , $l$ should be equal to $2$ . Because $n = 3 \Rightarrow {l_{\max }} = 2$ and ${m_l} = 2 \Rightarrow {l_{\min }} = 2$ . And since the spin quantum number is also given, there will exist only one electron corresponding to the given set of quantum numbers.