Question

What is the maximum number of orbitals that can be identified with the following quantum numbers?
$n = 3,l = + 1,{m_l} = 0$
A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

Hint: In order to solve this question we need to understand the structure of atoms. So atoms are considered to be a combination of heavy nucleus and electrons revolving around the nucleus. So in atoms electrons are distributed in various energy levels or shells like K shell, L shell, M shell. This definition was given by Bohr but later when the hydrogen atoms was studies by using quantum mechanics, it had been found that inside the shell electrons are actually revolving in orbitals for example s orbitals denote that electrons are revolving around nucleus in spherical shell, p orbitals denote that electrons are revolving in dumbbell shape which is oriented around three axis.

Complete step by step answer:
Since the n value is given as, $n = 3$ that means electrons are in third shell that is M shell.We know $l$ is azimuthal quantum number and it is defined for each orbital that is,
For “s” orbital, $l = 0$
For “p” orbital, $l = 1$
For “d” orbital, $l = 2$
For “f” orbital, $l = 3$
Given the problem, $l = 1$ that means that the orbital is $p$. Now, we know p orbital is oriented along three axes, so to identify them we use magnetic quantum number ${m_l}$.Given ${m_l} = 0$ which signifies that orbital is ${p_z}$. So $n = 3,l = + 1,{m_l} = 0$ signifies orbital $3{p_z}$. So only $1$ orbital is placed.

So the correct option is A.

Note: It should be remembered that the defined value of azimuthal quantum number is being calculated using Schrodinger equation of hydrogen atom. Actually the wave function for hydrogen atoms is divided into three parts, one is the radial wave function, other is angular wavefunction, and the last one is spin wavefunction. So by solving the angular wave function we can get the value of azimuthal quantum number.