Question

An atomic orbital has n = 3. What are the possible values of l and ${m_l}$?

Answer 1

Hint: The value ‘n’ designates the principal shell in which the electron is entering, The value of ‘l’ can be calculated from (n - 1). And further, the value of ‘${m_l}$’ which is a magnetic quantum number can be from +l to -l.

Complete step by step answer:
-First, let us understand what is n, l and ${m_l}$values. The ‘n’ alphabet is used to describe the Principal quantum number. These designate the main shell in which the electron entered in an atom. Its value can be any integer with a positive value starting from 1. The ‘l’ alphabet is used to describe the Azimuthal quantum number. These describe the shape of the given orbital. The value of azimuthal quantum number is obtained from principal quantum number. Its value can be obtained as- ‘l’ = n - 1.
-The ‘${m_l}$’ alphabet is used to describe the magnetic quantum number. These describe the total number of orbitals in a sub shell and their orientation. Its value can be obtained from the azimuthal quantum number. The value can be from +l to -l. Now, we have n = 3. The 3 indicates the principal quantum number.
For n = 3
‘l’ = n - 1
l = 3 - 1
‘l’ = 2
So, l can be 0, 1, 2.
For l = 0.
${m_l}$= 0
For l = 1.
${m_l}$= +l to -l
So, ${m_l}$= +1, 0, -1.
So, the value of ${m_l}$can be any from +1 to -1.
For l = 2.
${m_l}$= +2, +1, 0, -1, -2.
So, the value of ${m_l}$ can be any from +2 to -2.

Note: For l, we have fixed values. If l = 0; then s orbital
l = 1 ; then p orbital
l = 2 ; then d orbital
l = 3 ; then f orbital
l = 4 ; then g orbitals
The azimuthal quantum number can also be known as orbital angular momentum quantum number.