Write the probable values of #l# and #m_l# for the principal quantum number #n=3#. Why?

1 Answer
Mar 11, 2018

Answer:

Here's what I got.

Explanation:

The angular momentum quantum number, #l#, describes the energy subshell in which an electron is located in an atom.

The possible values that the angular momentum quantum number can take depend on the value of the principal quantum number, #n#, which describes the energy shell in which the electron is located.

The relationship between these two quantum numbers is given by

#l = {0, 1, ..., n-1}#

In your case, you have

#n = 3 -># the third energy shell

and so you can have

  • #l = 0 -># the #s# subshell
  • #l = 1 -># the #p# subshell
  • #l = 2 -># the #d# subshell

Basically, the energy shell in which the electron is located will determine the possible subshells that can hold the electron, i.e. the type of orbitals that can hold the electron.

Now, the magnetic quantum number, #m_l#, which describes the orientation* of the orbital that holds the electron.

The relationship between the angular momentum quantum number and the magnetic quantum number is given by

#m_l = {-l, - (l-1), ..., -1, 0, 1, ..., (l-1), l}#

In your case, you can have

  • #l = 0 => m_l = 0#
  • #l = 1 -> m_l = {-1, 0, 1}#
  • #l = 2 -> m_l = {-2, -1, 0, 1, 2}#

The number of values that the magnetic quantum number can take tells you the number of orbitals present in a given subshell.

  • #m_l = 0 -># one #s# orbital
  • #m_l = {-1, 0, 1} -># three #p# orbitals
  • #m_l = {-2, -1, 0, 1, 2} -> # five #d# orbitals

So, for example, you can have

#n =3, l =0, m_l= 0#

These quantum numbers describe an electron located in the third energy shell, in the #3s# subshell, in the #3s# orbital.

#n = 3, l =1, m_l = -1#

These quantum numbers describe an electron located in the third energy shell, in the #3p# subshell, in one of the three #3p# orbitals, let's say #3p_y#.