In a ground-state #"P"# atom in the gas phase, how many electrons have the quantum numbers #n = 3, l = 1, m_l = –1# ?

1 Answer

Answer:

One electron.

Explanation:

The key here is the electron configuration of a neutral atom of phosphorus, which looks like this

#"P: " 1s^2 2s^2 2p^6 3s^2 3p^3#

Now, the #p# subshell contains a total of #3# orbitals. According to Hund's Rule, every orbital present in a given subshell must be half filled before any one of the orbitals can be completely filled.

This means that all the three #p# orbitals present in the #p# subshell must contain one electron, i.e. be half-filled, before any one of these orbitals can contain two electrons, i.e. be completely filled.

Notice that phosphorus has #3# electrons in the #3p# subshell. This tells you that each #3p# orbital will hold a single electron.

http://www.chemprofessor.com/quantum.htm

Now, you know that the principal quantum number, #n#, tells you the energy shell in which an electron is located inside an atom.

In your case, you have

#n=3 -># the third energy shell

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

You know that you have

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

and so on. In your case, you're dealing with the #p# subshell.

Finally, the magnetic quantum number, #m_l#, tells you the orientation of the orbital in which the electron is located. For the #p# subshell, you have

#l = 1 implies m_l = {-1, 0 , +1}#

In your case, you have

#m_l = -1#

which refers to one of the three orbitals present in the #3p# subshell. Since all three #3p# orbitals contain a single electron, you can say that the incomplete quantum number set

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

describes one electron present in the third energy shell, in the #3p# subshell, in one of the three #3p# orbitals--let's say #3p_y#.