Question #21ca6

1 Answer
Dec 3, 2017

Answer:

Here's what I got.

Explanation:

For starters, you know that nitrogen's electronic configuration looks like this

#"N: " 1s^2 2s^2 2p^3#

Now, you know that the #p# subshell contains a total of #3# orbitals, each capable of holding a maximum of two electrons as stipulated by Pauli's Exclusion Principle.

As you can see, for a neutral atom of nitrogen, the #2p# subshell contains a total of #3# electrons. According to Hund's Rule, which states that for a given subshell, every orbital must be half-filled before any one of the orbitals present in said subshell can be completely filled.

This means that a neutral atom of nitrogen has #3# unpaired electrons, all #3# located in the #2p# subshell.

https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Electronic_Structure_of_Atoms_and_Molecules

In this case, the principal quantum number, which gives you the energy shell in which the electrons reside, is equal to #2# for all three electrons.

#n = 2 -># designates the second energy shell

The angular momentum quantum number, which gives you the identity of the subshell in which an electron is located, is equal to #1# for all three electrons.

#l = 1 -># designates the #p# subshell

The magnetic quantum number, which tells you the orientation of the orbital in which an electron is located, can take the following values for a #p# subshell

#m_l = (-1, 0, +1}#

Since all three orbitals located in the #2p# subshell contain an electron, you can assign a value to each electron present in this subshell.

Finally, the spin quantum number, which tells you the spin of the electron, can take one of two possible values

#m_s = {+1/2, - 1/2}#

By convention, the electrons added to an empty orbital are assigned a spin-up, so you can say that all three electrons will have

#m_s = +1/2#

This means that the complete quantum number sets for the three electrons are going to look like this

  • #n = 2, l=1, m_l = -1, m_s + 1/2#

  • #n = 2, l= 1, m_l = color(white)(-)0, m_s = +1/2#

  • #n=2, l = 1, m_l = +1, m_s = +1/2#

Assuming that you want the sum of all #12# quantum numbers, you will end up with

#sum (n, m, m_l, m_s) = 3 xx 2 + 3 xx 1 + [color(red)(cancel(color(black)(-1))) + 0 +(color(red)(cancel(color(black)(-1))))] + 3 xx 1/2#

#sum (n, m, m_l, m_s) = 21/2#