Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# An oxygen atom has a total of 8 elections. How do you write the 4 quantum numbers for each of the 8 electrons in the ground state ?

You know that a neutral oxygen atom has 8 electrons.

It's always helpful to write the for an element before trying to assign for each individual electron . As you can see in the image above, the for neutral oxygen is

"O": 1s^(2) 2s^(2) 2p^(4)

Assuming you know the basics behind , you can write the unique 4-number set for each electron belonging to oxygen like this

The first energy level:

"n=1", "l=0", "m"_l"=0", "m"_s"=-1/2" -> the electron located in the 1s-orbital, spin-down; "n=1", "l=0", "m"_l"=0", "m"_s"=+1/2" -> the electron located in the 1s-orbital, spin-up;

The second energy level:

"n=2", "l=0", "m"_l"=0", "m"_s"=-1/2" -> the electron lcoated in the 2s-orbital, spin-down; "n=2", "l=0", "m"_l"=0", "m"_s"=+1/2" -> the electron located in the 2s-orbital, spind-up;

"n=2", "l=1", "m"_l"=-1", "m"_s"=-1/2" -> the electron located in the "2p"_x"-orbital", spin-down; "n=2", "l=1", "m"_l"=-1", "m"_s"=+1/2" -> the electron located in the "2p"_x"-orbital", spin-up; "n=2", "l=1", "m"_l"=0", "m"_s"=+1/2" -> the electron located in the "2p"_y"-orbital", spin-up; "n=2", "l=1", "m"_l"=1", "m"_s"=+1/2" -> the electron located in the "2p"_z"-orbital", spin-up;