Question

Which of the following is a correct set of quantum numbers for outermost valence electron in a neutral atom in the ground state of Oxygen O 3,1,2 0 2,-1,1 0 3,0,0 0 2,1,-1

Answer 1

Answer:

2,1,-1

Explanation:

The electronic configuration of oxygen is $1s^22s^22p^4$

The valence electrons are in the second principle quantum number, which means that the value of n = 2.

For, n =2,

l can be 0 or 1 which corresponds to s and p orbital respectively.

If l = 1 , which is p orbitals and it exits in 3 degenerate orbitals which has values of m = -1 , 0 , +1

So, The possible set pf quantum number for the outermost valence electron of the oxygen atom is 2, 1, -1.