Chem problem please help me how to answer this?

Question

Give a possible set of values of the four quantum numbers for all the electrons in a carbon atom if it is in the ground state.

Answer 1

The four quantum numbers are n,L,m,s. (the "L" is usually a lower case L but could be confused with the number "1" so I am using a capital "L")

Possible values for these numbers are:

n = 1 to infinity (integers) and identifies the level the electron is in
L = 0 to (n-1) and identifies the orbital (s,p,d,f)
m = -L to +L (including 0) and identifies the orientation of the orbital
s = + 1/2 or - 1/2 and identifies the spin of the electron
(not to be confused with an s orbital)

carbon's electron configuration is: 1s2, 2s2, 2px2, 2py1, 2pz1
(this is using Hund's rule to spread out the electrons evenly in the p orbitals)

for the 2 electrons in the 1s orbital the four quantum numbers are:
n = 1 L = 0 m = 0 s = +1/2
n = 1 L = 0 m = 0 s = -1/2

for the 2 electrons in the 2s orbital the four quantum numbers are:
n = 2 L = 0 m = 0 s = +1/2
n = 2 L= 0 m = 0 s = -1/2

for the 2 electrons in the 2px orbital the four quantum numbers are:
n = 2 L = 1 m = -1 s = +1/2
n = 2 L= 1 m = -1 s = -1/2

for the 1 electron in the 2py orbital the four quantum numbers are:
n = 2 L= 1 m = 0 s = +1/2

for the 1 electron in the 2pz orbital the four quantum numbers are:
n = 2 L= 1 m = +1 s = +1/2

Answer 2

ohh wait sorry that's electron configuration numbers


1s2 2s2 2p2