In the ground state, the outermost shell (n = 1) of helium (He) is filled with e

Question

In the ground state, the outermost shell (n = 1) of helium (He) is filled with electrons, as is the outermost shell (n = 2) of neon (Ne). The full outermost shells of these two elements distinguish them as the first two so-called "noble gases." Suppose that the spin quantum number ms had three possible values, rather than two. If that were the case, which elements would be (a) the first and (b) the second noble gases? Assume that the possible values for the other three quantum numbers are unchanged, and that the Pauli exclusion principle still applies.

1. Hydrogen
2. Helium
3. Lithium
4. Beryllium
5. Boron
6. Carbon
7. Nitrogen
8. Oxygen
9. Fluorine
10. Neon
11. Sodium
12. Magnesium
13. Aluminum
14. Silicon
15. Phosphorus
16. Sulfur
17. Chlorine
18. Argon

Why would this be, I assumed the first one would be Ar b/c it's possble values are 0,1,2. . .

Explanation / Answer

If spin quantum number had 3 possible values, each orbital can have 3 electrons. So

Hydrogen = 1s1

elium = 1s2

Lithium = 1s3 -> Outermost shell n=1 filled

Be = 1s3 2s1

B = 1s3 2s2

C = 1s3 2s3

N = 1s3 2s3 2px1 2py0 2pz0

O = 1s3 2s3 2px1 2py1 2pz0

F = 1s3 2s3 2px1 2py1 2pz1

Ne = 1s3 2s3 2px2 2py1 2pz1

Na = 1s3 2s3 2px2 2py2 2pz1

Mg = 1s3 2s3 2px2 2py2 2pz2

Al = 1s3 2s3 2px3 2py2 2pz2

Si = 1s3 2s3 2px3 2py3 2pz2

P = 1s3 2s3 2px3 2py3 2pz3 -> n=2 filled

S = 1s3 2s3 2px3 2py3 2pz3 3s1

...

So noble elements would have been Lithium and Phosphorous.

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