Question

Suppose that spin quantum number did not exist, such that only one electron could each orbital of a many-electron atom will have. Now the atomic number of the noble gas atoms in this case is:

• A. 2
• B. 10
• C. 9
• D. none

Answer 1

The correct option is C 9

The atomic number of first three noble gases are 2, 10 and 18 . If spin quantum number does not exist and there is only one electron in one orbital.
K orbit has 1 orbital i.e., only 1 electron completes the shell.
L orbit has 4 orbitals to have K, L completely filled 5 electrons are needed.
Successively 9 electrons are needed for next inert gas.
Therefore the atomic numbers will be 1, 5 and 9.