Question

The electron configuration of atoms of a noble gas would be:

1s22s22p6
1s22s22p3
1s22s2
1s22s22p4
none of the above

Answer 1

Noble gas must have completely filled orbitals, which can be 2, 10, 18, 36 etc.
Here, only first one has completely filled i.e., 10 electrons, so it is inert.

In short, Your Answer would be Option A

Hope this helps!

Answer 2

Correct option is
`1s^22s^22p^6`. It is electronic configuration of Neon gas.

Explanation:

Noble gases are the gases which does not react with other elements because their outermost orbit is completely filled with electrons and these gases does not need any electron to complete its octet.

Example of noble gases with atomic number and electronic configuration

1. helium:
   `1s^(2)`
2. neon :
   `1s^(2) 2s^22p^6`
3. argon:
   `1s^(2) 2s^22p^63s^23p^6`
4. krypton:
   `1s^(2) 2s^22p^63s^23p^63d^(10)4s^24p^6`
5. xenon :
   `1s^(2) 2s^22p^63s^23p^63d^(10)4s^24p^64d^(10)5s^25p^6`
6. radon:
   `1s^(2) 2s^22p^63s^23p^63d^(10)4s^24p^64d^(10)5s^25p^64f^(14)5d^(10)6s^26p^6` It is clear that electronic configuration of all Noble gases except He end with
   `p^6` therefore only option first represent electronic configuration of a noble gas (neon)