Answer: Vanadium -23 has an electronic configuration of
[Ar]3d3 4s2 (e)
Explanation:
Vanadium has an atomic number of 23 . The total number of electrons in all the subshells will also be 23 .
For (a) [ Ar] has an 18 electrons already, adding 2 from 3s subshell 7 from 3p subshell makes it 27 electrons which is greater than 18. Also, it doesn't conform with the rule of electronic configuration.
(b) [ Ar] has an 18 electrons already, adding, adding 4 from 3s orbital and 5 from 3p orbital makes it 27 which is greater than 23.
(c) [ Ne] has an 10 electrons already, adding, 3 from 3s and 3 from 3p makes it 16. Which is less than 23
(d) [ Ne] has an 10 electrons already, adding, 4 from 3s and 1 from 3p makes it 14 . Which is less than 23.
(e)[ Ar] has an 18 electrons already, adding, 3 from 3d subshell and 2 from 4s subshell makes it 23.
Therefore [Ar]3d3 4s2 is the condesed electronic configuration of Vanadium.
