Question #483bb

1 Answer
Sep 1, 2017

Answer 3

Explanation:

The electronic configuration of Cl neutral atom is
1^(st) shell
s^(1)= 2e^(-)
s^(2)= 2e^(-)
p^(2)= 6e^(-)
s^(3)= 2e^(-)
p^(3)= 5e^(-)
sum up e^(-) to get 17 = atomic no. of chlorine

Now, since it is cation Cl^(2+)
it would have 15e^(-)

These e^(-) would be lost from p sub-shell (outermost shell) .

Thus , the p sub shell in chlorine cation will be 3 e^(-) in P sub-shell of 3^(rd) shell.
Hope this helps!