# Question #483bb

Sep 1, 2017

#### Explanation:

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

Now, since it is cation $C {l}^{2 +}$
it would have 15${e}^{-}$

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 ${3}^{r d}$ shell.
Hope this helps!