Question #483bb

1 Answer
Sep 1, 2017

Answer:

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 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^(rd)# shell.
Hope this helps!