Question
Which of the following electronic configuration of an atom has the lowest ionisation enthalpy?
A.
$$1{s^2},2{s^2}2{p^5}$$
B.
$$1{s^2},2{s^2}2{p^3}$$
C.
$$1{s^2},2{s^2}2{p^5},3{s^1}$$
D.
$$1{s^2},2{s^2}2{p^6}$$
Answer :
$$1{s^2},2{s^2}2{p^5},3{s^1}$$
Solution :
The electronic configuration $$1{s^2},2{s^2}2{p^5},3{s^1}$$ shows lowest ionisation energy because this configuration is unstable due to the presence of one electron in $$s$$-orbital. Hence, less energy is required to remove the electron.