With which of the following electronic configuration of an atom has the lowest ionization enthalpy:

Options

A. `1 s^(2) 2 s^(2) 2 p^(6)`
B. `1 s^(2) 2 s^(2) 2 p^(5)`
C. `1 s^(2) 2 s^(2) 2p^(3)`
D. `1 s^(2) 2 s ^(2) 2 p^(6) 3 s^(1)`

Answer

D

Explanation

Since 3s- orbital is father than 2p- orbital from the nucleus therefore `Delta_(i)H_(1)` of element with electronic configuration ( d) is correct.