Ionizarion enegry is the highest in

Options

A. `[Ne]3s^(1)`
B. `[Ne]3s^(2) 3p^(3)`
C. `[Ar]3d^(10) 4s^(2)4p^(3)`
D. `[Ne] 3s^(2)3p^(4)`

Answer

B

Explanation

Both `(2)` and `(3)` have exactly half-full `p` subshells, which are relatively difficult to ionize on account of the extra stability, but `(2)` has the highest `IE` on account of the lesser number of shells. The elements are
`1(Na), 2(P), 3(As)`, and `4(S)`
The correct order of ionization energies is
`Na(496kJ mol^(-1))lt As (947) lt S (1000) lt P (1012)`