Question

the first IE of S is lower than that of P, but the second IE of S is higher than P. explain

Answer 1

Sulphur(S) = 16

S = 1s²2s²2p6 3s²3p4

First IE = 1s²2s²2p6 3s²3p³

Second IE = 1s²2s²2p6 3s²3p²

Phosphorous (P) = 15

P = 1s²2s²2p6 3s² 3p3

First IE = 1s²2s²2p6 3s² 3p²

Second IE = 1s²2s²2p6 3s² 3p

We should remember, ionization enthalpy means the energy required to remove electrons

So when we arrange them in orbitals for first IE we see

1. In phosphorus the electrons have half filled 3p3 subshell however sulphur has 3p4 configuration so sulphur has one paired electron and 2 unpaired electrons so due to repulsion between paired electrons in sulphur, the electrons get easily removed from sulphur, so sulphur has low first ionization enthalpy than phosphorus

Then for second IE

1. In sulphur in first IE it has half filled p orbital while in phosphorus it has 3p² subshell in first IE so it is easy to remove electrons from phosphorous while we know its difficult to remove electrons from full filled and half filled orbitals. So in second IE sulphur has more IE than phosphorous