Consider this set of ionization energies. IE1 = 578 kJ/mol, IE2 = 1820 kJ/mol, IE3 =...

Question

Question

Consider this set of ionization energies. IE1 = 578 kJ/mol, IE2 = 1820 kJ/mol, IE3 = 2750

kJ/mol, IE4 = 11,600 kJ/mol. To which third-period element do these ionization energies

belong?

Answer 1

Answer #1

Solution:

The values of four ionization energies indicates the ionization energies of Al.

Al has electronic configuration:

Al (13) = 1s2 2s2 2p6 3s2 3p1

Thus, the removal of first electron from 3p subshell will requires less energy ( IE1 = 578 kJ/mol).

The removal of second electron from 3s subshell required higher energy due to full filled 3s subshell ( IE2 = 1820 kj/mol).

The removal of fourth electron required large amount of energy due to inert configuration like Ne. ( IE4 = 11600 kj/mol

Answer 2

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