Question

An element of the third period has the following ionization energies in KJ/mol: IE1 = 1012, IE2 = 1903, IE3 = 2910, IE4 = 4956, IE5 = 6278 and IE6 = 22,230. USing only this data determine which one is the element.

Answer 1

3rd period contain 8 elements = Na, Mg, Al, Si, P, S, Cl, Ar

Na = 11 = [Ne] 3s¹

Mg = 12 = [Ne] 3s²

Al = 13 = [Ne] 3s² 3p¹

Si = 14 = [Ne] 3s² 3p²

P = 15 = [Ne] 3s² 3p³

S = 16 = [Ne] 3s² 3p⁴

Cl = 17 = [Ne] 3s² 3p⁵

Ar = 18 = [Ne] 3s² 3p⁶

Given that Ionization energies are IE1 = 1012, IE2 = 1903, IE3 = 2910, IE4 = 4956, IE5 = 6278 and IE6 = 22,230

P = 15 = [Ne] 3s² 3p³

After removal of 5 electrons from Phosphorous , its electronic configuration is equal to inert gas configuration i.e. Neon.

Hence, removal of 6th electron requires a lot of energy i.e. IE6 = 22,230 kJ/mol.

Therefore,

the given element = Phosphorous