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1) - 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.

[ Hint: Look which electron requires the highest energy to be removed and remember the full sub- shell configurations require a much highes ionization energy than those with some electrons in it.]

Explanation / Answer

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

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