Solution: Arrange Na, Mg And K In Order Of Increasing First Ionization Energies

This means that the elements with the lowest ionization energies would be in the bottom left-hand corner of the periodic table. The change in ionization energies is also bigger going down the periodic table (by change within a group) than going across the periodic table (by change within a period).

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So let”s start from the bottom of the periodic table:#Pb# is the element that is in the lowest period at 6 (and lowest group at 14) in the periodic table; it”s the smallest ionization energy.

The period above (5) has two of the elements: Sn and Te. Well, since ionization energy increases across a period, Sn will have a smaller ionization energy than Te.#Pb, Sn, Te#

Now, let”s go to the third period, where #S# and #Cl# are. Since #S# is before #Cl,# #S# has a lower ionization energy than #Cl#.#Pb, Sn, Te, S, Cl#

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Ernest Z.
Nov 15, 2016

The order is #”Sn .

Explanation:

You have learned that ionization energy increases from top to bottom and from left to right in the Periodic Table.

You probably saw a diagram something like this.

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Here”s the portion of the Periodic Table that includes the elements in this question.

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(Adapted from ZON PENA)

You would naturally predict the order to be

#”Pb

This is almost correct, but the correct order is #”Sn , as shown in the image below.

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Why is this so?

The electron configuration of #”Sn”# is #” 5s”^2 “4d”^10 “5p”^2#.

The electron configuration of #”Pb”# is #” 6s”^2 “4f”^14 “5d”^10 “6p”^2#.

The #”4f”# electrons in #”Pb”# are poor at shielding the outermost electrons.

Thus the outer electrons experience a greater effective nuclear charge, and it is more difficult to remove them.

Hence #”Pb”# has a higher ionization than #”Sn”#, and the correct order is #”Sn .

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I hope that your instructor told you about this phenomenon before asking you to make a prediction.

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