This indicates that the aspects with the lowest ionization energies would be in the bottom left-hand edge of the periodic table. The adjust in ionization energies is likewise bigger going down the routine table (by change within a group) than going throughout the regular table (by adjust within a period).

You are watching: Rank the following elements according to their ionization energy.

So let"s start from the bottom of the regular table:#Pb# is the facet that is in the lowest duration at 6 (and lowest team at 14) in the routine table; it"s the smallest ionization power.

The period over (5) has actually two of the elements: Sn and Te. Well, given that ionization power rises across a duration, Sn will certainly have a smaller ionization energy than Te.#Pb, Sn, Te#

Now, let"s go to the 3rd duration, wbelow #S# and #Cl# are. Due to the fact that #S# is prior to #Cl,# #S# has actually a reduced ionization energy than #Cl#.#Pb, Sn, Te, S, Cl#


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Erswarm Z.
Nov 15, 2016

The order is #"Sn .


Explanation:

You have learned that ionization power increases from height to bottom and also from left to ideal in the Periodic Table.

You probably witnessed a diagram somepoint choose this.

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Here"s the percent of the Periodic Table that includes the aspects in this question.

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(Adapted from ZON PENA)

You would certainly naturally predict the order to be

#"Pb

This is nearly correct, however the correct order is #"Sn , as shown in the image listed 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 higher effective nuclear charge, and it is even more tough to remove them.

Hence #"Pb"# has actually a greater ionization than #"Sn"#, and the correct order is #"Sn .

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


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