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 .


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.


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

(Adapted from ZON PENA)

You would certainly naturally predict the order to be


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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