This means that the facets with the shortest ionization energies would certainly be in the bottom left-hand corner of the periodic table. The readjust in ionization energies is also bigger going down the periodic table (by adjust within a group) than going throughout the regular table (by change 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 aspect that is in the lowest duration at 6 (and lowest team at 14) in the regular table; it"s the the smallest ionization energy.

The period above (5) has two the the elements: Sn and Te. Well, due to the fact that ionization energy increases throughout a period, Sn will have actually a smaller sized ionization energy than Te.#Pb, Sn, Te#

Now, let"s walk to the third period, where #S# and #Cl# are. Because #S# is prior to #Cl,# #S# has actually a lower ionization energy than #Cl#.#Pb, Sn, Te, S, Cl#

Ernest Z.
Nov 15, 2016

The stimulate is #"Sn .

Explanation:

You have actually learned the ionization energy increases from height to bottom and from left to best in the regular Table.

You most likely saw a diagram something choose this.

Here"s the portion of the routine Table that consists of the elements in this question.

(Adapted from ZON PENA)

You would normally predict the order to be

#"Pb

This is almost correct, but the exactly order is #"Sn , as displayed in the photo below.

Why is this so?

The electron construction of #"Sn"# is #" 5s"^2 "4d"^10 "5p"^2#.

The electron construction 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 external electrons endure a greater effective nuclear charge, and also it is more daunting to eliminate them.

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

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