This method that the facets with the shortest ionization energies would certainly be in the bottom left-hand corner of the periodic table. The adjust in ionization energies is also bigger going down the periodic table (by adjust within a group) than going across the periodic table (by adjust within a period).

You are watching: Arrange these elements in order of decreasing first ionization energy:

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

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

Now, let"s go to the third period, wherein #S# and also #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 power increases from height to bottom and also from left to appropriate in the periodic Table.

You more than likely saw a chart something favor this.

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Here"s the part of the periodic Table that consists of the facets 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 exactly order is #"Sn , as displayed in the photo below.

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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"# electron in #"Pb"# are negative at shielding the outermost electrons.

Thus the outer electrons suffer a greater efficient nuclear charge, and it is more difficult to eliminate them.

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

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


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