This means that the elements with the shortest ionization energies would certainly be in the bottom left-hand corner of the regular table. The readjust in ionization energies is also bigger going under the regular table (by change within a group) 보다 going across the regular table (by change within a period).

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

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

Now, let"s walk to the third period, wherein #S# and also #Cl# are. Due to the fact that #S# is prior to #Cl,# #S# has a lower ionization energy than #Cl#.#Pb, Sn, Te, S, Cl#

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Ernest Z.
Nov 15, 2016

The bespeak is #"Sn .


You have actually learned the ionization energy increases from optimal to bottom and also from left to appropriate in the routine Table.

You probably saw a chart something favor this.


Here"s the section of the routine Table that has the elements in this question.

(Adapted native ZON PENA)

You would naturally predict the order come be


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

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Why is this so?

The electron configuration 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 experience a greater effective nuclear charge, and also it is more complicated to remove them.

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

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I hope the your instructor told you around this phenomenon prior to asking you to do a prediction.

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