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).

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

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#

Ernest Z.
Nov 15, 2016

The bespeak is #"Sn .

Explanation:

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.

You would naturally predict the order come be

#"Pb

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

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.