How would you arrange the following elements in order of increasing ionization energy: Te, Pb, Cl, S, Sn?

2 Answers
Oct 31, 2015

Answer:

Pb, Sn, Te, S, Cl

Explanation:

The ionization energy increases across a period but decreases down a group.
Chemwiki

This means that the elements with the lowest ionization energies would be in the bottom left-hand corner of the periodic table. The change in ionization energies is also bigger going down the periodic table (by change within a group) than going across the periodic table (by change within a period).

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

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

Now, let's go to the third period, where #S# and #Cl# are. Since #S# is before #Cl,# #S# has a lower ionization energy than #Cl#.
#Pb, Sn, Te, S, Cl#

Nov 15, 2016

Answer:

The order is #"Sn < Pb < Te < S < Cl"#.

Explanation:

You have learned that ionization energy increases from top to bottom and from left to right in the Periodic Table.

You probably saw a diagram something like this.

images.tutorcircle.com

Here's the portion of the Periodic Table that includes the elements in this question.

Periodic Table
(Adapted from ZON PENA)

You would naturally predict the order to be

#"Pb < Sn < Te < S< Cl"#

This is almost correct, but the correct order is #"Sn < Pb"#, as shown in the image below.

Ionization Energies
(From facebook)

Why is this so?

The electron configuration of #"Sn"# is #"[Kr] 5s"^2 "4d"^10 "5p"^2#.

The electron configuration of #"Pb"# is #"[Xe] 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 greater effective nuclear charge, and it is more difficult to remove them.

Hence #"Pb"# has a higher ionization than #"Sn"#, and the correct order is #"Sn < Pb < Te < S < Cl"#.

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