# Part B Arrange the following elements in order of decreasing …

Chemistry

##### Science
• Anatomy & Physiology

• Astronomy

• Astrophysics

• Biology

• Chemistry

• Earth Science

• Environmental Science

• Organic Chemistry

• Physics

• Algebra

• Calculus

• Geometry

• Prealgebra

• Precalculus

• Statistics

• Trigonometry

##### Humanities
• English Grammar

• U.S. History

• World History

##### … and beyond
• Socratic Meta

Topics

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

Tiffany Quach

Oct 31, 2015

Pb, Sn, Te, S, Cl

#### Explanation:

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

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:
$P b$ 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.
$P b , S n , T e$

Now, let’s go to the third period, where $S$ and $C l$ are. Since $S$ is before $C l ,$ $S$ has a lower ionization energy than $C l$.
$P b , S n , T e , S , C l$

Ernest Z.

Nov 15, 2016

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.

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

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.

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”#.

##### Related questions
• How does ionization energy relate to reactivity?

• What is ionization energy measured in?

• What are the first and second ionization energies?

• How does ionization energy increase?

• How does ionization energy change down a group?

• How do trends in atomic radius relate to ionization energy?

• Question #44238

• Question #fbdfe

• Question #1a23b

• Question #d7de9

See all questions in Periodic Trends in Ionization Energy