Problem 112 - Chemistry An Atoms First Approach

Question

In each of the following sets, which atom or ion has the smallest ionization energy? a. \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) b. \(\mathrm{K}, \mathrm{Mn}, \mathrm{Ga}\) c. \(\mathrm{N}, \mathrm{O}, \mathrm{F}\) d. \(S^{2-}, S, S^{2+}\) e. \(\mathrm{Cs}, \mathrm{Ge}, \mathrm{Ar}\)

Step-by-Step Solution

Answer

a. Ba b. K c. N d. \(S^{2-}\) e. Cs

Step 1: Set (a)

\(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) are all alkaline-earth metals in Group 2, and as we go down the group, atomic radius increases and electron shielding becomes more effective. Therefore, Ba has the smallest ionization energy. Answer: Ba

Step 2: Set (b)

\(\mathrm{K}, \mathrm{Mn}, \mathrm{Ga}\) are all in the same row of the periodic table, but they belong to different groups. As we move across the row from left to right, the atomic radius decreases but nuclear charge increases. Since K has the largest atomic radius and the lowest nuclear charge in this set, it has the smallest ionization energy. Answer: K

Step 3: Set (c)

\(\mathrm{N}, \mathrm{O}, \mathrm{F}\) belong to the same row of the periodic table, and as we move across the row from left to right, the atomic radius decreases and nuclear charge increases. However, for N, an electron is removed from a half-filled orbital, which has extra stability compared to a less than half-filled orbital in O, so ionization energy is lower for N. Answer: N

Step 4: Set (d)

\(S^{2-}, S, S^{2+}\) are all sulfur ions with different charges. As we increase the positive charge on sulfur, the atomic radius decreases and the nuclear charge experienced by the electrons increases, which leads to a higher ionization energy. Therefore, \(S^{2-}\) has the smallest ionization energy in this set. Answer: \(S^{2-}\)

Step 5: Set (e)

\(\mathrm{Cs}, \mathrm{Ge}, \mathrm{Ar}\) belong to different rows and groups of the periodic table, which makes it hard to directly compare them in terms of ionization energy. However, Cs is an alkali metal with a larger atomic radius and lower nuclear charge, making it more likely to have the smallest ionization energy in this set compared to Ge (a metalloid) and Ar (a noble gas with a full electron shell). Answer: Cs

Key Concepts

Understanding Atomic Radius

The atomic radius is a measure of the size of an atom's electron cloud. It is determined by the distance from the nucleus to the outer boundary of the surrounding cloud of electrons. As one might guess, a larger atomic radius means that an atom is larger, and this plays a crucial role in determining various properties of elements, including ionization energy.

How Does Atomic Radius Affect Ionization Energy?The ionization energy tends to decrease as the atomic radius increases because electrons farther from the nucleus are less strongly attracted to the center and are thus easier to remove. For instance, among the alkaline-earth metals (Group 2 elements) like Ca, Sr, and Ba, we would expect Ba to have the lowest ionization energy due to its largest atomic radius—precisely what we observe in the exercise.In summary:

• Larger atomic radius often leads to lower ionization energy.
• Elements down a group in the periodic table usually increase in size due to additional electron shells.
• This increase in size then typically causes a decrease in ionization energy.

Electron Shielding and its Effects

Electron shielding is a phenomenon where the inner shell electrons shield the outer shell electrons from the full charge of the nucleus. This is crucial because it affects the overall pull of the nucleus on the outermost electrons, influencing properties like ionization energy.

More Layers, More ShieldingAs we add layers of electrons (going down a group on the periodic table), the inner electrons shield the outer electrons more effectively. This increased shielding makes it easier for an atom to lose an electron, reducing ionization energy, which is reflected in the larger atomic radius of elements like Ba compared to Ca and Sr.Considerations include:

• More electron shells result in better shielding and thus, lower ionization energy.
• Within the same group, as electron shielding increases, the outermost electrons are less tightly held by the nucleus.

Electron shielding helps explain why Stends to have a higher ionization energy than St after losing two electrons, due to reduced shielding and a stronger nuclear pull on the remaining electrons.

Periodic Table Trends

The periodic table is organized in such a way that it reveals trends in the properties of elements, including ionization energy. These trends are due to atomic size, electron shielding, and the number of protons in the nucleus (nuclear charge).

Horizontal and Vertical TrendsWhen moving across a period (row), atomic size typically decreases, and nuclear charge increases, which tends to increase ionization energy—moving from Kto Mnto Gademonstrates this with Ga having a higher ionization energy than K. Vertically, as one moves down a group, increasing atomic size and shielding result in decreasing ionization energy.

Key patterns to remember:

• From left to right across a period, ionization energy generally increases due to increasing nuclear charge and decreasing atomic radius.
• From top to bottom of a group, ionization energy generally decreases because of increasing atomic size and electron shielding.

By understanding these periodic table trends, one can quickly estimate the relative ionization energies of elements based on their position on the table, just as with predicting that Cs will have a lower ionization energy compared to Ge and Ar.