Question

Arrange the following in order of increasing radius and increasing ionization energy. a. \(\mathrm{N}^{+}, \mathrm{N}, \mathrm{N}^{-}\) b. \(\mathrm{Se}, \mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\) c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)

Short answer

a. Increasing radius: \(\mathrm{N}^{+} < \mathrm{N} < \mathrm{N}^{-}\) and increasing ionization energy: \(\mathrm{N}^{-} < \mathrm{N} < \mathrm{N}^{+}\) b. Increasing radius: \(\mathrm{Cl}^{+} < \mathrm{Cl} < \mathrm{Se} < \mathrm{Se}^{-}\) and increasing ionization energy: \(\mathrm{Se}^{-} < \mathrm{Se} < \mathrm{Cl} < \mathrm{Cl}^{+}\) c. Increasing radius: \(\mathrm{Sr}^{2+} < \mathrm{Rb}^{+} < \mathrm{Br}^{-}\) and increasing ionization energy: \(\mathrm{Br}^{-} < \mathrm{Rb}^{+} < \mathrm{Sr}^{2+}\)

Step-by-step solution

Comparative atomic radius

All of these species have the same number of electron shells because they are different forms of nitrogen. However, \(\mathrm{N}^{+}\) has one less electron than \(\mathrm{N}\), and \(\mathrm{N}^{-}\) has one more electron than \(\mathrm{N}\). Therefore, the effective nuclear charge is highest for \(\mathrm{N}^{+}\) and lowest for \(\mathrm{N}^{-}\), so the order of increasing radius is \(\mathrm{N}^{+} < \mathrm{N} < \mathrm{N}^{-}\).

Comparative ionization energy

Ionization energy is the energy required to remove an electron from an atom or ion. In this case, it is directly related to the effective nuclear charge. Thus, \(\mathrm{N}^{+}\) has the highest ionization energy, and \(\mathrm{N}^{-}\) has the lowest ionization energy. The order of increasing ionization energy is \(\mathrm{N}^{-} < \mathrm{N} < \mathrm{N}^{+}\). b. \(\mathrm{Se}, \mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\)

Comparative atomic radius

Comparing \(\mathrm{Se}\) and \(\mathrm{Se}^{-}\), we see that \(\mathrm{Se}^{-}\) has an extra electron and thus a lower effective nuclear charge. Similarly, \(\mathrm{Cl}^{+}\) has one less electron than \(\mathrm{Cl}\), leading to a higher effective nuclear charge. Comparing \(\mathrm{Se}\) and \(\mathrm{Cl}\), we see that \(\mathrm{Se}\) is located below \(\mathrm{Cl}\) in the periodic table, indicating a larger atomic radius. Therefore, the order of increasing radius is \(\mathrm{Cl}^{+} < \mathrm{Cl} < \mathrm{Se} < \mathrm{Se}^{-}\).

Comparative ionization energy

Using the same logic as in Step 1, we can arrange the species in order of increasing ionization energy as: \(\mathrm{Se}^{-} < \mathrm{Se} < \mathrm{Cl} < \mathrm{Cl}^{+}\). c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)

Comparative atomic radius

Comparing \(\mathrm{Br}^{-}\) and \(\mathrm{Rb}^{+}\), the \(\mathrm{Br}^{-}\) has an extra electron and same number of shells as \(\mathrm{Rb}^{+}\). Also, \(\mathrm{Sr}^{2+}\) has two more protons and the same number of electron shells as \(\mathrm{Rb}^{+}\), which leads to a higher effective nuclear charge and a smaller atomic radius. The order of increasing radius is \(\mathrm{Sr}^{2+} < \mathrm{Rb}^{+} < \mathrm{Br}^{-}\).

Comparative ionization energy

Similarly, using the same logic as in Step 1, we can arrange the species in order of increasing ionization energy as: \(\mathrm{Br}^{-} < \mathrm{Rb}^{+} < \mathrm{Sr}^{2+}\).

Key concepts

Atomic Radius

The atomic radius is a fundamental concept in chemistry, reflecting the size of an atom. Essentially, it is a measure of the distance from the nucleus of an atom to the outer boundary of its electron cloud. The atomic radius can vary significantly between elements and is influenced by several factors.
The trend of atomic radius can be observed across periods and groups on the periodic table.

• Across a Period: Generally, as you move from left to right across a period, the atomic radius decreases. This is because electrons are added to the same electron shell, but the effective nuclear charge increases, pulling the electrons closer to the nucleus.
• Down a Group: As you move down a group, the atomic radius increases. More electron shells are added, creating a greater distance from the nucleus to the outermost electron shell, despite the addition of more protons.

When thinking about ions, the concept of atomic radius helps us understand why cations (positively charged ions) are smaller than their neutral atoms, and anions (negatively charged ions) are larger. In cations, the loss of electrons reduces electron-electron repulsion and the effective nuclear charge draws the remaining electrons closer. In anions, the addition of electrons increases repulsion, resulting in a larger electron cloud.

Ionization Energy

Ionization energy refers to the amount of energy required to remove an electron from an atom or ion in its gaseous state. It is a critical indicator of how strongly an atom holds onto its electrons. Ionization energy trends are quite consistent across the periodic table.

• Across a Period: As you move from left to right across a period, ionization energy generally increases. This happens because the effective nuclear charge increases while atomic size decreases, making it harder to remove an electron.
• Down a Group: As you descend a group, ionization energy tends to decrease. Although the nucleus holds more protons, the increased distance due to additional electron shells and the shielding effect make it easier to remove an electron.

For ions, cations typically have higher ionization energies compared to their neutral atoms because the effective nuclear charge on the remaining electrons is increased. Conversely, anions have lower ionization energies because the additional electron causes increased electron-electron repulsion.

Effective Nuclear Charge

Effective nuclear charge (Z_{ ext{eff}}) is a concept used to describe the net positive charge experienced by an electron in a multi-electron atom. It is a crucial factor that influences the size of the electron cloud and the energy required to ionize an atom.
The effective nuclear charge can be calculated using the formula:\(Z_{ ext{eff}} = Z - S\)where:

• \(Z\) is the atomic number, representing the total number of protons in the nucleus.
• \(S\) is the shielding constant, accounting for the effects of electron repulsion in intervening shells.

A higher effective nuclear charge means that electrons are held more tightly to the nucleus, resulting in a smaller atomic radius and a higher ionization energy. The concept of effective nuclear charge helps explain why a \(Cl^{+}\) ion is smaller than a neutral \(Cl\) atom, and why \(N^{+}\) exhibits higher ionization energy compared to neutral \(N\). As the effective nuclear charge increases, it becomes more challenging to remove an electron, reflecting in increased ionization energies and a reduced atomic radius.