Problem 34 - Chemistry: An Atoms-Focused Approach

Question

Place the following ions in order of increasing ionization energy: \(\mathrm{Na}^{+}, \mathrm{S}^{2-}, \mathrm{F}^{+},\) and \(\mathrm{Mg}^{+}\),

Step-by-Step Solution

Answer

Question: Arrange the following ions in order of increasing ionization energy: \(\mathrm{F}^{+}\), \(\mathrm{Na}^{+}\), \(\mathrm{Mg}^{+}\), and \(\mathrm{S}^{2-}\). Answer: The ions in order of increasing ionization energy are \(\mathrm{F}^{+}\), \(\mathrm{Na}^{+}\), \(\mathrm{Mg}^{+}\), and \(\mathrm{S}^{2-}\).

Step 1: Write down the atomic number and electron configuration for each ion

To determine the atomic number and electron configuration for each ion, we can refer to a periodic table or look up the information online. In the periodic table, the atomic number is the number of protons in the nucleus of an atom. - For \(\mathrm{Na}^{+}\), the atomic number is 11 (sodium) and its electron configuration is \(1s^2 2s^2 2p^6\). - For \(\mathrm{S}^{2-}\), the atomic number is 16 (sulfur) and its electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6\). - For \(\mathrm{F}^{+}\), the atomic number is 9 (fluorine) and its electron configuration is \(1s^2 2s^2 2p^4\). - For \(\mathrm{Mg}^{+}\), the atomic number is 12 (magnesium) and its electron configuration is \(1s^2 2s^2 2p^6 3s^1\).

Step 2: Determine the effective nuclear charge for each ion

The effective nuclear charge (\(Z_{eff}\)) is the net positive charge experienced by an electron in an atom. It is given by the formula \(Z_{eff} = Z - S\), where Z is the atomic number and S is the shielding constant, which accounts for the shielding effect causing electrons to partially cancel out the attractive force of the protons in the nucleus. Since we are comparing ions in the same period, electron shielding is relatively constant, so we can approximate \(Z_{eff}\) just by looking at the atomic number. - For \(\mathrm{Na}^{+}\), the \(Z_{eff} \approx 11\) - For \(\mathrm{S}^{2-}\), the \(Z_{eff} \approx 16\) - For \(\mathrm{F}^{+}\), the \(Z_{eff} \approx 9\) - For \(\mathrm{Mg}^{+}\), the \(Z_{eff} \approx 12\)

Step 3: Compare ionization energies and arrange ions in the required order

Higher effective nuclear charge increases ionization energy, so based on the \(Z_{eff}\) values calculated, we can predict the order of increasing ionization energy: \(\mathrm{F}^{+} < \mathrm{Na}^{+} < \mathrm{Mg}^{+} < \mathrm{S}^{2-}\) Thus, the ions in order of increasing ionization energy are \(\mathrm{F}^{+}\), \(\mathrm{Na}^{+}\), \(\mathrm{Mg}^{+}\), and \(\mathrm{S}^{2-}\).

Key Concepts

Effective Nuclear Charge

The concept of effective nuclear charge is crucial in understanding how atoms and ions attract electrons. Consider an ion or atom and its protons in the nucleus, pulling on the electrons. The strength of this pull is quantified as the effective nuclear charge, often abbreviated as \( Z_{eff} \). The formula to find this is simple: \( Z_{eff} = Z - S \). Here, \( Z \) represents the atomic number, or the total number of protons in the nucleus. \( S \) is the shielding constant, which accounts for the fact that electrons are partially "shielded" by those in inner shells. However, a simplified method for ions in the same period is to approximate the effective nuclear charge using just the atomic number, as their shielding is relatively uniform. So, for example:

• \( \mathrm{Na}^{+} \) approximates \( Z_{eff} \) at 11.
• \( \mathrm{S}^{2-} \) approximates \( Z_{eff} \) at 16.
• \( \mathrm{F}^{+} \) approximates \( Z_{eff} \) at 9.
• \( \mathrm{Mg}^{+} \) approximates \( Z_{eff} \) at 12.

High effective nuclear charge enhances the nucleus's hold on its electrons, typically resulting in higher ionization energies.

Electron Configuration

Electron configuration is a way of arranging electrons in atoms, detailing their position among various shells and subshells.Electrons occupy atomic orbitals, which are designated in the sequence of \( 1s, 2s, 2p, 3s, 3p, \) and so forth. Each type of orbital can hold a specific number of electrons:

• \( s \) orbitals hold up to 2 electrons.
• \( p \) orbitals accommodate up to 6 electrons.

Consider the exercise's ions for clear examples:

• \( \mathrm{Na}^{+} \): \( 1s^2 2s^2 2p^6 \)
• \( \mathrm{S}^{2-} \): \( 1s^2 2s^2 2p^6 3s^2 3p^6 \)
• \( \mathrm{F}^{+} \): \( 1s^2 2s^2 2p^4 \)
• \( \mathrm{Mg}^{+} \): \( 1s^2 2s^2 2p^6 3s^1 \)

Understanding electron configurations helps predict chemical behavior. For instance, having a full outer shell (like \( \mathrm{Na}^{+} \) with the noble gas configuration of \( \mathrm{Ne} \): \( 1s^2 2s^2 2p^6 \)) can contribute to the stability of ions.

Periodic Table Trends

The periodic table organizes elements in a way that reveals trends and patterns in their properties. One such pattern involves ionization energy, which is the energy required to remove an electron from an ion or atom.A few key points related to periodic trends are:

• Ionization energy generally increases across a period (left to right) because the effective nuclear charge increases. Thus, the nucleus's pull on the electrons becomes stronger, making it harder to remove them.
• Ionization energy tends to decrease down a group (top to bottom) as additional electron shells are added, increasing the distance between the nucleus and the electrons, which weakens the pull.

In the context of the exercise, ions from the same period (like \( \mathrm{Na}^{+} \) and \( \mathrm{Mg}^{+} \)) can be compared directly. Their order of increasing ionization energy agrees with these principles: - More protons mean a higher effective nuclear charge.- \( \mathrm{Mg}^{+} \) has a higher ionization energy than \( \mathrm{Na}^{+} \) as expected from their positions in the periodic table.Recognizing these trends helps in predicting and understanding various chemical behaviors and reactivities of elements.