Question

Given that the ions in \(\mathrm{LiF}\) and in \(\mathrm{MgO}\) are of similar size, which compound has stronger ionic bonding? Use Coulomb's law in your explanation.

Short answer

\(\text{MgO}\) has stronger ionic bonding because the product of the charges (\(+2\) and \(-2\)) is greater in magnitude than that of \(\text{LiF}\) (\(+1\) and \(-1\)).

Step-by-step solution

- Understand Coulomb's Law

Coulomb's law states that the force of attraction between two ions is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for Coulomb's law is given by: \[ F = k_e \frac{q_1 q_2}{r^2} \] where \( F \) is the force between the charges, \( k_e \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.

- Identify Charges on the Ions

Determine the charges of the ions in \(\text{LiF}\) and \(\text{MgO}\). In \(\text{LiF}\), lithium (Li) has a charge of +1 and fluoride (F) has a charge of -1. In \(\text{MgO}\), magnesium (Mg) has a charge of +2 and oxide (O) has a charge of -2.

- Compare the Products of the Charges

Using Coulomb's law, compare the products of the charges for both compounds. For \(\text{LiF}\), the product of the charges is \( (+1) \times (-1) = -1 \). For \(\text{MgO}\), the product of the charges is \( (+2) \times (-2) = -4 \).

- Analyze the Effect of Charge Magnitude

The strength of the ionic bond is directly proportional to the absolute value of the product of the charges. Since \(\text{MgO}\) has a product of -4, which has a larger magnitude compared to -1 in \(\text{LiF}\), \(\text{MgO}\) has stronger ionic bonding.

- Final Answer

Thus, considering the charges and Coulomb's law, \(\text{MgO}\) has the stronger ionic bonding compared to \(\text{LiF}\).

Key concepts

Coulomb's Law

Coulomb's Law is crucial for understanding the strength of ionic bonds. This law states that the force of attraction (or repulsion) between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance separating them.

The formula for Coulomb’s Law is:




Let's break the formula down:

• \( F \) is the electrostatic force.
• \( k_e \) is Coulomb's constant.
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the distance between the charges.

This relationship indicates that higher charges (\( q_1 \) and \( q_2 \)) and shorter distances (\( r \)) lead to stronger forces, thereby suggesting stronger ionic bonds.

Ion Charges

The charge on an ion is fundamental to determining the bond strength via Coulomb's Law.

Charges are simpler to find if you consider the common oxidation states of elements:

• Lithium (Li) typically has a +1 charge.
• Fluoride (F) typically has a -1 charge.
• Magnesium (Mg) typically has a +2 charge.
• Oxide (O) typically has a -2 charge.

In our example, we have two compounds: \( \mathrm{LiF} \) and \( \mathrm{MgO} \). In \( \mathrm{LiF} \), the charges on the ions are \( +1 \) and \( -1 \). For \( \mathrm{MgO} \), the charges are \( +2 \) and \( -2 \).

The product of these charges gives us insight into the strength of the bond:

• For \( \mathrm{LiF} \), the product is \( (+1) \times (-1) = -1 \).
• For \( \mathrm{MgO} \), the product is \( (+2) \times (-2) = -4 \).

The larger magnitude of \( -4 \) in \( \mathrm{MgO} \) suggests a stronger ionic bond compared to \( -1 \) in \( \mathrm{LiF} \).

Bond Strength

Bond strength in ionic compounds is largely dependent on the charges of the ions involved and their separation distance. Since we're considering that the ions in \( \mathrm{LiF} \) and \( \mathrm{MgO} \) are of similar size, the separation distance (\( r \)) can be considered constant.

This directs our focus to the magnitude of the ion charges as per Coulomb’s Law.

A higher product of the ionic charges leads to a stronger electrostatic force, which implies a stronger bond:

• Product for \( \mathrm{LiF} \): \( -1 \)
• Product for \( \mathrm{MgO} \): \( -4 \)

The greater absolute value of \( -4 \) in \( \mathrm{MgO} \) confirms that \( \mathrm{MgO} \) has a stronger ionic bond compared to \( \mathrm{LiF} \).

To summarize, when comparing the bond strength of ionic compounds, always consider the products of the ionic charges and their relative distances. In our example, \( \mathrm{MgO} \) with its higher product of ion charges demonstrates a stronger bond than \( \mathrm{LiF} \).