Question

Sodium ions, \(\mathrm{Na}^{+}\), form ionic compounds with fluoride ions, \(\mathrm{F}^{-},\) and iodide ions, \(\mathrm{I}^{-}\). The radii of these ions are as follows: \(\mathrm{Na}^{+}=116 \mathrm{pm}\) \(\mathrm{F}^{-}=119 \mathrm{pm} ;\) and \(\mathrm{I}^{-}=206 \mathrm{pm} .\) In which ionic compound, NaF or NaI, are the forces of attraction between cation and anion stronger? Explain your answer.

Short answer

NaF has stronger forces of attraction due to smaller ion distance.

Step-by-step solution

Understand Coulomb's Law

Coulomb's Law explains the force of attraction between two charged particles. The formula is \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \), where \( F \) is the force, \( k \) is the constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the centers of the charges.

Calculate Distance Between Ions

For \( \text{NaF} \), the distance \( r \) is the sum of the radii of \( \text{Na}^+ \) and \( \text{F}^- \): \( 116 \, \text{pm} + 119 \, \text{pm} = 235 \, \text{pm} \). For \( \text{NaI} \), \( r = 116 \, \text{pm} + 206 \, \text{pm} = 322 \, \text{pm} \).

Analyze the Impact of Distance

According to Coulomb's Law, a smaller distance \( r \) results in a stronger force \( F \). Comparing \( \text{NaF} \) with \( \text{NaI} \), \( \text{NaF} \) has the smaller \( r \).

Conclusion

The forces of attraction are stronger in \( \text{NaF} \) because it has a smaller distance between the ions, leading to a greater force according to Coulomb's Law.

Key concepts

Coulomb's Law

Coulomb's Law is a foundational principle in understanding ionic compounds. It explains the force of attraction between two charged particles, such as ions. The law is expressed with the formula: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]Here, \( F \) represents the force of attraction or repulsion, \( k \) is a constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges on the ions, and \( r \) is the distance between the centers of the two ions.
This equation shows that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This means that:

• A larger charge leads to a stronger force.
• A smaller distance between ions results in a stronger attraction.

Using this law, we can compare different ionic compounds to understand which has the stronger attraction between its ions.

Ion Radii

Ion radii refer to the size of ions in an ionic compound. The radius of an ion affects how close it can get to another ion, influencing the ionic bond's strength. In our context:

• The radius of \( \mathrm{Na}^+ \) is 116 pm.
• The radius of \( \mathrm{F}^- \) is 119 pm.
• The radius of \( \mathrm{I}^- \) is significantly larger at 206 pm.

The sum of the radii of the cation and anion gives us the distance \( r \) between the ions' centers. This is crucial in applying Coulomb's Law.
For example, a smaller ionic radius means ions can get closer, reducing \( r \), and resulting in a stronger force of attraction. Thus, knowing the ion radii helps predict and compare the strength of ionic bonds in compounds.

Ionic Bond Strength

Ionic bond strength is determined by the attractive forces between ions in a compound. The strength depends on several factors:

• Charge of the ions: More highly charged ions create stronger bonds.
• Distance between ions: Smaller distances result in stronger attractions, as per Coulomb's Law.

For example, in sodium fluoride (NaF), the smaller distance between \( \mathrm{Na}^+ \) and \( \mathrm{F}^- \) contributes to a stronger ionic bond compared to sodium iodide (NaI), where the larger \( \mathrm{I}^- \) results in greater separation.
Understanding ionic bond strength is essential for predicting a compound's properties, such as melting point, solubility, and hardness. Stronger bonds often lead to compounds with higher melting points and less solubility.

Sodium Fluoride

Sodium fluoride (NaF) is an ionic compound formed by the attraction between sodium ions (\( \mathrm{Na}^+ \)) and fluoride ions (\( \mathrm{F}^- \)). In NaF, both ions have relatively small radii:

• \( \mathrm{Na}^+ \) has a radius of 116 pm
• \( \mathrm{F}^- \) has a radius of 119 pm

The sum of these radii results in a distance of 235 pm between the centers of the ions.
According to Coulomb's Law, this shorter distance allows for a stronger force of attraction.
NaF is known for its high melting point and efficiency in forming strong ionic bonds, making it useful in various applications such as toothpaste for preventing cavities.

Sodium Iodide

Sodium iodide (NaI) is another ionic compound, consisting of sodium ions (\( \mathrm{Na}^+ \)) and iodide ions (\( \mathrm{I}^- \)). Compared to fluoride ions, iodide ions have a much larger radius:

• \( \mathrm{Na}^+ \) has a radius of 116 pm
• \( \mathrm{I}^- \) has a radius of 206 pm

This results in a larger distance of 322 pm between the ions in NaI.
Because of the greater separation, the force of attraction is weaker compared to NaF.
Despite the weaker attraction, sodium iodide is still important in certain fields, such as medicine and photography, due to its unique properties that extend beyond bond strength.