Chapter 10, Problem 10.1VP

Interpretation Introduction

Interpretation: The substance which has the stronger cation-anion attractive forces and higher melting point is to be stated.

Concept introduction: The attractive force between the charged objects is calculated by the coulombs law. The cation is a positively charged species while anion is a negatively charged species. When cationic species comes under the influence of anionic species attraction takes place. The strength of this attraction depends upon the magnitude of the charges present on the ions and the charge separation between the ions.

To determine: The substance which has the stronger cation-anion attractive forces and higher melting point.

Answer to Problem 10.1VP

Solution

The substance which has the stronger cation-anion attractive forces and higher melting point is $\text{KF}$.

Explanation of Solution

Explanation

The given $\text{KF}$ and $\text{KI}$ compounds are ionic compounds and exist in solid form at room temperature. The strength of force of attraction between ionic compounds is calculated by the coulomb’s law.

According to this law the strength of attractive force between two ions is directly proportional to the product of the charges on cation and anion and inversely proportional to distance between the ions that is bond distance.

Mathematically it is represented as,

$U\propto \frac{q_1{q}_2}{\left(r_{+}+r_{-}\right)}$

Where,

• $U$ is the strength of attractive force.
• $q_1$ is the charge on cation.
• $q_2$ is the charge on anion.
• $r_{+}$ is the radius of cation.
• $r_{-}$ is the radius of anion.

The sum of radius $\left(r_{+}+r_{-}\right)$ is also known as bond distance and is denoted by $d$. The above expression is denoted as,

$U=\frac{q_1{q}_2}{d}$

From the above relation it is clear that the strength of attractive force between the two opposite ions increases as the charge on ions increases and size or distance decreases.

In the given figure two ionic molecules are given.

In these two molecules the size of $\text{KF}$ molecule is quite smaller than the size of $\text{KI}$ molecule. The charge on both the ions is same. For such situation the above formula is modified as,

$U\propto \frac{1}{d}$

This relation states that smaller the distance, higher is the strength of attractive force. The compound $\text{KF}$ consists of two ions named as ${\text{K}}^{+}$ and ${\text{F}}^{-}$ and the compound $\text{KI}$ also consists of two ions named as ${\text{K}}^{+}$ and ${\text{I}}^{-}$.

As in the given compounds the fluoride ion ${\text{F}}^{-}$ is smaller than the iodide ion ${\text{I}}^{-}$. Hence, the attractive force in $\text{KF}$ molecule is higher than the $\text{KI}$ molecule.

Now, on the basis of attractive forces the melting points are also explained. The given compounds are ionic in nature. The ionic interactions are quite stronger interactions and these are depends upon the size of the compound.

Smaller the size higher will be the interactions. Hence, the compound $\text{KF}$ due to smaller size has the higher melting point than the compound $\text{KI}$.

Conclusion

The substance which has the stronger cation-anion attractive forces and higher melting point is $\text{KF}$.