Question

Which one of the following statements about halogens is incorrect? (1) The electron affinities are in order \(\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}\) (2) \(\mathrm{F}_{2}\) has lower bond dissociation energy than \(\mathrm{Cl}_{2}\) (3) \(\mathrm{HF}\) is the strongest hydrothalic acid (4) Van der Waals forces are the weakest in iodine (a) 1,2 and 4 (b) 1,3 and 4 (c) 1,2 and 3 (d) 2,3 and 4

Short answer

Option (b) 1,3 and 4 are incorrect.

Step-by-step solution

Analyze Statement (1)

Examine the electron affinities order: \( \mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I} \). While fluorine is more electronegative, chlorine actually has a higher electron affinity due to less repulsion between electrons in a larger atom. Hence, statement (1) is incorrect.

Analyze Statement (2)

Assess the bond dissociation energy: \( \mathrm{F}_{2} \) has a lower bond dissociation energy than \( \mathrm{Cl}_{2} \) because the smaller size of fluorine leads to greater electron-electron repulsion. Therefore, statement (2) is correct.

Analyze Statement (3)

Check the strength of hydrohalic acids: \( \mathrm{HI} \), not \( \mathrm{HF} \), is the strongest hydrohalic acid because larger halides form weaker bonds with hydrogen, making \( \mathrm{HI} \) the strongest. Thus, statement (3) is incorrect.

Analyze Statement (4)

Evaluate Van der Waals forces: Iodine, being the largest molecule with more electrons, actually has the strongest Van der Waals forces, not the weakest. So, statement (4) is incorrect.

Select the Incorrect Statements

Combine the incorrect statements: Statements (1), (3), and (4) are incorrect, which matches option (b).

Key concepts

Electron Affinity

Electron affinity is a vital concept in understanding halogens. It's the amount of energy released when an electron is added to a neutral atom in the gaseous state. Halogens are known for their high electron affinity because they need just one more electron to achieve a stable octet configuration.

Usually, we might expect electron affinity to increase with electronegativity. However, in the case of halogens, chlorine (\(\mathrm{Cl}\)) actually has a higher electron affinity than fluorine (\(\mathrm{F}\)). This may seem counterintuitive since fluorine is more electronegative. Yet due to the smaller size of fluorine, there is increased electron-electron repulsion, which slightly lowers its electron affinity compared to chlorine. As a result, the correct order is: \(\mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I}\).

This order highlights the importance of size and repulsion factors in influencing electron affinity, rather than just electronegativity alone.

Bond Dissociation Energy

Bond dissociation energy (BDE) is the energy required to break the bond between two atoms in a molecule. For diatomic molecules like \(\mathrm{F}_2\) and \(\mathrm{Cl}_2\), this concept is crucial in understanding their reactivity.

Fluorine (\(\mathrm{F}_2\)) has a surprisingly low bond dissociation energy compared to chlorine (\(\mathrm{Cl}_2\)). Although fluorine forms very strong bonds with other atoms, the bond between two fluorine atoms is affected by the small size of the atoms, which leads to increased repulsion between the electron pairs. This means it takes less energy to break a \(\mathrm{F}_2\) bond than a \(\mathrm{Cl}_2\) bond.

This lower BDE of \(\mathrm{F}_2\) contributes to its higher reactivity, making fluorine a highly reactive and powerful oxidizing agent. Thus, the understanding of BDE is not just about bond strength, but also about the interplay of atomic size and electron repulsion.

Hydrohalic Acids

Hydrohalic acids are hydrogen halides that dissolve in water to produce acidic solutions. Among them, hydrofluoric acid (\(\mathrm{HF}\)) is often mistakenly thought to be the strongest acid due to fluorine's high electronegativity.

Actually, the strongest hydrohalic acid is hydrogen iodide (\(\mathrm{HI}\)). The strength of these acids is primarily dictated by the bond strength between the hydrogen atom and the halogen atom. Since the bond strength decreases with an increase in the size of the halogen, the \(\mathrm{H-I}\) bond is weaker than the \(\mathrm{H-F}\) bond.

As bonds weaken, they are easier to dissociate in water, leading to greater acid strength. Therefore, \(\mathrm{HI}\) is strong because its bond is weaker, allowing for more complete ionization, compared to \(\mathrm{HF}\) which forms a very stable bond and is less dissociated.

Van der Waals Forces

Van der Waals forces are weak intermolecular attractions that occur in all molecules but are especially important in nonpolar molecules like halogen molecules. These forces increase with the number of electrons and the size of the atom or molecule.

Iodine (\(\mathrm{I_2}\)), being larger and having more electrons than its counterparts, exhibits the strongest Van der Waals forces among the halogens. This is because molecules with more electrons have larger electron clouds which can become polarized, thereby inducing temporary dipoles.

This misconception in the exercise that iodine has the weakest Van der Waals forces is common. Instead, \(\mathrm{I_2}\)’s large size and extensive electron cloud lead to substantial Van der Waals attractions, contributing to properties like its higher boiling point compared to other halogens. Understanding Van der Waals forces helps elucidate these physical characteristics of molecules.