Problem 113 - A Complete Resource Book in Chemistry for JEE Main

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) 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

Step-by-Step Solution

Answer

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

Step 1: Understanding Electron Affinity

The electron affinity is the energy change when an electron is added to a neutral atom. Generally, electron affinity decreases as you move down a group in the periodic table. This is because the added electron is further from the nucleus in larger atoms. Therefore, the order is generally expected as \( \mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I} \), contradicting statement 1.

Step 2: Analyzing Bond Dissociation Energy of \(\mathrm{F}_2\) and \(\mathrm{Cl}_2\)

The bond dissociation energy is lower for \(\mathrm{F}_2\) than for \(\mathrm{Cl}_2\) because of the smaller size and electron-electron repulsion in \(\mathrm{F}_2\). Thus, statement 2 is correct.

Step 3: Evaluating HF as a Hydrothalic Acid

Hydrothalic acids are stronger down the group due to weaker H-X bonds, where X is a halogen. HF is not the strongest; in fact, it is weaker than HCl or HI owing to stronger H-F bond. Thus, statement 3 is incorrect.

Step 4: Exploring Van der Waals Forces in Iodine

Van der Waals forces are stronger in larger molecules. Iodine, being the largest halogen molecule in this series, has the strongest Van der Waals forces, contradicting statement 4.

Step 5: Selecting the Incorrect Statements

From the analysis above, the incorrect statements are 1, 3, and 4. Thus, the correct option is (b) 1,3 and 4.

Key Concepts

Electron Affinity

Electron affinity refers to the amount of energy released when an electron is added to a neutral atom, converting it into a negatively charged ion. Halogens, located in group 17 of the periodic table, generally have high electron affinities due to their strong desire to gain an electron and achieve a stable, full outer shell.
Generally, electron affinity decreases as you move down the halogen group. This happens because as atoms increase in size, the added electron is further away from the nucleus, and hence experiences less nuclear attraction. For halogens, the electron affinity trend traditionally follows:

• Chlorine (Cl) generally ranks highest due to a balance of size and nuclear charge, making it more energetically favorable for Cl to gain an electron compared to other halogens.
• Fluorine (F) has a slightly lower electron affinity than Chlorine, despite being more electronegative, due to its small size that leads to greater electron-electron repulsions.
• The order then continues with Bromine (Br) and Iodine (I), each showing decreased affinity due to increased size and reduced nuclear attraction effect.

This challenges the misconception that fluorine should automatically have the highest electron affinity simply because it is the most electronegative element.

Bond Dissociation Energy

Bond dissociation energy is the energy required to break a bond in a molecule and separate it into atomic fragments. When we compare elements like \(\mathrm{F}_2\) and \(\mathrm{Cl}_2\), it is important to consider the repulsions and attractions within the molecule.
Fluorine has a very small atomic radius, causing its electron clouds to come very close together. This closeness results in strong electron-electron repulsions which makes the \(\mathrm{F}_2\) bond easier to break than that of \(\mathrm{Cl}_2\). This explains why the bond dissociation energy for \(\mathrm{F}_2\) is actually lower compared to \(\mathrm{Cl}_2\) despite the fact that individual F-F bonds might seem inherently strong due to high electronegativity.
Thus, whenever examining bond strength, always consider internal repulsion forces, such as those caused by overlapping electron clouds, which can significantly affect overall bond dissociation energy.

Van der Waals Forces

Van der Waals forces are weak intermolecular forces that arise from transient dipoles arising when electron clouds form temporary polarizations. These forces increase with surface area and molecular size.
In the case of halogens, iodine ( I ) being the largest atom in the series, exhibits the strongest Van der Waals forces among the halogens, contrary to some misconceptions.
The larger surface area of iodine molecules means more opportunities for induced polarizations, resulting in stronger intermolecular attractions as compared to smaller halogens like fluorine or chlorine.
This concept plays a key role in determining physical properties like melting and boiling points, where a greater sum of Van der Waals forces translates to higher temperatures needed to change states of matter.

Hydrohalic Acids

Hydrohalic acids are binary acids formed when hydrogen bonds with halogens, following the formula H-X, where X is a halogen. The strength of these acids typically increases down the group.
This is primarily attributed to the bond strength between hydrogen and the halogen atom; the weaker the H-X bond, the stronger the acid. This is because weaker bonds more readily dissociate in solution, increasing the concentration of ions in the solution.
Although HF is often wrongly assumed to be the strongest due to fluorine’s impressive electronegativity, it forms a very strong bond with hydrogen, limiting its ionization in solution. In contrast, acids like HCl, HBr, and HI are actually much stronger because the bonds can be broken easily, allowing greater ionization.

• HCl and HBr are stronger acids due to their weaker H-X bonds compared to HF.
• HI, having the weakest H-I bond among common hydrohalic acids, is the strongest acid in this series.

Understanding this trend is crucial in predicting the behavior and applications of different hydrohalic acids in chemical reactions and solutions.