Problem 37 - Chemistry: The Central Science

Question

Explain each of the following observations: (a) At room temperature \(\mathrm{I}_{2}\) is a solid, \(\mathrm{Br}_{2}\) is a liquid, and \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) are both gases. (b) \(\mathrm{F}_{2}\) cannot be prepared by electrolytic oxidation of aqueous \(\mathrm{F}^{-}\) solutions. (c) The boiling point of \(\mathrm{HF}\) is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\)

Step-by-Step Solution

Answer

The physical states of I2, Br2, Cl2, and F2 at room temperature can be attributed to the differences in their intermolecular forces, mainly Van der Waals forces, which depend on the number of electrons in a molecule. F2 is difficult to prepare by electrolytic oxidation of aqueous F- solutions because the reduction potential of water is greater than that of the F-/F2 couple, and F2 readily reacts with water. The higher boiling point of HF compared to other hydrogen halides is due to the presence of hydrogen bonding between HF molecules, resulting from the high electronegativity of fluorine. The decreasing oxidizing power of halogens (F2 > Cl2 > Br2 > I2) is because of their decreasing electronegativity values, with fluorine being the most electronegative element and possessing the highest affinity for electrons.

Step 1: (Observation a: Physical States of Halogens at Room Temperature)

To understand the different physical states of I2, Br2, Cl2, and F2 at room temperature, recognize that their intermolecular forces become weaker with decreasing molecular weight. I2 has the highest molecular weight, followed by Br2, Cl2, and F2. These forces are mainly Van der Waals forces, which depend on the number of electrons in a molecule. I2, having more electrons than Br2, Cl2, and F2, has stronger Van der Waals forces, making it a solid at room temperature. Br2, with weaker Van der Waals forces than I2, is a liquid. Cl2 and F2 have even weaker forces, causing them to exist as gases.

Step 2: (Observation b: F2 Preparation Difficulty)

The reason why F2 cannot be prepared by the electrolytic oxidation of aqueous F- solutions is that the reduction potential of water is greater than that of the F-/F2 couple. When an electric current is passed through aqueous F- solutions, water gets oxidized to form O2 gas instead of F- ions getting oxidized to F2 gas. Moreover, F2 is a highly reactive and strong oxidizing agent, and it would readily react with water to form hydrofluoric acid (HF) and oxygen (O2).

Step 3: (Observation c: Higher Boiling Point of HF)

The boiling point of a substance is related to the strength of the intermolecular forces between its molecules. HF has a much higher boiling point than the other hydrogen halides because of the presence of hydrogen bonding between its molecules. The relatively high electronegativity of fluorine (3.98) compared to other halogens leads to a highly polarized H-F bond, creating a partial positive charge on hydrogen and a partial negative charge on fluorine. This causes HF molecules to form strong hydrogen bonds with one another, resulting in a higher boiling point compared to other hydrogen halides.

Step 4: (Observation d: Decreasing Oxidizing Power of Halogens)

Oxidizing power refers to the ability of a substance to take electrons from other substances and get reduced in the process, ultimately leading to the oxidation of the electron-donating substance. The halogens' oxidizing power decreases in the order F2 > Cl2 > Br2 > I2 because of their decreasing electronegativity values. Fluorine, being the most electronegative element, possesses the highest affinity for electrons, making F2 the strongest oxidizing agent among halogens. As we move down the group of halogens, the electronegativity value and electron affinity decrease, thus also decreasing the oxidizing power.