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. \((\mathbf{c})\) The boiling point of \(\mathrm{HF}\) is much higher than those of the other hydrogen halides. (d) The halugems decrease in uxidizing puwer in the urder \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\).

Short answer

(a) Intermolecular dispersion forces determine the states of I₂, Br₂, Cl₂, and F₂ at room temperature. The forces increase with size and molar mass, thus I₂ is a solid, Br₂ is a liquid, and Cl₂ and F₂ are gases. (b) F₂ cannot be prepared by electrolytic oxidation of aqueous F⁻ solutions because water is a weaker oxidizing agent than F₂ and preferentially decomposes into H₂ and O₂ rather than oxidizing F⁻ ions. (c) HF has a much higher boiling point than other hydrogen halides due to the strong hydrogen bonding caused by fluorine's high electronegativity and small size. (d) The decreasing oxidizing power in the order F₂ > Cl₂ > Br₂ > I₂ is due to decreasing electronegativity and increasing atomic size, resulting in weaker halogen-halide bonds and reduced electron attraction.

Step-by-step solution

Observation (a)

To explain the different states of I₂, Br₂, Cl₂, and F₂ at room temperature, we need to understand the intermolecular forces at play. In general, the stronger the intermolecular forces, the higher the boiling point and melting point. Since halogens are diatomic molecules, they experience dispersion (London) forces. The dispersion forces depend on the molar mass and the size of the molecules: as the size increases, the dispersion forces become stronger. In the case of halogens, the size and molar mass increase as we move down the group - the order becomes I₂ > Br₂ > Cl₂ > F₂. Therefore, I₂ has the strongest dispersion forces, making it a solid, followed by Br₂ which is a liquid, and finally Cl₂ and F₂ exist as gases due to their weak dispersion forces at room temperature.

Observation (b)

To electrolytically oxidize aqueous F⁻ solutions to form F₂, a strong oxidizing agent is required. However, water is a weaker oxidizing agent than F₂ and would prefer to decompose into H₂ and O₂ rather than oxidize the F⁻ ions. The standard electrode potential of the reaction 2F⁻(aq) → F₂(g) + 2e⁻ is +2.87 V, whereas the standard electrode potential of the reaction 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ is +1.23 V. Therefore, it is easier for water to get oxidized in preference to the F⁻ ions, making the preparation of F₂ from F⁻ electrolytically very difficult.

Observation (c)

The boiling point of a molecule depends on its intermolecular forces. In the case of hydrogen halides, these forces are mainly the stronger hydrogen bonds and the weaker dispersion forces. HF exhibits the strongest hydrogen bonding of all hydrogen halides, due to fluorine's high electronegativity and small size. This strong hydrogen bond results in a much higher boiling point for HF as compared to the other hydrogen halides.

Observation (d)

To explain the decreasing oxidizing power in the order F₂ > Cl₂ > Br₂ > I₂, we can look at the electronegativity and atomic size of each halogen. Fluorine has the highest electronegativity, which means it attracts electrons more easily than the other halogens, making it a strong oxidizing agent. As we move down the group, the electronegativity decreases and the atomic size increases, both of which contribute to a decrease in oxidizing power. The larger atomic size results in weaker halogen-halide bonds, making it more difficult for the halogen to attract electrons and subsequently diminishing its oxidizing power.

Key concepts

Intermolecular Forces

The state of matter of a substance at room temperature can be largely attributed to the nature and strength of the intermolecular forces present. When discussing halogens such as Iodine (\( ext{I}_2 \)), Bromine (\( ext{Br}_2 \)), Chlorine (\( ext{Cl}_2 \)), and Fluorine (\( ext{F}_2 \)), we find that they are non-polar molecules. This non-polarity means that their primary interaction is through dispersion or London forces.

• Dispersion forces are weak forces that arise because of temporary fluctuations in electron distribution.
• The strength of these forces is directly related to the size and molar mass of the molecules.

As you go down the halogen group in the periodic table, the size and molar mass increase significantly.
This results in stronger dispersion forces, thereby impacting their physical state.
Iodine, being the heaviest, exhibits the strongest dispersion forces, thus it remains a solid.
Bromine, with slightly weaker forces, exists as a liquid. Chlorine and Fluorine with even weaker forces are gases at room temperature.

Electrolytic Reactions

Electrolytic reactions are processes that use electrical energy to drive non-spontaneous chemical reactions. When dealing with the production of \( ext{F}_2 \) from aqueous \( ext{F}^- \) ions through electrolysis, one must consider competitive reactions occurring at the electrodes.
This involves the oxidation of different species which releases electrons.

• The standard electrode potential values for two possible reactions tell us the likelihood of them occurring; \( ext{2F}^- ightarrow ext{F}_2 + 2e^- \) has a high potential of +2.87 V, whereas \( 2 ext{H}_2 ext{O} ightarrow ext{O}_2 + 4 ext{H}^+ + 4e^- \) is only +1.23 V.
• Water has a lower activation energy required to oxidize, thus it often decomposes into H₂ and O₂ instead of forming \( ext{F}_2 \).

In summary, for electrolytic oxidation, reactions with lower electrode potentials occur more readily.
This is why forming \( ext{F}_2 \) from fluoride ions is more challenging, as water oxidation naturally prevails over fluoride ion oxidation.

Oxidizing Agents

Halogens are well known for their ability to act as oxidizing agents. An oxidizing agent is a substance that can accept electrons from other molecules, causing oxidation. The ability of a halogen to oxidize other substances depends on two main factors:

• Electronegativity: A higher electronegativity means a stronger attraction for electrons.
• Atomic Size: Smaller atomic size leads to stronger bonds with electron pairs.

Fluorine sits at the top of both spectra—it's the most electronegative and has the smallest atomic size.
These properties translate into \( ext{F}_2 \) being a formidable oxidizing agent.
As we move down the halogen group, from \( ext{F}_2 \) to \( ext{I}_2 \), the atomic radius increases, and electronegativity decreases.
Consequently, their oxidizing power diminishes.
This order manifests as \( ext{F}_2 > ext{Cl}_2 > ext{Br}_2 > ext{I}_2 \), meaning \( ext{F}_2 \) is the strongest oxidizer among them, capable of grabbing electrons more effectively than \( ext{I}_2 \).

Molecular Polarity

Molecular polarity is critical in understanding intermolecular interactions and the resultant properties like boiling point. Polarity in molecules results from differences in electronegativity between atoms. If a molecule has a significant separation of electric charge, it is polar.
Within hydrogen halides, particularly with \( ext{HF} \), this concept becomes more prominent due to hydrogen bonding.

• Hydrogen bonds are a special, very strong type of dipole-dipole attraction between molecules.
• Fluorine's small size and high electronegativity mean it draws a significant electron cloud towards itself, resulting in a dipole.

This strong hydrogen bond in \( ext{HF} \) leads to an unusually high boiling point compared to other hydrogen halides.
The inherent polarity of \( ext{HF} \) from these bonds is stronger than in other hydrogen halides where larger, less electronegative atoms like chlorine, bromine, and iodine are present.
Thus, the boiling point differences among hydrogen halides highlight the significance of molecular polarity and hydrogen bonding in molecular interactions.