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

Expert verified
(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

01

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.
02

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.
03

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.
04

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.

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