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Explain the following observations: (a) For a given oxidation state, the acid strength of the oxyacid in aqueous solution decreases in the order chlorine \(>\) bromine \(>\) iodine. (b) Hydrofluoric acid cannot be stored in glass bottles. (c) HI cannot be prepared by treating NaI with sulfuric acid. (d) The interhalogen \(\mathrm{ICl}_{3}\) is known, but \(\mathrm{BrCl}_{3}\) is not.

Short Answer

Expert verified
The decrease in acid strength of oxyacids in the order chlorine > bromine > iodine is due to differences in electronegativity, with chlorine being the most electronegative. Hydrofluoric acid cannot be stored in glass bottles because it reacts with silicon dioxide, causing glass corrosion. The preparation of HI using NaI and sulfuric acid is not possible because a redox reaction occurs instead, forming iodine and hydrogen sulfide. ICl3 is known to exist, while BrCl3 does not because iodine's larger size can accommodate more than one chlorine atom, whereas bromine's smaller size prevents the formation of BrCl3.

Step by step solution

01

Reason for the Decrease in Acid Strength

The decrease in acid strength in the order chlorine > bromine > iodine can be explained by the electronegativity of these elements. Chlorine, being more electronegative than both bromine and iodine, attracts electron pairs in the O-H bonds more effectively. This leads to a weaker O-H bond, and therefore, easier release of H+ ions from chlorine-containing oxyacids. Consequently, chlorine-containing oxyacids are stronger acids than those containing bromine and iodine. (b) Hydrofluoric acid cannot be stored in glass bottles.
02

Reason for Hydrofluoric Acid's Interaction with Glass

Hydrofluoric acid (HF) cannot be stored in glass bottles because it reacts with the silicon dioxide (SiO2) present in the glass. When HF comes into contact with SiO2, it forms silicon tetrafluoride (SiF4) and water (H2O), which leads to the corrosion of the glass container. (c) HI cannot be prepared by treating NaI with sulfuric acid.
03

Reason for the Inability to Prepare HI from NaI and H2SO4

HI cannot be prepared by treating NaI with sulfuric acid because the reaction between NaI and H2SO4 does not proceed as expected. Instead of forming HI, a redox reaction occurs between iodide ions (I-) and the sulfate ions (SO4^2-). This results in the formation of molecular iodine (I2) and hydrogen sulfide (H2S) as the primary products, rather than the desired hydrogen iodide (HI). (d) The interhalogen ICl3 is known, but BrCl3 is not.
04

Reason for the Existence of ICl3 and Absence of BrCl3

Interhalogen compounds such as ICl3, BrCl3 forms when a less electronegative halogen combines with a more electronegative one. In case of ICl3, iodine (I) has a larger size, and its valence electrons are in the 5th shell which is further away from the nucleus. Thus, it can accommodate more than one Cl atom surrounding it, leading to the formation of ICl3. However, the smaller size of bromine (Br) and its valence electrons being in the 4th shell makes it incapable of accommodating multiple Cl atoms around it, and thus BrCl3 does not form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxyacids
Oxyacids are a type of acid that consists of hydrogen, oxygen, and another element, usually a non-metal. The strength of an oxyacid is frequently influenced by the central atom's electronegativity. In general, as the electronegativity of the central atom increases, the strength of the oxyacid also increases. This is because more electronegative atoms pull electron density away from the hydrogen in the O-H bond, making it easier for the hydrogen ion, or proton, to dissociate.

When comparing the oxyacids of chlorine, bromine, and iodine, we observe a decrease in acid strength in the order chlorine > bromine > iodine. This occurs because chlorine is more electronegative than bromine and iodine, causing the O-H bond in chlorine-based oxyacids to be weaker. Consequently, it is easier for chlorine oxyacids to release the hydrogen ion, making them stronger acids. This trend helps us understand why, when comparing similar oxyacids, those with more electronegative central atoms will typically be stronger acids.
Electronegativity
Electronegativity refers to an atom's ability to attract shared electrons within a chemical bond. It plays a crucial role in determining the behavior and properties of compounds, such as their acidity, polarity, and reactivity.

In the context of oxyacids, electronegativity is key in explaining why some acids are stronger than others. For example, elements like chlorine are highly electronegative, allowing them to create weaker O-H bonds by pulling electron density away from the hydrogen atom. This makes it easier for the acid to release hydrogen ions, contributing to higher acid strength.

Furthermore, electronegativity differences between elements also dictate the formation of interhalogen compounds. These compounds are formed when one halogen is less electronegative than the other, resulting in different bonding patterns. Understanding electronegativity trends helps predict various chemical behaviors and is essential in forecasting how different elements will interact in compounds.
Interhalogen Compounds
Interhalogen compounds are molecules formed from the combination of different halogen elements. These compounds offer fascinating insights into chemical bonding and reactivity, due to the differences in electronegativity and atomic size of the halogens involved.

Interhalogen compounds are typically more reactive than diatomic halogens. They can exist in various stoichiometries, such as XY, XY3, XY5, and XY7, where X and Y represent different halogen atoms. The availability of these stoichiometries depends on the sizes and electronegativities of the atoms involved.

A well-known example is ICl3, an interhalogen compound of iodine and chlorine. The larger atomic radius of iodine allows it to accommodate multiple chlorine atoms, forming the compound. In contrast, smaller halogen atoms, like bromine, with only slightly larger sizes than chlorine, may not form stable analogous compounds like BrCl3 because they cannot adequately incorporate multiple chlorine atoms around them.

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Most popular questions from this chapter

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 \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}\).

A sulfuric acid plant produces a considerable amount of heat. This heat is used to generate electricity, which helps reduce operating costs. The synthesis of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) consists of three main chemical processes: (a) oxidation of \(S\) to \(\mathrm{SO}_{2}\), (b) oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3},\) (c) the dissolving of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and the subsequent reaction with water to form \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If the third process produces \(130 \mathrm{~kJ} / \mathrm{mol}\), how much heat is produced in preparing a mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from a mole of \(S\) ? How much heat is produced in preparing \(2000 \mathrm{~kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

(a) How does the structure of diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) differ from that of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ?\) (b) Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as "hydridic"?

Borazine, \((\mathrm{BH})_{3}(\mathrm{NH})_{3},\) is an analog of \(\mathrm{C}_{6} \mathrm{H}_{6},\) benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from \(2.00 \mathrm{~L}\) of ammonia at STP, assuming diborane is in excess?

Select the member of group 14 that best fits each description: \((\mathbf{a})\) has the greatest tendency to form multiple bonds with itself, (b) forms polymeric structures with oxygen, (c) is a metal with +2 and +4 oxidation states.

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