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Chapter 22: Problem 36

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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Most popular questions from this chapter

An aqueous solution of \(\mathrm{SO}_{2}\) reduces (a) aqueous \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnSO}_{4}(a q),(\mathbf{b})\) acidic aqueous \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to aqueous \(\mathrm{Cr}^{3+}\) (c) aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) to mercury metal. Write balanced equations for these reactions.

One method proposed for removing \(\mathrm{SO}_{2}\) from the flue gases of power plants involves scrubbing with an alkali solid such as calcium carbonate to form calcium sulfite and carbon dioxide gas. (a) Write a balanced chemical equation for the reaction. (b) What mass of \(\mathrm{CaCO}_{3}\) would be required to remove the \(\mathrm{SO}_{2}\) formed by burning \(1000 \mathrm{~kg}\) of coal containing \(8.0 \% \mathrm{~S}\) by mass? (c) What volume of \(\mathrm{CO}_{2}\) is formed under standard temperature and pressure? Assume that all reactions are \(100 \%\) efficient.

Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 16 elements in each: (a) potassium peroxide, (b) potassium thiosulfate, \((\mathbf{c})\) selenium oxychloride, \((\mathbf{d})\) sodium telluride, (e) magnesium sulfite, (f) selenium hexafluoride.

(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"?

Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. ( \(\mathbf{b}\) ) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, \(\mathrm{PbS}(s)\) reacts with ozone to form \(\mathrm{PbSO}_{4}(s)\) and \(\mathrm{O}_{2}(g) .\) (d) When heated in air, \(\mathrm{ZnS}(s)\) is converted to \(\mathrm{ZnO}\). (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carbonate and \(\mathrm{O}_{2}\). (f) Oxygen is converted to ozone in the upper atmosphere.
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