Chapter 22: Problem 29
Why does xenon form stable compounds with fluorine, whereas argon does not?
Chapter 22: Problem 29
Why does xenon form stable compounds with fluorine, whereas argon does not?
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Get started for freeWrite a balanced equation for each of the following reactions: (a) hydrolysis of \(\mathrm{PCl}_{5},(\mathbf{b})\) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, (c) reaction of \(\mathrm{P}_{4} \mathrm{O}_{10}\) with water.
Although the \(\mathrm{ClO}_{4}^{-}\) and \(\mathrm{IO}_{4}^{-}\) ions have been known for a long time, \(\mathrm{BrO}_{4}^{-}\) was not synthesized until \(1965 .\) The ion was synthesized by oxidizing the bromate ion with xenon difluoride, producing xenon, hydrofluoric acid, and the perbromate ion. (a) Write the balanced equation for this reaction. (b) What are the oxidation states of Br in the Br-containing species in this reaction?
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 L of ammonia at STP, assuming diborane is in excess?
Write the chemical formula for each of the following, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) calcium hypobromite, (b) bromic acid, (c) xenon trioxide, (d) perchlorate ion, (e) iodous acid,(f) iodine pentafluoride.
Consider the elements \(\mathrm{O}, \mathrm{Ba}, \mathrm{Co}, \mathrm{Be}, \mathrm{Br}\) , and Se. From this list, select the element that (a) is most electronegative, (b) exhibits a maximum oxidation state of \(+7,(\mathrm{c})\) loses an electron most readily, (\mathbf{d} ) ~ f o r m s ~ \(\pi\) bonds most readily, (e) is a transition metal, \((\mathbf{f})\) is a liquid at room temperature and pressure.
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