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Argon was useful in the manufacture of gas-filled electric light bulbs, where its lower heat conductivity and chemical inertness made it preferable to nitrogen for inhibiting the vaporization of the tungsten filament and prolonging the life of the bulb. Fluorescent tubes commonly contain a mixture of argon and mercury vapor. Argon is the third most abundant gas in dry air.

Krypton-xenon flash tubes are used to take high-speed photographs. An electric discharge through such a tube gives a very intense light that lasts only 1 50,000 of a second. Krypton forms a difluoride, KrF 2 , which is thermally unstable at room temperature.

Stable compounds of xenon form when xenon reacts with fluorine. Xenon difluoride, XeF 2 , forms after heating an excess of xenon gas with fluorine gas and then cooling. The material forms colorless crystals, which are stable at room temperature in a dry atmosphere. Xenon tetrafluoride, XeF 4 , and xenon hexafluoride, XeF 6 , are prepared in an analogous manner, with a stoichiometric amount of fluorine and an excess of fluorine, respectively. Compounds with oxygen are prepared by replacing fluorine atoms in the xenon fluorides with oxygen.

When XeF 6 reacts with water, a solution of XeO 3 results and the xenon remains in the 6+-oxidation state:

XeF 6 ( s ) + 3H 2 O ( l ) XeO 3 ( a q ) + 6HF ( a q )

Dry, solid xenon trioxide, XeO 3 , is extremely explosive—it will spontaneously detonate. Both XeF 6 and XeO 3 disproportionate in basic solution, producing xenon, oxygen, and salts of the perxenate ion, XeO 6 4− , in which xenon reaches its maximum oxidation sate of 8+.

Radon apparently forms RnF 2 —evidence of this compound comes from radiochemical tracer techniques.

Unstable compounds of argon form at low temperatures, but stable compounds of helium and neon are not known.

Key concepts and summary

The most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds.

Chemistry end of chapter exercises

Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.

(a) XeF 2

(b) XeF 4

(c) XeO 3

(d) XeO 4

(e) XeOF 4

(a) sp 3 d hybridized; (b) sp 3 d 2 hybridized; (c) sp 3 hybridized; (d) sp 3 hybridized; (e) sp 3 d 2 hybridized;

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What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry.

(a) XeF 2

(b) XeF 4

(c) XeO 3

(d) XeO 4

(e) XeOF 4

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Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry.

(a) XeF 2

(b) XeF 4

(c) XeO 3

(d) XeO 4

(e) XeOF 4

(a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar

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What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry.

(a) XeO 2 F 2

(b) KrF 2

(c) XeF 3 +

(d) XeO 6 4−

(e) XeO 3

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A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon.

The empirical formula is XeF 6 , and the balanced reactions are:
Xe ( g ) + 3F 2 ( g ) Δ XeF 6 ( s ) XeF 6 ( s ) + 3H 2 ( g ) 6HF ( g ) + Xe ( g )

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Basic solutions of Na 4 XeO 6 are powerful oxidants. What mass of Mn(NO 3 ) 2 •6H 2 O reacts with 125.0 mL of a 0.1717 M basic solution of Na 4 XeO 6 that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate?

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Questions & Answers

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ln(k two/k one)= (- activation energy/R)(1/T two - 1/T one) T is tempuratue in Kelvin, K is rate constant but you can also use rate two and one instead, R is 8.314 J/mol×Kelvin
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if ur solving for k two or k one you will need to use e to cancel out ln
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An atom is the smallest particle of an element which can take part in a chemical reaction..
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Formula for equilibrium
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Practice Key Terms 2

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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