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Chapter 22: Problem 29

Why does xenon form stable compounds with fluorine, whereas argon does not?

Short Answer

Expert verified
Xenon forms stable compounds with fluorine due to its lower ionization energy and significant difference in electronegativity compared to argon. This allows for the formation of a chemical bond between the two elements. In contrast, argon does not form compounds with fluorine because of its higher ionization energy and similar electronegativity value, which makes it difficult for a bond to form between them.

Step by step solution

01

Verify Electronic Configurations

For Xenon (Xe) and Argon (Ar), we need to find their electronic configurations. Xe is in period 5 and group 18, while Ar is in period 3 and group 18. Their electronic configurations are: Xe: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\] Ar: \[1s^2 2s^2 2p^6 3s^2 3p^6\]
02

Understand Ionization Energy and Electronegativity

The ionization energy of an atom is the energy required to remove an electron from the outermost shell. The electronegativity of an atom is its tendency to attract a bonding pair of electrons. Generally, elements that have lower ionization energy and higher electronegativity tend to form compounds more easily.
03

Compare Ionization Energies and Electronegativity

Xenon has a lower ionization energy than argon. This means that it is easier to remove an electron from xenon than from argon. The electronegativity values for fluorine, xenon, and argon are 3.98, 2.6, and 3.24, respectively. Fluorine being highly electronegative, it can attract electrons from xenon, which has a relatively lower ionization energy. In contrast, argon has a higher ionization energy and its electronegativity value is closer to fluorine. As a result, argon does not readily lose electrons to form a bond with fluorine.
04

Relate Ionization Energy and Electronegativity to the Formation of Chemical Bonds

The difference in ionization energy and electronegativity between xenon and fluorine allows for the transfer of electrons between the two elements, which leads to the formation of the compound. On the other hand, the similar ionization energies and electronegativity values for argon and fluorine prevent the formation of a chemical bond between them.
05

Conclusion

Xenon forms stable compounds with fluorine because it has a lower ionization energy and a significant difference in electronegativity compared to argon. This allows for the formation of a chemical bond between the two elements. Argon does not form compounds with fluorine because of its higher ionization energy and similar electronegativity value, making it difficult for the elements to form a bond.

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Most popular questions from this chapter

Write the formulas for the following compounds, and indicate the oxidation state of the group 4 A element or of boron in each: (a) silicon dioxide, (b) germanium tetra chloride, (c) sodium borohydride, (d) stannous chloride, (e) diborane, (f) boron trichloride.

Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\) . (e) Hydrazine reduces \(\mathrm{CrO}_{4}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}^{-}\) in base (hydrazine is oxidized to \(\mathrm{N}_{2} )\) .

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) . In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

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Name the following compounds and assign oxidation states to the halogens in them: (a) \(\mathrm{KClO}_{3},\) (b) \(\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}\) ,\((\mathbf{c}) \mathrm{AlCl}_{3},(\mathbf{d}) \mathrm{HBrO}_{3},(\mathbf{e}) \mathrm{H}_{5} \mathrm{IO}_{6},(\mathbf{f}) \mathrm{XeF}_{4}\)
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