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Chapter 8: Problem 150

The only confirmed compound of radon is radon difluoride, \(\operatorname{RnF}_{2}\). One reason that it is difficult to study the chemistry of radon is that all isotopes of radon are radioactive so it is dangerous to handle the substance. Can you suggest another reason why there are so few known radon compounds? (Hint: Radioactive decays are exothermic processes.)

Short Answer

Expert verified
Aside from its radioactivity, another reason for the scarcity of radon compounds could be the combination of radon being a noble gas, therefore having low reactivity with other elements, and the exothermic nature of its decay process, which means any compound it does form is likely to break apart due to the energy released during the decay process.

Step by step solution

01

Understanding Radioactivity and Exothermic Reactions

Radioactivity involves the spontaneous emission of radiation, either directly from unstable atomic nuclei or as a consequence of a nuclear reaction. Radon is a radioactive element. An exothermic reaction is a chemical reaction that releases energy through light or heat. The hint suggests that the radioactive decay of Radon is an exothermic process, meaning it releases energy.
02

Recalling Radon Property

Radon is a noble gas. Noble gases are notable for their low levels of reactivity with other elements. This is due to their full valence electron shells which make them quite stable. Radon, being a noble gas, would naturally be resistant to forming compounds.
03

Connecting the Facts

The combination of these properties - radioactivity, exothermic decay, and inherent stability of radon - could provide another reason for the scarcity of radon compounds. Being a noble gas, radon resists forming compounds; being radioactive, any compound it does form is likely to break apart due to the exothermic nature of the radioactive decay process.

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Most popular questions from this chapter

Recent theoretical calculations suggest that astatine may be a monoatomic metal rather than a diatomic molecule like the other halogens. (a) Rationalize this prediction based on astatine's position in the periodic table. (b) The energy required to remove an electron from one At atom was determined by laser ionization to be \(9.3175 \mathrm{eV}\). Given that \(1 \mathrm{eV}=1.602\) \(\times 10^{-19} \mathrm{~J},\) calculate the first ionization energy of astatine in \(\mathrm{kJ} / \mathrm{mol} .\) (c) Comment on whether or not the following first ionization energies support your answer to part (a): \(\mathrm{Pb}, 715.6 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{Bi}, 702.9 \mathrm{~kJ} / \mathrm{mol} ;\) \(\mathrm{Po}, 811.8 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{Rn}, 1037 \mathrm{~kJ} / \mathrm{mol}\)

Little is known of the chemistry of astatine, the last member of Group 7A. Describe the physical characteristics that you would expect this halogen to have. Predict the products of the reaction between sodium astatide (NaAt) and sulfuric acid. (Hint: Sulfuric acid is an oxidizing agent.)

Explain which of the following anions is larger, and why: \(\mathrm{Se}^{2-}\) or \(\mathrm{Te}^{2-}\)

Arrange the following atoms in order of decreasing atomic radius: \(\mathrm{Na}, \mathrm{Al}, \mathrm{P}, \mathrm{Cl}, \mathrm{Mg} .\)

How does atomic radius change (a) from left to right across a period and (b) from top to bottom in a group?
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