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Chapter 22: Problem 101

The bond energies of \(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\) are 243 and \(159 \mathrm{kJ} \mathrm{mol}^{-1},\) respectively. Use these data to explain why \(\mathrm{XeF}_{2}\) is a much more stable compound than \(\mathrm{XeCl}_{2} .[\text { Hint: Recall that Xe exists as a monatomic gas. }]\).

Short Answer

Expert verified
\(\mathrm{XeF}_{2}\) is more stable than \(\mathrm{XeCl}_{2}\) because the F-F bonds in \(\mathrm{XeF}_{2}\) are weaker and require less energy to break compared to the Cl-Cl bonds in \(\mathrm{XeCl}_{2}\). The bond energy for F-F is 159 kJ/mol which is less than the bond energy for Cl-Cl which is 243 kJ/mol. Hence, compounds with lower bond energies are more stable.

Step by step solution

01

Consider Bond Energies

The given bond energies are 243 kJ/mol for \(\mathrm{Cl}_{2}\) and 159 kJ/mol for \(\mathrm{F}_{2}\). This means that it requires 243 kJ of energy to break a single mole of Cl-Cl bonds and 159 kJ of energy to break a single mole of F-F bonds.
02

Determine the Stability

The lower the bond energy, the less energy it requires to break the bond and hence the less stable the molecule. Hence, a compound with F-F bonds would be less stable than a compound with Cl-Cl bonds because F-F has a lower bond energy.
03

Apply to \(\mathrm{XeF}_{2}\) and \(\mathrm{XeCl}_{2}\)

The bond energies of F-F and Cl-Cl suggest that \(\mathrm{XeF}_{2}\) must be more stable than \(\mathrm{XeCl}_{2}\), because the F-F bonds in \(\mathrm{XeF}_{2}\) are weaker (lower bond energy) than the Cl-Cl bonds in \(\mathrm{XeCl}_{2}\). Hence, \(\mathrm{XeF}_{2}\) is more stable because it contains the weaker F-F bonds, which require less energy to break.

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Most popular questions from this chapter

Which of the following reactions are likely to go to completion or very nearly so? (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1)\) (b) \(\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1)+4 \mathrm{Cl}^{-}(\mathrm{aq}) \stackrel{-}{\longrightarrow}\) \(2 \mathrm{Cl}_{2}(\mathrm{g})+4 \mathrm{OH}^{-}(\mathrm{aq})\) (c) \(\mathrm{O}_{3}(\mathrm{g})+\mathrm{Pb}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\mathrm{PbO}_{2}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})\) (d) \(\mathrm{HO}_{2}^{-}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(3 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(1)\)

In water, \(\mathrm{O}^{2-}\) is a strong base. If \(50.0 \mathrm{mg}\) of \(\mathrm{Li}_{2} \mathrm{O}\) is dissolved in \(750.0 \mathrm{mL}\) of aqueous solution, what will be the pH of the solution?

In the electrolysis of a sample of water \(22.83 \mathrm{mL}\) of \(\mathrm{O}_{2}(\mathrm{g})\) was collected at \(25.0^{\circ} \mathrm{C}\) at an oxygen partial pressure of \(736.7 \mathrm{mmHg} .\) Determine the mass of water that was decomposed.

In \(1968,\) before pollution controls were introduced, over 75 billion gallons of gasoline were used in the United States as a motor fuel. Assume an emission of oxides of nitrogen of 5 grams per vehicle mile and an average mileage of 15 miles per gallon of gasoline. How many kilograms of nitrogen oxides were released into the atmosphere in the United States in \(1968 ?\)

Zn can reduce \(\mathrm{NO}_{3}^{-}\) to \(\mathrm{NH}_{3}(\mathrm{g})\) in basic solution. (The following equation is not balanced.) $$\begin{aligned}\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})+& \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \\\&\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{g})\end{aligned}$$ The \(\mathrm{NH}_{3}\) can be neutralized with an excess of \(\mathrm{HCl}(\mathrm{aq})\) Then, the unreacted HCl can be titrated with NaOH. In this way a quantitative determination of \(\mathrm{NO}_{3}^{-}\) can be achieved. A 25.00 mL sample of nitrate solution was treated with zinc in basic solution. The \(\mathrm{NH}_{3}(\mathrm{g})\) was passed into \(50.00 \mathrm{mL}\) of \(0.1500 \mathrm{M} \mathrm{HCl} .\) The excess \(\mathrm{HCl}\) required \(32.10 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for its titration. What was the \(\left[\mathrm{NO}_{3}\right]\) in the original sample?
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