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Chapter 22: Problem 69

All the group 15 elements form trifluorides, but nitrogen is the only group 15 element that does not form a pentafluoride. (a) Suggest a reason why nitrogen does not form a pentafluoride. (b) The observed bond angle in \(\mathrm{NF}_{3}\) is approximately \(102.5^{\circ} \mathrm{C} .\) Use VSEPR theory to rationalize the structure of the \(\mathrm{NF}_{3}\) molecule.

Short Answer

Expert verified
Nitrogen does not form a pentafluoride because its outer shell can only accommodate eight electrons. The \(\mathrm{NF}_{3}\) molecule forms a trigonal pyramidal shape due to three bonded electron pairs and one lone pair of electrons around the nitrogen atom, resulting in an observed bond angle of approximately \(102.5^{\circ} \mathrm{C}\).

Step by step solution

01

Analyze the Electron Configuration of Nitrogen

Nitrogen belongs to group 15 of the periodic table and has 5 electrons in its outermost shell. According to the octet rule, atoms tend to combine in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas.
02

Reason for Nitrogen Not Forming Pentafluoride

To form a pentafluoride, nitrogen would need to share five of its electrons with five fluorine atoms, which would result in a total of ten electrons around the nitrogen atom. However, nitrogen's outer shell can only accommodate eight electrons, so it does not form a pentafluoride.
03

Analyze the VSEPR Theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory states that the electron pairs around a central atom arrange themselves as far apart as possible to minimize electron-electron repulsion.
04

Apply the VSEPR Theory to NF3

In the \(\mathrm{NF}_{3}\) molecule, nitrogen is the central atom. It is surrounded by three bonded electron pairs (due to the three fluorine atoms) and one lone pair of electrons. According to the VSEPR theory, this forms a trigonal pyramidal shape, resulting in a bond angle of approximately \(102.5^{\circ} \mathrm{C}\), which is observed in \(\mathrm{NF}_{3}\).

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Most popular questions from this chapter

All of the following substances are bases except for (a) \(\mathrm{H}_{2} \mathrm{NNH}_{2} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{HN}_{3} ;\) (d) \(\mathrm{NH}_{2} \mathrm{OH}\); (e) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\).

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?

The structures of the \(\mathrm{NH}_{3}\) and \(\mathrm{NF}_{3}\) molecules are similar, yet the dipole moment for the \(\mathrm{NH}_{3}\) molecule is rather large (1.47 debye) and that of the NF \(_{3}\) molecule is rather small (0.24 debye). Provide an explanation for this difference in the dipole moments.

In the extraction of bromine from seawater (reaction 22.3), seawater is first brought to a pH of 3.5 and then treated with \(\mathrm{Cl}_{2}(\mathrm{g}) .\) In practice, the \(\mathrm{pH}\) of the seawater is adjusted with \(\mathrm{H}_{2} \mathrm{SO}_{4},\) and the mass of chlorine used is \(15 \%\) in excess of the theoretical. Assuming a seawater sample with an initial pH of 7.0 a density of \(1.03 \mathrm{g} \mathrm{cm}^{-3}\), and a bromine content of 70 ppm by mass, what masses of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{Cl}_{2}\) would be used in the extraction of bromine from \(1.00 \times 10^{3} \mathrm{L}\) of seawater?

Each of the following compounds produces \(\mathrm{O}_{2}(\mathrm{g})\) when strongly heated: (a) \(\mathrm{HgO}(\mathrm{s}) ;\) (b) \(\mathrm{KClO}_{4}(\mathrm{s})\) (c) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{s}) ;\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) .\) Write a plausible equation for the reaction that occurs in each instance.
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