/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 The nitrogen atoms in \(\mathrm{... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 52

The nitrogen atoms in \(\mathrm{N}_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has a stronger \(\mathrm{N}-\mathrm{N}\) bond?

Short Answer

Expert verified
The Lewis structures are N鈮 for N2 and H2N-NH2 for N2H4. The nitrogen atoms in N2 are sp hybridized, while those in N2H4 are sp3 hybridized. The N-N bond in N2 (triple bond) is stronger than the N-N bond in N2H4 (single bond).

Step by step solution

01

Draw Lewis Structures

To draw the Lewis structures for N2 and N2H4, start by counting the total number of valence electrons in each molecule. Nitrogen has 5 valence electrons, and hydrogen has 1. N2: 2 nitrogen atoms * 5 electrons each = 10 valence electrons N2H4: 2 nitrogen atoms * 5 electrons each + 4 hydrogen atoms * 1 electron each = 14 valence electrons Now, draw the Lewis structures. For N2: 1. Place two nitrogen atoms together, with one single bond. 2. Add 3 more pairs of electrons to each nitrogen atom to complete the octet. 3. To minimize formal charge, turn one lone pair into a bonding pair, resulting in a triple bond. Lewis Structure for N2: N鈮 For N2H4: 1. Place two nitrogen atoms together with a single bond. 2. Add 2 hydrogen atoms to each nitrogen atom, with single bonds. 3. Add 2 remaining electrons (1 lone pair) to each nitrogen atom. Lewis Structure for N2H4: H2N-NH2
02

Determine Hybridization

To find the hybridization, count the electron groups around each nitrogen atom. Electron groups include single bonds, double bonds, triple bonds, and lone pairs. For N2: Each nitrogen atom has 1 triple bond and 1 lone pair, which corresponds to 2 electron groups. This suggests sp hybridization. For N2H4 (hydrazine): Each nitrogen atom has 2 single bonds to hydrogen atoms, 1 single bond to another nitrogen atom, and 1 lone pair, which corresponds to 4 electron groups. This suggests sp3 hybridization. Nitrogen in N2: sp hybridization Nitrogen in N2H4: sp3 hybridization
03

Compare N-N Bond Strength

The N-N bond strength generally increases in the order single bond < double bond < triple bond. As a result, the N-N bond in N2 (triple bond) is stronger than the N-N bond in N2H4 (single bond). The molecule with a stronger N-N bond is N2.

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Most popular questions from this chapter

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \mathrm{~A}\) (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5\), is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-} ?\) (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-} .(\mathrm{e}) \mathrm{It}\) is often observed that \(\sigma\) bonds that involve an sp hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

(a) What conditions must be met if a molecule with polar bonds is nonpolar? (b) What geometries will give nonpolar molecules for \(\mathrm{AB}_{2}, \mathrm{AB}_{3}\), and \(\mathrm{AB}_{4}\) geometries?

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \mathrm{~A}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ}\). The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

The \(\mathrm{PF}_{3}\) molecule has a dipole moment of \(1.03 \mathrm{D}\), but \(\mathrm{BF}_{3}\) has a dipole moment of zero. How can you explain the difference?

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\mathrm{a}} ; \mathrm{PCl}_{3}\) \(100.3^{\circ} ; \mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{Pl}_{3}, 102.0^{\circ} .\) The trend is generally at- tributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(X-P-X\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCI}_{4}\)
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