/*! 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

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of NO? (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+}\), and \(\mathrm{NO}^{-}\), and describe the magnetic properties of each (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion, and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\) (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) Why are the \(\pi_{2 p}\) MOs lower in energy than the \(\pi_{2 p}^{*}\) MOs?

Use average bond enthalpies (Table 8.4) to estimate \(\Delta H\) for the atomization of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) : $$ \mathrm{C}_{6} \mathrm{H}_{6}(g) \longrightarrow 6 \mathrm{C}(g)+6 \mathrm{H}(g) $$ Compare the value to that obtained by using \(\Delta H_{f}^{\circ}\) data given in Appendix \(C\) and Hess's law. To what do you attribute the large discrepancy in the two values?

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN}\), (b) \(\mathrm{SO}_{3}{ }^{2-}\), (c) \(S F_{4}\) (d) \(\mathrm{PF}_{6}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).
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