Problem 99b
Sodium azide is a shock-sensitive compound that releases N2 upon physical impact. The compound is used in automobile airbags. The azide ion is N3-. (b) State the hybridization of the central N atom in the azide ion.
Problem 100d
In ozone, O3, the two oxygen atoms on the ends of the molecule are equivalent to one another. (d) How many electrons are delocalized in the p system of ozone?
Problem 101a
Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:
(a) Predict the bond angles around each of the carbon atoms and sketch the molecule.
Problem 101b
Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:
(b) From left to right, what is the hybridization of each carbon atom in butadiene?
Problem 101c
Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:
(c) The middle C¬C bond length in butadiene (1.48 Å) is a little shorter than the average C¬C single bond length (1.54 Å). Does this imply that the middle C¬C bond in butadiene is weaker or stronger than the average C¬C single bond?
Problem 102a
The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero.
Problem 102e
The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? Would you expect the molecule to be planar for both Lewis structures?
Problem 103c
The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as π2p-π2p*, σs-σ2s*, and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (c) The electronic absorption spectrum of the N2 molecule has the lowest energy peak at 170 nm. To what orbital transition does this correspond?
Problem 104a
One of the molecular orbitals of the H2− ion can be sketched as follows:
a. Is the molecular orbital a 𝜎 or 𝜋 MO? Is it bonding or antibonding?
Problem 104d
One of the molecular orbitals of the H2− ion can be sketched as follows:
d. Compared to the H—H bond in H2, the H—H bond in H2− is expected to be which of the following?
i. shorter and stronger
ii. longer and stronger
iii. shorter and weaker
iv. longer and weaker or
v. the same length and strength
Problem 105
Place the following molecules and ions in order from smallest to largest bond order: H2+,B2,N2+,F2+, and Ne2.
Problem 107b2
Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C12H10N2. A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are
(Recall the shorthand notation used for benzene.) (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? How many unhybridized atomic orbitals are there on the N and the C atoms in hydrazobenzene?
Problem 107c
Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C12H10N2. A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are
(Recall the shorthand notation used for benzene.) (c) Predict the N¬N¬C angles in each of the substances.
Problem 108d
Carbon monoxide, CO, is isoelectronic to N2. (d) Would you expect the p2p MOs of CO to have equal atomic orbital contributions from the C and O atoms? If not, which atom would have the greater contribution?
Problem 109a
The energy-level diagram in Figure 9.40 shows that the sideways overlap of a pair of p orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the to the molecular orbital. a. Assuming this electronic transition corresponds to the HOMO–LUMO transition, what is the HOMO in ethylene?
Problem 109b
The energy-level diagram in Figure 9.40 shows that the sideways overlap of a pair of p orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the to the molecular orbital. b. Assuming this electronic transition corresponds to the HOMO–LUMO transition, what is the LUMO in ethylene?
Problem 109c
The energy-level diagram in Figure 9.40 shows that the sideways overlap of a pair of p orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the to the molecular orbital. c. Is the bond in ethylene stronger or weaker in the excited state than in the ground state? Why?
- A compound composed of 6.7% H, 40.0% C, and 53.3% O has a molar mass of approximately 60 g>mol. (c) What is the geometry and hybridization of the C atom that is bonded to 2 O atoms?
Problem 110
Problem 111a
Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur tetrafluoride monoxide 1OSF42 according to the following unbalanced reaction: SF41g2 + O21g2¡OSF41g2 The O atom and the four F atoms in OSF4 are bonded to a central S atom. (a) Balance the equation.
Problem 111b
Sulfur tetrafluoride (SF4) reacts slowly with O2 to form sulfur tetrafluoride monoxide (OSF4) according to the following unbalanced reaction: SF4(g) + O2(g) → OSF4(g) The O atom and the four F atoms in OSF4 are bonded to a central S atom. (b) Write a Lewis structure of OSF4 in which the formal charges of all atoms are zero.
Problem 111c
Sulfur tetrafluoride (SF4) reacts slowly with O2 to form sulfur tetrafluoride monoxide (OSF4) according to the following unbalanced reaction: SF4(g) + O2(g) → OSF4(g) The O atom and the four F atoms in OSF4 are bonded to a central S atom. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic?
Problem 111e
Sulfur tetrafluoride 1SF42 reacts slowly with O2 to form sulfur
tetrafluoride monoxide 1OSF42 according to the following
unbalanced reaction:
SF41g2 + O21g2¡OSF41g2
The O atom and the four F atoms in OSF4 are bonded to a
central S atom.
(e) For each of the molecules you drew in part (d), state how many
fluorines are equatorial and how many are axial.
Problem 113a
Methyl isocyanate, CH3NCO, was made infamous in 1984 when an accidental leakage of this compound from a storage tank in Bhopal, India, resulted in the deaths of about 3800 people and severe and lasting injury to many thousands more. (a) Draw a Lewis structure for methyl isocyanate.
Ch.9 - Molecular Geometry and Bonding Theories