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Problem 87c
Consider the following XF4 ions: PF4-, BrF4-, ClF4+, and AlF4-. (c) Which of the ions will have an octahedral electron-domain geometry?
Problem 88a
Consider the molecule PF4Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry.
Problem 88c
Consider the molecule PF4Cl. (c) Predict the molecular geometry of PF4Cl. How did your answer for part (b) influence your answer here in part (c)?
Problem 88d
Consider the molecule PF4Cl. (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?
- Although I3- is a known ion, F3- is not. (d) Yet another classmate says F3- does not exist because F is too small to make bonds to more than one atom. Is this classmate possibly correct?
Problem 89
Problem 90
Fill in the blank spaces in the following chart. If the molecule column is blank, find an example that fulfills the conditions of the rest of the row. Molecule Electron-Domain Hybridization Dipole Geometry of Central Atom Moment? Yes or No CO2 sp3 Yes sp3 No Trigonal planar No SF4 Octahedral No sp2 Yes Trigonal bipyramidal No XeF2
- From their Lewis structures, determine the number of σ and π bonds in each of the following molecules or ions: (a) hydrazine, N2H4 (b) hydrogen cyanide, HCN (c) sulphur trioxide, SO3
Problem 91
Problem 92e
Ethyl propanoate, CH3CH2COOCH2CH3, gives a fruity pineapple-like smell. (e) What are the approximate bond angles around each carbon atom in the molecule?
Problem 93c
An AB5 molecule adopts the geometry shown here. (c) Suppose the B atoms are halogen atoms. Of which group in the periodic table is atom A a member: (i) Group 15, (ii) Group 16, (iii) Group 17, (iv) Group 18, or (v) More information is needed?
- The O-H bond lengths in the water molecule 1H2O are 96 pm, and the H-O-H angle is 104.5°. The dipole moment of the water molecule is 1.85 D. (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?
Problem 95
Problem 96a
a) Predict the electron-domain geometry around the central S atom in SF2, SF4, and SF6.
Problem 96b
(b) The anion IO4- has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of IO65-.
Problem 100a
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-. (a) Draw the Lewis structure of the azide ion that minimizes formal charge (it does not form a triangle). Is it linear or bent?
Problem 100b
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.
- For one of the resonance forms of ozone, O3, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons?
Problem 101
Problem 101d
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 102a
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 102b
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 102c
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 103a
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 103c
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. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of B and N, do the formal charges seem favorable or unfavorable? What are the formal charges on the atoms in the Lewis structure from part (b)?
Problem 103e
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?
- 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 π→π*, σ→σ*, 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. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for N₂. (b) Is N₂ paramagnetic or diamagnetic in its first excited state? (d) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N≡N bond in the first excited state stronger or weaker compared to that in the ground state?
Problem 104
Problem 104c
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 105a
One of the molecular orbitals of the H2- ion is sketched below:
(a) Is the molecular orbital a s or p MO? Is it bonding or antibonding?
Problem 105d
One of the molecular orbitals of the H2- ion is sketched below: (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 106
Place the following molecules and ions in order from smallest to largest bond order: N22+, He2+, Cl2 H2-, O22-.
- Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMOLUMO gap; see Exercise 9.104). Suppose you have two samples, one is lycopene which is responsible for the red color in tomato, and the other is curcumin which is responsible for the yellow color in turmeric. Which one has the larger HOMO-LUMO gap?
Problem 108
Problem 109b2
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 109c
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.