10. C
This question addresses the issue of when the octet rule can be violated. All atoms that are in the third or higher period have d-orbitals, each of which can hold 10 electrons. The typical eight “octet” electrons reside in s- and p-orbitals. (A) and (B) do not explain why more than eight electrons can be held. Electronegativity is irrelevant to whether or not an atom can exceed the octet rule, making (D) incorrect.
11. C
All of the listed types of forces dictate interactions among different types of molecules. However, noble gases are entirely uncharged and do not have polar covalent bonds, ionic bonds, or dipole moments. Recognize that the only types of forces listed that could relate to noble gases could be van der Waals forces (A) or dispersion forces (C). Of these two, van der Waals forces is an umbrella term that includes both dispersion and dipole–dipole interactions (which you’ve already eliminated). Because not all van der Waals forces apply, rule that out as the correct answer and stick with dispersion forces. Dispersion forces are a specific type of interaction that occurs among all bonded atoms due to the unequal sharing of electrons at any given moment in the electron’s orbit. This unequal sharing allows for instantaneous partial positive and partial negative charges within the molecule. Though these interactions are small, they are necessary for liquefaction.
12. B
The key to this question is understanding the pattern of filling for s- and d-orbitals among the transition metals. There are 24 electrons in chromium, 6 more than are present in argon. Where will these six electrons lie? It is advantageous for d-orbitals to be half filled. Therefore, one electron will fill each of the five 3d-orbitals. It will take less energy to put the sixth electron into an s-orbital than it would to add it to one of the d-orbitals that already has an electron (and force it to be no longer half filled).
13. A
In this Lewis diagram, the PO43- molecule has an overall formal charge of -3. The four oxygens each would be assigned a formal change of -1, based on the following formula: Formal charge = V (valence electrons in the free atom) -½ Nbonding (electrons shared in bonds) -Nnonbonding (lone pairs/free electrons). For each oxygen, we calculate: FC = 6 - ½ (2) - 6 = -1. For the central phosphorus, assume then that with a total formal charge of -3 and four oxygens with a change of -1 each, the phosphorus must have a formal charge of +1. Alternatively, calculate its formal charge as FC = 5 - ½ (8) - 0 = +1. Considering this molecule’s other resonance structures, you’d come to the same conclusion—that phosphorus is the most positive atom.
14. C
The reaction in this question shows a water molecule, which has two lone pairs of electrons on the central oxygen, combining with a free hydrogen ion. The resulting molecule, H3O+, has formed a new bond between H+ and H2O. This bond is created through the sharing one of oxygen’s lone pairs with the free H+ ion. This is essentially a donation of a shared pair of electrons from a Lewis base (H2O) to a Lewis acid (H+, electron acceptor). The charge in the resulting molecule is +1, and it is mostly present on the central oxygen, which now only has one lone pair. This type of bond, formed from a Lewis acid and Lewis base, is called a coordinate covalent bond.
15. B
NH3 has three hydrogen atoms bonded to the central nitrogen and one lone pair on the central nitrogen. These four groups—three atoms, one lone pair—lead NH3 to be sp3 hybridized. By hybridizing all three p-orbitals and the one s-orbital, four groups are arranged about the central atom, maximizing the distances between the groups to minimize the energy of the configuration. NH3’s hybridization leads to its tetrahedral electronic geometry yet trigonal pyramidal molecular geometry. In contrast, BF3 has three atoms but no lone pairs, resulting in sp2 hybridization. Its shape is called trigonal planar. (A) is incorrect; although BF3 does has three bonded atoms and no lone pairs, its geometry is not trigonal pyramidal. (C) is tempting because NH3 has a lone pair, but this answer choice is not complete enough to explain the differences in molecular geometry. Finally, although (D) is also true, the polarity of the molecules does not explain their geometry; rather, the molecules’ different geometries contribute to the overall polarity of the molecules.
16. B
Most atoms require eight valence electrons to follow the octet rule. However, some atoms, like beryllium, can have fewer than eight valence electrons (suboctet). As a result, when bonding with chloride, beryllium is likely to form only two bonds, using its own two outer valence electrons and one from each chloride to form BeCl2, as drawn in (B). (A) is incorrect because Be cannot both bond to chloride and have lone pairs. (C) assumes that all three atoms need their octets filled, and (D) places 10 electrons on each chlorine.
17. B
This answer requires an understanding of the trends that cause higher or lower bond energies. Bonds of high energy are those that are difficult to break. These bonds tend to have more shared pairs of electrons and, thus, cause a stronger attraction between the two atoms in the bonds. This stronger attraction also means that the bond length of a high-energy, high-order bond (i.e., a triple bond) is shorter than that of its lower-energy counterparts (i.e., single or double bonds). Thus, (A) is incorrect; bond energy increases with decreasing bond length. Single bonds are longer than double bonds. Thus, (C) is incorrect; single bonds are easier to break than double bonds. Finally, as previously discussed, bond energy is inversely related to bond length, making (D) incorrect.
18. A
Polarity is precisely described by the first answer choice. Dipole moments describe the relationship of shared electrons between two bonded atoms. If there is a dipole moment between two atoms, then the electrons in their shared bond are preferentially centered on one atom, typically the more electronegative of the two. The polarity of a molecule is then defined as the vector sum of these dipole moments in the molecule’s three-dimensional configuration. For example, although individual carbon–chloride bonds have dipole moments with a partial negative charge on chloride, the molecule CCl4 has a tetrahedral configuration. Thus, the four polarized bonds cancel each other out for no net dipole moment, creating a nonpolar molecule. Based on this description, it is clear that (B) is incorrect, because molecular geometry is an essential component of determining molecular polarity. For the same reasons described previously, it is possible for a nonpolar molecule to contain one or more polar bonds, making (C) incorrect. Finally, it is possible for a molecule to contain at least one nonpolar bond and yet be a polar molecule. For example, CH3CH2Cl has a nonpolar bond (C–C), yet a molecular dipole exists. In its tetrahedral arrangement, the one polar bond between C–Cl will have a dipole moment, creating a net polarity of the molecule in the direction of the chloride atom.
Chapter 4: Compounds and Stoichiometry