MCAT General Chemistry Review - Alexander Stone Macnow, MD 2019-2020
Bonding and Chemical Interactions
Explanations to Discrete Practice Questions
1. BCarbon monoxide, CO, has a triple bond between carbon and oxygen, with the carbon and oxygen each retaining one lone pair. In polar covalent bonds, the difference in electronegativity between the bonded atoms is great enough to cause electrons to move disproportionately toward the more electronegative atom but not great enough to transfer electrons completely. This is the case for CO. Oxygen is significantly more electronegative than carbon, so electrons will be disproportionately carried on the oxygen, leaving the carbon atom with a slight positive charge.
2. BTo answer this question, one must understand the contribution of resonance structures to average formal charge. In (B), there are three possible resonance structures. Each of the three oxygen atoms carries a formal charge of −1 in two out of the three structures. This averages to approximately charge on each oxygen atom, which is more negative than in the other answer choices. Both water and formaldehyde, (A) and (D), have no formal charge on the oxygen. Ozone, (C), has a on two of the three oxygens and a +1 charge on the central oxygen.
3. CThe two greatest contributors are structures I and II. Resonance structures are representations of how charges are shared across a molecule. In reality, the charge distribution is a weighted average of contributing resonance structures. The most stable resonance structures are those that minimize charge on the atoms in the molecule; the more stable the structure, the more it will contribute to the overall charge distribution in the molecule. Structures I and II minimize formal charges, so will be the largest contributors to the resonance hybrid.
4. DThe key to answering this question is to understand the types of intermolecular forces that exist in each of these molecules because larger intermolecular forces correspond to higher boiling points. Kr is a noble gas with a full octet, so the only intermolecular forces present are London dispersion forces, the weakest type of intermolecular forces. Acetone and isopropyl alcohol are both polar, so both have dipole—dipole interactions, which are stronger than dispersion forces. However, isopropyl alcohol can also form hydrogen bonds, increasing its boiling point. Finally, the strongest interactions are ionic bonds, which exist in potassium chloride.
5. CThe central carbon in carbonate has no lone pairs. It has three resonance structures, each of which involves a double bond between carbon and one of the three oxygens. Having made four bonds, carbon has no further orbitals for bonding or to carry lone pairs. This makes carbonate’s geometry trigonal planar. Alternatively, ClF3 also has three bonds; however, chloride still maintains two extra lone pairs. These lone pairs each inhabit one orbital, meaning that the central chloride must organize five items about itself: three bonds to fluorides and two lone pairs. The best configuration for maximizing the distance between all of these groups is trigonal bipyramidal. (A) and (B) are true statements but do not account for the difference in geometry.
6. AThe best way to approach this problem is to draw the structure of each of these molecules, then consider the electronegativity of each bond as it might contribute to an overall dipole moment. HCN is the correct answer because of the large differences in electronegativity aligned in a linear fashion. There is a strong dipole moment in the direction of nitrogen, without any other moments canceling it out. Water, (B), has two dipole moments, one from each hydrogen pointing in the direction of oxygen. The molecule is bent, and the dipole moments partially cancel out. There is a molecular dipole, but it is not as strong as in HCN. Sulfur dioxide, (C), has a similar bent configuration, and its dipole will again be smaller than that of HCN. Further, oxygen and sulfur do not have as large a difference in electronegativity, so even the individual bond dipoles are smaller than those in the other molecules. CCl4, (D), has tetrahedral geometry. Although each of the individual C—Cl bonds is highly polar, the orientation of these bonds causes the dipoles to cancel each other out fully, yielding no overall dipole moment.
7. DBond lengths decrease as the bond order increases, and they also decrease with larger differences in electronegativity. In this case, because both C2H2 and HCN have triple bonds, we cannot compare the bond lengths based on bond order. We must then rely on other periodic trends. The bond length decreases when moving to the right along the periodic table’s rows because more electronegative atoms have shorter atomic radii. The nitrogen in HCN is likely to hold its electrons closer, or in a shorter radius, than the carbons in C2H2.
8. CElectronegative atoms bonded to hydrogen disproportionately pull covalently bonded electrons toward themselves, which leaves hydrogen with a partial positive character. That partial positive charge is attracted to nearby negative or partial negative charges, such as those on other electronegative atoms.
9. CFirst recall that ammonium is NH4+, while ammonia is NH3. Ammonium is formed by the association of NH3, an uncharged molecule with a lone pair on the nitrogen, with a positively charged hydrogen cation. In other words, NH3 is a Lewis base, while H+ is a Lewis acid. This type of bonding between a Lewis acid and base is a coordinate covalent bond.
10. CAll atoms in the third period or greater have d-orbitals, which can hold an additional 10 electrons. The typical “octet” electrons reside in s- and p-orbitals, but elements in period 3 or higher can place electrons into these d-orbitals.
11. CAll of the listed types of forces describe interactions between different types of molecules. However, noble gases are entirely uncharged and do not have polar covalent bonds, ionic bonds, or dipole moments. Therefore, the only intermolecular forces experienced by noble gases are London dispersion forces. Although these interactions are small in magnitude, they are necessary for condensation into a liquid.
12. AIn this Lewis diagram, the phosphate molecule has an overall formal charge of −3. The four oxygen atoms would each be assigned a formal charge of −1. Given the overall charge of −3 and the −1 charge on each oxygen, the phosphorus must have a formal charge of +1.
13. CThe reaction in this question shows a water molecule, which has two lone pairs of electrons on the central oxygen, combining with a free hydrogen cation. The resulting molecule, H3O+ has formed a new bond between H+ and H2O. This bond is created via the sharing of one of oxygen’s lone pairs with the free H+ ion. This represents the donation of a shared pair of electrons from a Lewis base (H2O) to a Lewis acid (H+, electron acceptor). This type of bond is called a coordinate covalent bond.
14. CNH3 has three hydrogen atoms bonded to the central nitrogen, which also has a lone pair. These four groups—three atoms, one lone pair—lead NH3 to assume tetrahedral electronic geometry yet trigonal pyramidal molecular geometry. The nitrogen in ammonia is 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 molecule with a tetrahedral configuration. In contrast, BF3 has three atoms and no lone pairs, resulting in sp2-hybridization. Its shape is called trigonal planar.
15. BThis 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 such as a triple bond is shorter than that of its lower-energy counterparts such as single or double bonds.