Kaplan MCAT General Chemistry Review

The key to answering this question is to understand the types of intermolecular forces that exist in each of these molecules. Kr (III) is a noble gas with a full octet, so the only intermolecular forces present are dispersion forces (also called London forces), the weakest type of intermolecular forces. This means that these molecules are held together extremely loosely and will be the easiest to transition from the organized liquid phase to the disorganized gaseous phase. In fact, because Kr is a noble gas, assume that it’s usually in the gaseous state and therefore has a low boiling point. Next, decide between two polar molecules: acetone (I) and isopropyl alcohol (IV). Both have the benefit of dipole–dipole forces, which are stronger than dispersion forces. Dipole–dipole forces arrange polar molecules such that the positive ends associate with their neighbors’ negative ends, which keeps these molecules in close proximity to one another. So which boils at a higher temperature? Alcohols are known for their ability to hydrogen bond, which significantly raises boiling point. Hydrogen bonding occurs when an electronegative atom (fluorine, oxygen, or nitrogen) binds to hydrogen, stripping it of electron density and leaving it with a partial positive charge. That partial positive charge can then associate with nearby partial negative charges. These forces are even stronger than dipole–dipole interactions, so isopropyl alcohol will boil at a higher temperature than acetone. Finally, the strongest interactions are ionic, and these exist in compounds such as potassium chloride (II).

5. C

The central atom in CO₃, carbon, 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 CO₃’s geometry trigonal planar. Alternatively, ClF₃ also has three bonds, one to each of three fluoride atoms. However, chloride still maintains two extra lone pairs (without which the formal charge on the central chloride atom is +4; with the two lone pairs it is zero, a more stable configuration). 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) is true but does not account for the difference in geometry. Similarly, although (B) is true, the charge does not explain a difference in geometry. (D) is incorrect; CO₃ has no lone pairs on its central atom, while ClF₃ has two lone pairs.

6. A

The 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 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. H₂O 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 HCN’s. Sulfur dioxide has a similar bent configuration, so its dipole will be smaller than that of HCN. Another consideration for sulfur dioxide is that oxygen and sulfur do not have a large difference in electronegativity, so even the individual bond dipoles are smaller than those we have seen so far. CCl₄ has a 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. D

Bond lengths decrease as the bond order increases, and they also decrease in a trend moving up the periodic table’s columns or to the right across to the periodic table’s rows. In this case, because both C₂H₂ and NCH have triple bonds, we cannot compare the bond lengths based upon 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 NCH is likely to hold its electrons closer, or in a shorter radius, than the carbon in C₂H₂. (B) is incorrect because there are no significant resonance structures contributing to the character of either triple bond. (C) expresses a true statement but one that is irrelevant to the length of the carbon–carbon triple bond.

8. C