3. C
The 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 that it will contribute to the overall charge distribution in the molecule. Structure I has a -1 charge on one oxygen and a single positive charge on the central nitrogen atom. Structure II has the exact same distribution, except the negative charge is on the other oxygen. It is equally as stable as structure I, so they both contribute equally to the hybrid. Structure III involves two negative charges (one on each oxygen) and a +2 charge on the central nitrogen. With so many charges, it’s less stable than structures I and II and, thus, is not an important resonance structure for NO2.
4. D
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
(C) correctly describes the underlying forces of hydrogen bonding. Electronegative atoms bound 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. (A) is not true; hydrogen has little electronegativity and does not hold its valence electrons closely. (B) is a true statement but not descriptive enough to explain hydrogen bonding. (D) is not correct; although these bonds are highly polarized, they are not ionic.
9. D
First recall that ammonium is NH4+, while ammonia (commonly confused) is NH3. It helps to associate the suffix -ium with a charged form of the molecule. Once you remember that ammonium is NH4+, eliminate any answer choice that only accounts for three bonds [answer choice (A)]v. Next, it helps to recall that ammonium is formed by the association of NH3 (uncharged, with a lone pair on the nitrogen) with a positively charge hydrogen cation (no lone pairs). In other words, NH3 is a Lewis base, while H+ is a Lewis acid. This type of bonding, between Lewis acid and base, is a coordinate covalent bond. Thus, you know that there is one coordinate covalent bond in this molecule, making (D) the correct answer.