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Kaplan MCAT General Chemistry Review

According to the passage, molecules most likely to cross the blood-brain barrier are hydrophobic or nonpolar. Thus, ion-dipole and dipole-dipole interactions [(A) and (B)] are unlikely to be the most important forces governing intermolecular interactions among these molecules (because they require charge and/or polarity). Although hydrogen bonding (C) might have a role in these molecules, hydrogen bonds, because they occur between atoms in otherwise polarized bonds (a hydrogen with a partial negative charge and a lone pair or otherwise electronegative atom with a negative charge), would be unlikely to facilitate transport in a hydrophobic environment. Dispersion forces (D) refer to the unequal sharing of electrons that occurs among nonpolar molecules as the result of rapid polarization and counterpolarization; these are likely to be the prevailing intermolecular forces affecting molecules that move easily through the blood-brain barrier.

5. D

All three items are false. A polar molecule can still have a formal charge of zero, because the molecule’s formal charge is the sum of the formal charges of the individual atoms. Each atom could individually still have a positive or negative formal charge, calculated by the formula:

FC = valence electrons - ½ bonding electrons - nonbonding electrons

Thus, a molecule with a formal charge of zero could have multiple polarized bonds and/or multiple atoms with positive/negative formal charges, making it unlikely to permeate through the nonpolar blood-brain barrier. Similarly, having a negative formal charge on the molecule overall would be unfavorable to move through a nonpolar barrier. Finally, two molecules, both with formal charges of zero, could have very different characteristics, making them more or less likely to permeate the blood-brain barrier. For example, one compound could be comprised entirely of nonpolar bonds, with all atoms of formal charge of zero, both of which would make it favorable to pass through the blood-brain barrier. Size also plays a key role, as large molecules do not readily pass through the barrier. A separate molecule with a formal charge of zero could, as described above, contain positive or negative formal charges on different atoms and/or have polar bonds, making it pass through the barrier less readily.

6. A

In cisplatin, a molecule with square planar geometry, two chloride atoms and two ammonia molecules each are bonded directly to the central platinum, without any remaining lone pairs on the central platinum. The NH₃ groups bond to the central platinum by donating a lone pair of electrons into an unfilled orbital of the platinum atom. As such, NH₃ is acting as a Lewis base, and platinum is acting as a Lewis acid, and they form a coordinate covalent bond. A polar covalent bond (B) is not formed from this type of donation of a lone pair; an example of a polar covalent bond would be the N–H bonds in the NH₃ group, with partial negative charge on the nitrogen and partial positive charge on the hydrogen. In (C) and (D), the Pt–N bond is formed by a Lewis acid/ base relationship, which is not the case for a nonpolar covalent bond or ionic bond.

7. B

Although the geometry of this carbon atom is likely to be changed somewhat by the ring strain on the adjacent ring structures, it is still most likely to approximate a tetrahedral geometry. This carbon is bonded to four groups: the two phenyl groups, a nitrogen, and the additional carbon atom in the adjacent carboxyl group. The tetrahedral geometry (B) maximizes the space among these four groups. Octahedral geometry (D) typically refers to a central atom surrounded by six groups, not four.

8. C

Resonance structures serve to spread out formal charge. The most important resonance structures minimize or eliminate the formal charge on individual atoms. As a result, polarity in any single bond might be minimized. (A) is essentially the opposite of this argument. (B) is incorrect; it is possible for molecules with or without resonance to be polar, and this is too broad of a generalization to be true. (D) is incorrect because polar bonds will intrinsically place a partial negative charge on the more electronegative atom.

Important resonance structures would likely further accentuate this inclination, placing extra electrons on more electronegative atoms. Counteracting this effect and essentially “removing” electrons from highly electronegative atoms to counterbalance the natural polarity of the bond would form an extremely high-energy, unfavorable, and thus unimportant, resonance structure.

PASSAGE II

9. B

The order of this reaction can be determined by the equation r = k[A]^x [B]^y. You must divide r₃ by r₁ to get: = . This simplifies to 4 = 4^x, so x = 1. This means that H⁺ is a first-order reactant.