Kaplan MCAT General Chemistry Review

This question is simple if you recall that “periods” name the horizontal rows of the periodic table, while “families” refer to its columns. Within the same period, an additional valence electron is added with each step towards the right side of the table.

11. B

This question requires knowledge of the trends of electronegativity within the periodic table. Electronegativity increases as one moves from left to right across periods for the same reasons that effective nuclear force increases. Electronegativity decreases as one moves down the periodic table, because there are more electron shells separating the nucleus from the outermost electrons. The noble gases, however, also have extremely low electronegativities because they already have full valence shells and do not desire additional electrons. The most electronegative atom in the periodic table is fluorine. The answer choice closest to fluorine is (B), chlorine. Although iodine (D) will be fairly electronegative, its higher atomic radius and position farther down on the periodic table make it less electronegative than Cl. The remaining answer choices, Mg and Li, are elements with very low electronegativities. Because they have only two and one valence electrons, respectively, they are more likely to lose these electrons in a bond than to gain electrons; the loss of electrons would leave them with a full octet. Metals like these are often called electropositive.

12. B

Electron affinity is related to several factors, including atomic size (radius) and filling of the valence shell. As atomic radius increases, the distance between the inner protons in the nucleus and the outermost electrons increases, thereby decreasing the attractive forces between protons and electrons. Additionally, as more electron shells are added from period to period, these shells shield the outermost electrons increasingly from inner protons. As a result, increased atomic radius will lead to lower electron affinity. Because atoms are in a low-energy state when their outermost valence electron shell is filled, atoms needing only one or two electrons to complete this shell will have high electron affinities. In contrast, atoms with already full valence shells (a full octet of electrons) will have very low electron affinities, because adding an extra electron would require a new shell. With this information, it is clear that (C) and (D) will likely have lower electron affinities than (A) and (B) because there is an extra electron shell “shielding” the nucleus from the outer electrons. Answer (A) is incorrect because its valence electron shell is already full with a complete octet, granting it extremely low electron affinity. Finally, (B) has one electron missing from its outermost shell, as does (D). This valence electron configuration is conducive to wanting to accept electrons readily—or to having a high electron affinity. (B) is the configuration of chlorine, while (D) is bromine. (B) is a better answer than (D), however, because the additional shell of electrons shielding the nucleus in (D) will decrease its electron affinity when compared with (B).

13. C

Electronegativity is the only property listed that has a consistent inverse correlation with atomic radius. Highly electronegative atoms hold bonding electrons tightly, while atoms with low electronegativity hold bonding electrons loosely. In atoms with large atomic radii, the distance between the outermost electrons used for bonding and the central positively charged nucleus is large. This increased distance means that the positively charged nucleus has little ability to attract new, bonding electrons toward it. In comparison, if the atomic radius is small, the force from the positively charged protons will have a stronger effect, because the distance through which they have to act is decreased. (A) is incorrect because atomic radius decreases when moving from left to right across periods in the periodic table. (B) is incorrect because atomic radius alone does not give enough information for one to ascertain the second ionization energy; it is also significant to consider the valence electron configuration. Additionally, all atoms have high second ionization energies. Finally, there is insufficient information supporting (D).

14. D