This article focuses on a couple of questions from the 2013 CHEM 13 NEWS Exam. The first is question #9, which is reproduced below. Almost all students responded (94%), with 34% of them choosing the correct response (B), and about the same percentage (32%) choosing answer C. This question was, from a statistical perspective, among those that best discriminated between the “stronger” students and the “weaker” students. By this I mean that, generally speaking, students who answered question #9 correctly had better test scores than those who answered it incorrectly.
Which of the following ions has the smallest ionic radius?
I suspect a significant fraction of students chose answer C because they focused on the atoms from which these ions are derived, and their relative positions within the periodic table, without taking into consideration the effect of adding or removing electrons from these atoms. If the question had instead been “Which of the following atoms has the smallest radius?”, with the choices given as K, Ca, Cl, Br and P, then the correct answer would have been Cl. Students would deduce that Cl has the smallest atomic radius presumably because they have been taught that, for the main group elements at least, atomic radii decrease from left to right within a period and increase from top to bottom within a group.1
To answer question #9, it is probably helpful to first eliminate the Br− ion from further consideration by comparing it to the Cl− ion. These two ions have the same charge and are derived from atoms of the same group. The valence electrons in Br− and Cl− occupy orbitals in, respectively, the n = 4 and n = 3 shells. Their ground state electronic configurations are, respectively, [Ar]3d10 4s2 4p6 and [Ne] 3s2 3p6. The valence electrons are, on average, much further from the nucleus in the Br− ion than in the Cl− ion. Consequently, the Br− ion is larger than the Cl− ion. With the Br− ion eliminated from consideration, we can now focus on the ions K+, Ca2+, Cl− and P3−. These ions have 18 electrons each and are isoelectronic with each other. Although the total number of electrons is the same for each of these ions, the nuclear charge is different: +19 for K+; +20 for Ca2+; +17 for Cl− and +15 for P3−. For this set of isoelectronic ions, the greater the nuclear charge, the more closely the electrons are drawn in, the more compact the orbitals and the smaller the radius. Thus, when these ions are ranked in order of increasing radius, the order is Ca2+ < K+ < Cl− < P3−. We cannot include the Br− ion in this ranking because, although we know that Br− is larger than Cl−, it is difficult to establish whether Br− is larger or smaller than P3−. It turns out that P3− has a larger ionic radius (about 212 pm for P3− and 196 pm for Br−). Thus, the addition of 3 electrons to the n = 3 shell of the P atom causes a larger increase in radius than does the addition of just one electron to the n = 4 shell of the Br atom.2
The following two guidelines are helpful when considering the relative sizes of monatomic ions.
- Cations are smaller than the atoms from which they are formed. Anions are larger than the atoms from which they are formed.
- For isoelectronic ions, the more positive the charge, the smaller the ionic radius.
The second question we’ll examine (#38) is reproduced below. Most students (87%) entered a response, but less than 13% chose the correct response (E) and almost 60% chose answer D. Because the success rate was so low, this question was not as effective at discriminating between the “stronger” and “weaker” students.
If the reaction W + X ⇄ Y + Z is exothermic in the forward direction, then what is the effect of an increase in temperature on the forward rate, the reverse rate and the equilibrium constant?
I think that many students recognize that an increase in temperature generally causes an increase in reaction rate and, for a system initially at equilibrium, net reaction in the endothermic direction. The results of this question suggest that many students do not appreciate that an increase in temperature causes increases in the rates of both the forward and reverse reactions (i.e., the rates of converting reactants into products and of converting products into reactants), or have difficulty reconciling the increases in these rates with a shift in the equilibrium position. Thinking back on the many conversations I’ve had with students, I suspect that many students might reason through this problem as follows:
- Because the reaction is exothermic in the forward direction, an increase in temperature would cause the equilibrium position to shift to the left, thereby producing a smaller value for the equilibrium constant.3
- The shift to the left actually occurs because the rate of the forward reaction decreases whereas the rate of the reverse reaction increases. This reasoning breaks down in step (2). The shift to the left actually occurs because the rate of the reverse reaction increases by a greater factor than that of the forward reaction. For example, if an increase in temperature causes the forward reaction rate to increase by a factor of 10 and the reverse reaction rate to increase by a factor of 100, there will be net reaction to the left even though both reactions are occurring faster than they were initially.
I encourage teachers, when discussing chemical equilibrium, to emphasize that although an increase in temperature may cause the equilibrium constant K to increase or decrease, depending on whether the reaction is endothermic or exothermic, the rates of the forward and reverse reactions always increase with temperature.4
- Although many students know that atomic radii decrease across a period and increase down a group, relatively few are able to provide a solid explanation.
- The atomic radius of P is about 110 pm and that of Br is about 114 pm.
- Actually, it is more correct to say that there is a net reaction to the left because, for an exothermic reaction, the equilibrium constant decreases as the temperature increases.
- There are some reactions for which the rate decreases as temperature increases, so the use of “always” is not totally justified.