MCAT Question A Day - 2/15/14 - Answer!
Which one of the following has the largest ionic radius?
A. Na+
B. K+
C. Mg2+
D. Al3+
E. Cl–
E. In general, a cation is smaller than a neutral atom of the same element since the removal of an electron or electrons to form the cation leaves fewer electrons to shield one another from the attractive force of the nucleus and to repel one another. The remaining electrons on the cation are drawn closer to the nucleus, resulting in a reduction in size. By this same reasoning, it follows that the greater the positive charge, the smaller the ion. Conversely, anions are generally larger than the corresponding neutral atoms due to an increased number of electrons. A size trend is also observed in ions of equal charge: as one moves down a group, or column of the Periodic Table, the atomic and ionic radii increase due to the larger size of the highest occupied orbital. Therefore sodium atoms are smaller than potassium atoms, and Na+ ions are smaller than K+ ions. If we apply these facts and place the ions into a partial order by size, we have:
K+ > Na+ > Mg2+ > Al3+
Thus choices A, C, and D can be eliminated, and we must now compare the remaining species, K+ and Cl–. When potassium, atomic number 19, gives up an electron to form the cation K+, it is left with 18 electrons and is thus isoelectronic to argon, atomic number 18. Chlorine, atomic number 17, takes on an electron to form Cl–, and thus it likewise becomes isoelectronic to argon. Each species is isoelectronic to the other, and the relative radii will be affected most by the attractive force exerted on the outer electrons by the positive charge of the nucleus. It follows that the more protons the nucleus contains, i.e., the larger the atomic number is, the smaller the ionic radius will be. The final comparison is thus Cl– > K+, verifying choice E as the credited choice.
