Atomic radius trend..kaplan error?

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

jillzhou

Full Member
10+ Year Member
Joined
Jun 24, 2010
Messages
91
Reaction score
7
For the section tests on the Kaplan course, the question states: place the following species in order of increasing atomic radius: Cl-, Ar, K+, Br-

The answer is: K+<Ar<Cl-<Br- and Kaplan's rationale is as it gains an electron the atomic radius increases. I thought as you go top to bottom and from right to left in the periodic table, atomic radius increases?! So confused, help!!:scared:
 
That's true for neutral atoms, but you're dealing with ions here. Additional electrons make anions larger due to electron repulsion. Removing electrons make cations smaller due to the increased effective nuclear charge.
 
K+, Ar, and Cl- are all isoelectronic (same number of electrons). As mentioned above, more electrons= less of an effective nuclear charge. But when you're dealing with isoelectronic ions, just remember that the species with the most protons has the smallest radius. The # of shells and electrons are the same, so it just comes down to effective nuclear charge. That gives you K+<Ar<Cl-. Br- has a larger radius than all of them because it has an extra shell of electrons that the others don't have (including K+ because it lost the K 4s1 electron), reducing attraction to the nucleus.

This is one of those cases where you have to think about why the trend is "Atomic radius increases to the left and downward", because the ions are tricky, and because K would have a larger radius than Br-, but K+ doesn't.
 
Top