In which anion do you expect a shorter O-O distance? BaO2 or KO2
Scroll down to the Synthesis, structure, etc and the bond distances for O2, O2^- and O2^-1 are given.
http://en.wikipedia.org/wiki/Superoxide
To determine which anion, BaO2 or KO2, is expected to have a shorter O-O distance, we first need to understand a few concepts.
BaO2 and KO2 are both ionic compounds consisting of a metal cation (Ba2+ or K+) and a peroxide anion (O22-). The peroxide anion consists of two oxygen atoms connected by a single bond.
Ionic compounds are formed through the transfer of electrons from the metal cation to the nonmetal anion. The resulting ionic bond forms due to the electrostatic attraction between the oppositely charged ions.
The size of the metal cation plays a critical role in the overall structure of the ionic compound. As we move across a period on the periodic table, from left to right, the atomic radius of metal cations decreases. This is due to increasing nuclear charge and stronger attraction between the electrons and nucleus.
Now, let's compare BaO2 and KO2. Ba2+ belongs to the alkaline earth metal group, which is located in the lower left corner of the periodic table. K+ belongs to the alkali metal group, which is located in the second column from the left. Hence, Ba2+ is larger than K+.
When the metal cation is larger, it can accommodate larger anions more easily without significant distortion of the crystal structure. In this case, the Ba2+ cations can accommodate the larger O22- anions more effectively than the smaller K+ can.
Therefore, we expect that BaO2 will have a shorter O-O distance compared to KO2 due to the larger size of the Ba2+ cation, which allows for closer proximity between the oxygen atoms in the peroxide anion.