**The Kullback-Leibler divergence is unbounded.** Indeed, since there is no lower bound on the $q(i)$'s, there is no upper bound on  the $p(i)/q(i)$'s. For instance, the Kullback-Leibler divergence between a Normal $N(\mu_1,\sigma^2)$ and a Normal $N(\mu_2,\sigma^2)$ with equal variance is 
$$\frac{1}{2\sigma^{2}}(\mu_1-\mu_2)^2$$which is clearly unbounded.

[Wikipedia][1] [which has been known to be wrong!] indeed states

> "...a Kullback–Leibler divergence of 1 indicates that the two
> distributions behave in such a different manner that the expectation
> given the first distribution approaches zero."

which makes no sense (expectation of which function? why 1 and not 2?)

A more satisfactory explanation from the same 
[Wikipedia][1] page is that the Kullback–Leibler divergence

> "...can be construed as measuring the expected number of extra bits
> required to code samples from P using a code optimized for Q rather
> than the code optimized for P."


  [1]: https://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence