As you pointed out, with 4 possibilities all equally likely, the entropy is 2, which is the maximum for 4 possible outcomes. This is because entropy is defined as:
$$
H(X)=-\sum_ip(x_i)\log_bp(x_i)
$$
In your example, you're using base 2 for the logarithm (i.e. $b=2$), which is a common choice and leads to an answer that has units of bits. For 4 equally likely outcomes, we have $p(x_i)=1/4$ for all $i$, so substituting that into the equation we get $H(X) = -\sum_i\frac{1}{4}log_2\frac{1}{4}=-4\times\frac{1}{4}\times-2=2$.
If the possible outcomes (base pairs) aren't (a priori) equally likely for a given position in the sequence, you can get an entropy anywhere between 0-2 bits. For instance, an entropy of 1 can result from knowing that only 2 base pairs are possible at that position (e.g. only A or C), and equally likely. Those get probability $1/2$, while the others have probability $0$, which results in: $H(X)=-2\times\frac{1}{2}\times\log_2\frac{1}{2}-2\times0\times\log_20=-1\times-1=1$.
Fractional numbers are also entirely possible. For instance, if you know that only 3 base pairs are possible, one with probability $1/2$ and the other two both with probability $1/4$, you get an entropy for that position of about 0.801.
In general, numbers closer to 2 in your example indicate that the probabilities of your outcomes are close to uniform, while numbers closer to 0 mean that the probability is more concentrated (with an entropy of 0 meaning only one base pair was possible at that location to begin with, so you have 0 surprise at observing that outcome).
