Picard number $1$ is the most frequent case among all varieties, so you cannot expect a classification. It's quite the opposite, you might stand a chance to classify those (within some class) that have Picard number larger than $1$. For instance a general $K3$ surface has Picard number $1$ and the locus of those with a given Picard number becomes smaller as the Picard number increases.

If the Picard number is larger than $1$, that usually means that the variety admits some non-trivial maps which gives you a handle on them or a starting point if you will. If the Picard number is $1$, it is hard to get any traction to get some way to study the object.

On the other hand that means that you can get lots of examples with Picard number $1$. Chances are, if you choose a variety at random it will have Picard number $1$.
You can get lots of examples that are not complete intersections by the simple observation that for a complete intersection of dimension $d$, the middle cohomology groups of the structure sheaf vanish, that is, $H^i(X,\mathscr O_X)=0$ for $0<i<d$.
This is actually another way to see that an abelian variety of dimension at least $2$ cannot be a complete intersection.

In particular, you can find lots of examples among surfaces. Surfaces with $H^1(X,\mathscr O_X)\neq 0$ are known as *irregular*, so any irregular surface with Picard number $1$ gives and example that you want. One way to ensure that the Picard number is $1$ is to make sure that $\mathrm{rk}\, H^2(X,\mathbb{Z})=1$. In other words, any surface with $q \ne 0$ and $b_2=1$ gives you an example.

Of course, you would want an explicit example. Unfortunately, I can't think of one at the moment, but I am fairly certain, that a general surface with $H^1(X,\mathscr O_X)\neq 0$ has Picard number one, so that should give you plenty of examples. In fact, one could argue that that's why I can't give an explicit one, because they are the general ones (and anything explicit is not general).