Rank one adjoint operators on a Lie algebra Let $\mathfrak{g}$ be a (finite dimensional) semi-simple Lie algebra over a field $k$ and let $x \in \mathfrak{g}$. By definition, we have the equivalence:
$$ \mathrm{rk}(\mathrm{ad}_x) = 0 \iff x = 0,$$
where $\mathrm{rk}(\mathrm{ad}_x)$ is the rank of $\mathrm{ad}_x$ seen as an element of $\mathrm{End}(\mathfrak{g})$. I would like to know if there is a classification of elements $x \in \mathfrak{g}$ such that $\mathrm{rk}(\mathrm{ad}_x) \leq 1$? I am primarily interested in the case where $k = \mathbb{C}$ and $\mathfrak{g}$ is simple of classical type.
 A: $\DeclareMathOperator\ad{ad}\DeclareMathOperator\rk{rk}$I claim that it is impossible to have $\rk(\ad_x) = 1$.
I'll assume that $\mathfrak g$ is $k$-split, which seems to be OK since you are interested in the case $k = \mathbb C$.  (Or you could just tensor up to $\overline k$, which does not change the rank of the adjoint operator.)  Let $\mathfrak h$ be a split Cartan subalgebra.
Suppose that $x \in \mathfrak g$ satisfies $\rk(\ad_x) = 1$.
If $x$ lies in $\mathfrak h$, $\beta$ is any root that does not vanish at $x$, and $E_\beta$ and $E_{-\beta}$ are non-$0$ vectors in the appropriate root spaces, then $\ad_x(E_\beta) = \beta(x)E_\beta$ and $\ad_x(E_{-\beta}) = -\beta(x)E_{-\beta}$ are non-$0$ elements of different root spaces, hence are linearly independent, which is a contradiction.
Thus, $x$ does not lie in $\mathfrak h$.  Let $x = \sum x_\alpha$ be the root-space decomposition with respect to $\mathfrak h$, where $\alpha$ runs over the roots of $\mathfrak h$ in $\mathfrak g$ and $0$, and choose a root $\alpha$ such that $x_\alpha$ is non-$0$.
We have that $\ad_x(h)$ equals $-\sum \alpha(h)x_\alpha$ for all $h \in \mathfrak h$.  Since all these elements lie on a common line, if $\beta \ne \pm\alpha$ is a root, then, since $\{\alpha, \beta\}$ and $\{\alpha, -\beta\}$ are linearly independent subsets of $\mathfrak h^*$, we have $x_\beta = 0$ and $x_{-\beta} = 0$.
Let $h$ be an element of $\mathfrak h$ not annihilated by $\alpha$.
We have that $\ad_x(h) = -\alpha(h)x_\alpha + \alpha(h)x_{-\alpha}$ and $\ad_x(x_\alpha)$ equals $\alpha(x_0)x_\alpha + \ad_{x_{-\alpha}}(x_\alpha)$.  Noting that $\ad_{x_{-\alpha}}(x_\alpha)$ lies in $\mathfrak h$ and comparing root-space decompositions of these linearly dependent elements shows that $\alpha(h)x_{-\alpha}$ equals $0$, so that $x_{-\alpha}$ equals $0$.  That is, $\ad_x(h)$ equals $-\alpha(h)x_\alpha$, hence is a non-$0$ element of the $\alpha$-root space.
If $x_0$ is non-$0$, then we can choose a root $\beta \ne -\alpha$ (possibly $\beta = \alpha$) such that $\beta(x_0) \ne 0$.  If $E_\beta$ is a non-$0$ element of the $\beta$-root space, then $\ad_x(E_\beta) = \beta(x_0)E_\beta + \ad_{x_\alpha}(E_\beta)$ has projection $\beta(x_0)E_\beta \ne 0$ on $\mathfrak h$, hence is linearly independent of $\ad_x(h)$, which is a contradiction.
Thus $x$ lies in the $\alpha$-root space.  If $E_{-\alpha}$ is a non-$0$ element of the $(-\alpha)$ root space, then $\ad_x(E_{-\alpha})$ is a non-$0$ element of $\mathfrak h$, hence is again linearly independent of $\ad_x(h)$, which is a contradiction.
A: Another approach.
To show it's impossible (the rank can't be 1), it is enough to show this when the field (assumed of char 0) is algebraically closed, and in turn it's enough to show the result in case $\mathfrak{g}$ is simple. If $x$ has $\mathrm{ad}(x)$ of rank 1, $x$ has centralizer of codimension 1. It is known (see e.g. this MathSE answer) that $\mathfrak{g}$ has no subalgebra of codimension 1, unless $\mathfrak{g}$ is 3-dimensional. But for $\mathfrak{sl}_2$, the operator $\mathrm{ad}_x$ has rank 2 for every nonzero $x$ (alternatively, all 2-dimensional subalgebras have a trivial centralizer, so can't be centralizer of an element).
A similar approach can be used (with a little further work) to classify which $\mathfrak{g}$ admit an $x$ with $\mathrm{ad}(x)$ of rank 2, and classify such elements $x$.
