Discussion in the comments has revealed that the intended question is whether $\operatorname{var}(x) > \operatorname{var}(y)$ implies that $\operatorname{range}(x) > \operatorname{range}(y)$. The answer is no; for an extreme counterexample, let $\begin{align} x &= [\underbrace{-1, -1, \ldots, -1}_{\text{$n/2$ elements}}, \underbrace{1, 1, \ldots, 1}_{\text{$n/2$ elements}}],\\ y &= [-k, \underbrace{0, 0, \ldots, 0}_{\text{$n-2$ elements}}, k], \end{align}$
Then $x$ has variance $1$ and range $2$, while $y$ has variance $2k^2/n$ and range $2k$. Pick any $k$ between $1$ and $\sqrt{n/2}$, and you get $\operatorname{var}(x) > \operatorname{var}(y)$ but $\operatorname{range}(x) < \operatorname{range}(y)$. The point is that the variance averages over all the elements, while the range only looks at the two most extreme ones, so they can disagree when the two extreme points are not representative of the rest.
(In the counterexample in my comment, I had switched $x$ and $y$ by mistake.)