4. Given that $m, n$ are real numbers, and satisfy $m^{2}+2 n^{2}+m-\frac{4}{3} n+\frac{17}{36}=0$. Then the square root of $-m n^{2}$ is ( ).
(A) $\frac{\sqrt{2}}{6}$
(B) $\pm \frac{\sqrt{2}}{6}$
(C) $\frac{1}{6}$
(D) $\pm \frac{1}{6}$
4.B.

From the problem, we know that $\left(m+\frac{1}{2}\right)^{2}+2\left(n-\frac{1}{3}\right)^{2}=0$.
Solving, we get $m=-\frac{1}{2}, n=\frac{1}{3}$. Then $-m n^{2}=\frac{1}{18}$.
Therefore, the square root of $-m \pi^{2}$ is $\pm \sqrt{\frac{1}{18}}= \pm \frac{\sqrt{2}}{6}$.