No, let $X_{n,k}$ be iid $N(0,1)$ for $k \ne n$, and $X_{n,n} = n$. The hypothesis is satisfied but if we take $r_n = n$ then $$\frac{1}{\sqrt{n}}(X_{n,1} + \dots + X_{n,n}) \sim N(\sqrt{n}, \frac{n-1}{n})$$ which is not even tight.

No. Take any sequence $r_n \to \infty$ and let $X_{n,k}$ be iid $N(0,1)$ for $k \ne r_n$, and $X_{n,r_n} = r_n$. The hypothesis is satisfied because $(X_{n,1}, \dots, X_{n,k})$ are iid $N(0,1)$ as soon as $n$ is so large that $r_n > k$. But $$\frac{1}{\sqrt{r_n}}(X_{n,1} + \dots + X_{n,r_n}) \sim N(\sqrt{r_n}, \frac{r_n-1}{r_n})$$ which is not even tight.