Suppose we have that a random variable sequence $(X_n)_n$ converges in distribution to a law with mean $\bar{\mu}$ and variance $\bar{\sigma}^2$, or formally $X_n \stackrel{d}{\to} \mathcal{L}(\bar{\mu}, \bar{\sigma})$. Further assume the moments of the sequence and the limiting distribution are bounded: $\bar{\mu}, \bar{\sigma} < \infty$ and $\mu_n, \sigma_n < \infty$.
Do we have that $\mathrm{E}[X_n] = \mu_n \to \bar{\mu}$ and $\mathrm{Var}[X_n] = \sigma_n^2 \to \bar{\sigma}^2$ holds ?
My intuition is as follows: for $n$ large, we have that $X_n$ is approximately distributed as $ \mathcal{L}(\bar{\mu}, \bar{\sigma})$, so when we compute its expectation $\mu_n$ we should find approximately $\bar{\mu}$.
Is there a formal statement which corroborates or invalidates this intution of mine ?