It is a known fact that any perfectly normal (countably) compact space $X$ is ccc. The proof is what you would try: start with an uncountable cellular family $\mathcal{U}$, choose a point $p_U \in U$ for each $U \in \mathcal{U}$ and consider the closed set $C=X \setminus \bigcup \mathcal{U}$. Since the space is perfectly normal, $C$ is a $G_\delta$ (say $C= \bigcup_nV_n$) and since $\mathcal{U}$ is uncountable you can find $n$ such that infinitely many of the $p_U$'s are outside of $V_n$. Now this infinite set of $p_U$´s has no limit points, contradicting the compactness of $X$.

Since $I^2_{lex}$ is compact and not ccc (both standard facts), it cannot be perfectly normal.

It is a known fact that any perfectly normal (countably) compact space $X$ is ccc. The proof is what you would try: start with an uncountable cellular family $\mathcal{U}$, choose a point $p_U \in U$ for each $U \in \mathcal{U}$ and consider the closed set $C=X \setminus \bigcup \mathcal{U}$. Since the space is perfectly normal, $C$ is a $G_\delta$ (say $C= \bigcap_nV_n$) and since $\mathcal{U}$ is uncountable you can find $n$ such that infinitely many of the $p_U$'s are outside of $V_n$. Now this infinite set of $p_U$´s has no limit points, contradicting the compactness of $X$.

Since $I^2_{lex}$ is compact and not ccc (both standard facts), it cannot be perfectly normal.