$G_1$ is a catacondensed benzenoid graph and $G_2$ a phenylene, then it holds $$\begin{aligned}
M_2(G_1) & = & 33h(G_1) + s(G_1) + b(G_1) - 10, \\
M_2(G_2) & = & 60h(G_2) + s(G_2) + b(G_2) - 37.\end{aligned}$$

In this proof, an edge $e=xy$ will be called an edge of *type 1* if $x$ and $y$ both have degree two. Moreover, if one of $x$ and $y$ has degree two and the other one has degree three, we say that $e$ is of *type 2*. Finally, $e$ is an edge of *type 3* if $x$ and $y$ both have degree three.

We notice that a terminal hexagon of $G_1$ or $G_2$ has three edges of type 1, two edges of type 2, and one edge of type 3. A linear hexagon has four edges of type 2 and two edges of type 3. An angular hexagon has one edge of type 1, two edges of type 2, and three edges of type 3. Finally, a branched hexagon has six edges of type 3.

If $G_1$ has only one hexagon, the result obviously holds. Otherwise, in a catacondensed benzenoid graph $G_1$ there are $h(G_1)-1$ edges of type 3 that appear on two 