= Solution
{c}
The <primitive unit cell> contains two <carbon> atoms and four edge-shared <oxygen> atoms, each contributing one half: $n=2+4/2=4$ atoms. Equivalently, one molecular repeat contributes its two carbon atoms but only half of each of its four shared oxygen atoms. There are therefore $3n=12$ <phonon branches>.
At the <Brillouin zone> centre, three rigid translations give three <acoustic phonon branches>; the remaining $3n-3$ are <optical phonon branches>:
$$
\boxed{3\text{ acoustic}+9\text{ optical}=12\text{ phonon branches}.}
$$
A free layer has two in-plane acoustic modes and an out-of-plane flexural acoustic mode. A two-dimensional lattice does not imply that its atoms can move only in two directions. Artificially suppressing all out-of-plane motion would instead define a different model with two acoustic and six optical branches.
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