Where is the mirror plane in 1-bromo-1-chloroethene? I am studying point groups in inorganic chemistry and having a very difficult time understanding some of these concepts. 

In my book, Inorganic Chemistry: Fifth Edition by Miessler, Fischer, &amp; Tarr, I am presented with a table showing that $\ce{CHFClBr}$ is in the $C_1$ point group and has no symmetry other than the identity operation.



$\ce{H2C=CClBr}$ is in the $C_\mathrm s$ point group and its symmetry is listed as having only one mirror plane.


I'm having difficulty seeing this mirror plane. I would have listed this compound as being in the $C_1$ point group. Does this have something to do with the molecule being flat?
For all symmetry operations, you should assume the atoms reduced to a point size. Then, if you have a perfectly flat molecule (like in the case of your disubstituted ethene), all atoms will be in the same mathematical plane meaning there is nothing above or below the plane.

Therefore, the plane is an element of symmetry that puts the nothing above the plane on the nothing below the plane and leaves the atoms themselves (in the plane of symmetry) untouched.