Answer
All structures including F atoms at two of the vertices of a tetrahedron and H atoms at the remaining two are superimposable, i.e. placed over each other yet all are visible. Only one molecule has the formula $\mathrm{CH_2F_2}$. There are two possibilities for four atoms at the corners of a square in which there are two F atoms on one side (cis) or on opposite corners (trans).
Work Step by Step
See answer above.
