Step 1: Link the shape to a steric number.
A trigonal bipyramidal electron geometry corresponds to a steric number of 5, which means five electron pairs ($\text{sp}^3\text{d}$ hybridisation) around the central atom.
Step 2: Set up the counting rule.
For a central atom, steric number $= \dfrac{1}{2}\,[\,\text{valence electrons} + \text{number of monovalent atoms}\,]$ for neutral molecules. We look for the option giving 5.
Step 3: Check $\text{SeCl}_2$.
Se has 6 valence electrons, plus 2 chlorines gives 8 electrons, so steric number $= 4$. Not trigonal bipyramidal.
Step 4: Check $\text{SeF}_4$.
Se has 6 valence electrons, plus 4 fluorines gives 10 electrons, so steric number $= 5$. This gives a trigonal bipyramidal arrangement (one lone pair makes the molecular shape see-saw).
Step 5: Check $\text{SF}_6$.
S has 6, plus 6 fluorines gives 12 electrons, steric number $= 6$, which is octahedral. Not the answer.
Step 6: Check $\text{TeF}_6$.
Te also has 6 valence electrons, plus 6 fluorines gives 12, steric number $= 6$, again octahedral.
Step 7: Conclude.
Only $\text{SeF}_4$ has steric number 5 and the trigonal bipyramidal framework.
\[ \boxed{\text{SeF}_4\text{, option (2)}} \]