To make the antecedent true at least one thing that is or is not in the extension of $F$ must also be or not be in the extension of $G$.
But then this seems to automatically make at least one case of the second bi-conditional, true. And to find an interpretation where the first is true and the second false, I need there to be no cases where what is or is not $F$ is not the same as what is or is not $G$.
One idea I had was to set $F$'s extension to {1} and $G$'s extension to $\{∅\}$.
In general I am not clear on what it takes to make the antecedent false. For the bi-conditioanl to be false, the truth of '$\exists x(Fx)$' and '$\exists x(Gx)$' need to mismatch. And the truth of either requires at least one member of the domain to be in the extension of the predicate letter. So if I satisfy that requirement, then to get the mismatch that falsifies the bi-conditional I need to have nothing from the domain in the extension of the other predicate letter.
Thanks for any help.
That's correct.
With $x \mapsto 2$, there is an object of which neither $F$ nor $G$ is true, so the biconditional is true, so $\exists x (F(x) \leftrightarrow G(x))$ is true.
However, with $x \mapsto 1$ there is an object of which $F$ is true, so $\exists x F(x)$ is true, but since the extension of $G$ is empty, $\exists x G(x)$ is false, so $\exists x F(x) \leftrightarrow \exists x G(x)$ is false, as desired.
If by antecedent you mean the left-hand side of the biimplication (typically the antecedent/succedent terminology is only used with one-directional implication), $\exists x F(x)$, then the requirement would be that the extension of $F$ is empty. But you don't necessarily need the left-hand side false, but, as you say, a mismatch between the two sides, which is achieved by having $\exists x F(x)$ true but $\exists x G(x)$ false (or vice versa).