$$FeX_2$$ and $$FeY_3$$ are known when $$x$$ and $$y$$ are:

Iron forms two types of halides: $$FeX_2$$ (ferrous halides, where Fe is in +2 state) and $$FeY_3$$ (ferric halides, where Fe is in +3 state).

For $$FeX_2$$ (iron(II) halides): All four ferrous halides are known — $$FeF_2$$, $$FeCl_2$$, $$FeBr_2$$, and $$FeI_2$$. Iron in the +2 state is not a strong enough oxidising agent to oxidise any of the halide ions, so all halides are stable.

For $$FeY_3$$ (iron(III) halides): $$FeF_3$$, $$FeCl_3$$, and $$FeBr_3$$ are known, but $$FeI_3$$ does not exist. This is because $$Fe^{3+}$$ is a sufficiently strong oxidising agent to oxidise $$I^-$$ to $$I_2$$ (since the reduction potential of $$Fe^{3+}/Fe^{2+}$$ is +0.77 V, which is greater than that of $$I_2/I^-$$ at +0.54 V). So $$Fe^{3+}$$ oxidises $$I^-$$ to $$I_2$$ and gets reduced to $$Fe^{2+}$$, making $$FeI_3$$ unstable.

Therefore, $$x$$ = F, Cl, Br, I and $$y$$ = F, Cl, Br, which corresponds to option (1).