We need to identify for which halogens iron forms:

• $\mathrm{FeX_2}$
• $\mathrm{FeY_3}$

Iron commonly shows oxidation states $+2$ and $+3$.

• In $\mathrm{FeX_2}$, iron is in the $+2$ state.
• In $\mathrm{FeY_3}$, iron is in the $+3$ state.

Now check iron halides:

Ferrous halides, $\mathrm{FeX_2}$:

• $\mathrm{FeF_2}$ exists
• $\mathrm{FeCl_2}$ exists
• $\mathrm{FeBr_2}$ exists
• $\mathrm{FeI_2}$ exists

So, for $x$, all halogens $\mathrm{F, Cl, Br, I}$ are possible.

Ferric halides, $\mathrm{FeY_3}$:

• $\mathrm{FeF_3}$ exists
• $\mathrm{FeCl_3}$ exists
• $\mathrm{FeBr_3}$ exists
• $\mathrm{FeI_3}$ does not exist as a stable compound

Reason: $\mathrm{I^-}$ is a good reducing agent and reduces $\mathrm{Fe^{3+}}$ to $\mathrm{Fe^{2+}}$, while itself gets oxidized to iodine: $2\mathrm{Fe^{3+}} + 2\mathrm{I^-} \rightarrow 2\mathrm{Fe^{2+}} + \mathrm{I_2}$ Hence $\mathrm{FeI_3}$ is not known.

Therefore:

• $x = \mathrm{F, Cl, Br, I}$
• $y = \mathrm{F, Cl, Br}$

Matching with the options:

• Option C: $x = \mathrm{F, Cl, Br, I}$ and $y = \mathrm{F, Cl, Br}$

So, Option C is correct.