Given below are two statements :
Statement I : $$F_2O < H_2O < Cl_2O$$ is the correct trend in terms of bond angle.
Statement II : SiF$$_4$$, SnF$$_4$$ and PbF$$_4$$ are ionic in nature. In the light of the above statements, choose the correct answer from the options given below :

For (F_2O), the highly electronegative fluorine atoms attract the bonding electron pairs towards themselves, reducing the bond pair-bond pair repulsion around the central oxygen atom. As a result, the bond angle is approximately (103^\circ).

In (H_2O), the bond angle is about (104.5^\circ), which is the typical value for a bent molecule with two lone pairs on an (sp^3)-hybridized oxygen atom.

In (Cl_2O), chlorine is less electronegative and larger in size than fluorine. The increased steric repulsion between the bulky chlorine atoms leads to a larger bond angle of about (111^\circ).

Hence, the bond angle order is

$$F_2O<H_2O<Cl_2O.$$

Therefore, Statement I is true.

(SiF_4) is a covalent molecular compound. Similarly, (SnF_4) and (PbF_4) are predominantly covalent tetrahalides rather than ionic compounds. According to Fajans' rule, these tetrafluorides possess significant covalent character and cannot be classified as ionic.

Therefore, Statement II is false.

Hence, Statement I is true, Statement II is false, and the correct answer is Option C.