Which one of the following correctly represents the order of stability of oxides, X$$_2$$O; (X = halogen)?

We have to compare the relative stability of the monoxides $$\mathrm{X_2O}$$ in which the halogen atom X is present in the $$+1$$ oxidation state. The factors that mainly decide the stability are

1. The ease with which the halogen can exist in a positive oxidation state.
2. The strength of the X-O bond that actually holds the molecule together.

First, let us examine the ease of attaining the $$+1$$ state. For a halogen the tendency to show positive oxidation states increases as its electronegativity decreases, because a less-electronegative atom can more readily part with electron density to oxygen. The Pauling electronegativities are

$$\chi(\mathrm{Cl}) = 3.0,\; \chi(\mathrm{Br}) = 2.8,\; \chi(\mathrm{I}) = 2.5.$$

Thus the stability arising only from the oxidation state follows

$$\mathrm{I} > \mathrm{Br} > \mathrm{Cl}.$$

Now we must consider the second factor, X-O bond strength. The X-O bond is formed by the lateral overlap of the $$p$$ orbitals of X and O. A larger atomic radius leads to poorer overlap and therefore a weaker bond. The atomic radii increase down the group, so the bond strength order is

$$\text{X-O bond energy:}\;\; \mathrm{Cl-O} > \mathrm{Br-O} > \mathrm{I-O}.$$

A stronger bond clearly enhances stability; hence, by this factor alone, the order would be

$$\mathrm{Cl} > \mathrm{Br} > \mathrm{I}.$$

To obtain the overall trend we combine both arguments. For iodine, although the I-O bond is somewhat weaker, the very large gain in stability obtained by having iodine comfortably in the $$+1$$ state more than compensates for the weaker bond. For bromine the I-O advantage does not exist, yet its Br-O bond is weaker than the Cl-O bond, making dibromine monoxide the least stable among the three. Hence the net order becomes

$$\boxed{\mathrm{I}_2\mathrm{O}\; >\; \mathrm{Cl}_2\mathrm{O}\; >\; \mathrm{Br}_2\mathrm{O}}.$$

So iodine monoxide is the most stable, chlorine monoxide comes next, and bromine monoxide is the least stable.

Hence, the correct answer is Option C.