The correct order of basicity of oxides of vanadium is

We need to determine the correct order of basicity of vanadium oxides: $$V_2O_3$$, $$V_2O_4$$, and $$V_2O_5$$.

First, determine the oxidation states of vanadium in each oxide.

- In $$V_2O_3$$: Let oxidation state of V be $$x$$. Then $$2x + 3(-2) = 0$$, so $$x = +3$$.

- In $$V_2O_4$$: $$2x + 4(-2) = 0$$, so $$x = +4$$.

- In $$V_2O_5$$: $$2x + 5(-2) = 0$$, so $$x = +5$$.

Next, apply the rule relating oxidation state to acid-base character.

For transition metal oxides, there is a well-established trend:

- Lower oxidation states produce more basic oxides. This is because the metal ion with lower charge has more ionic character in its bonding with oxygen, and ionic metal oxides are basic (they produce OH$$^-$$ in water).

- Higher oxidation states produce more acidic oxides. Higher charge on the metal leads to more covalent M-O bonds, and the oxide becomes acidic (it can accept OH$$^-$$ or donate H$$^+$$).

Now, apply the trend to vanadium oxides.

- $$V_2O_3$$ (V in +3): Most basic -- behaves as a basic oxide

- $$V_2O_4$$ (V in +4): Amphoteric -- intermediate character

- $$V_2O_5$$ (V in +5): Most acidic -- behaves as an acidic oxide

Therefore, the order of decreasing basicity is:

$$ V_2O_3 > V_2O_4 > V_2O_5 $$

The correct answer is Option 1: $$V_2O_3 > V_2O_4 > V_2O_5$$.