Which of the following can reduce decomposition of H$$_2$$O$$_2$$ on exposure to light

Hydrogen peroxide (H$$_2$$O$$_2$$) is unstable and decomposes when exposed to light, heat, or catalysts:

$$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$$

Light (particularly UV light) provides energy that breaks the weak O-O bond in H$$_2$$O$$_2$$ (bond energy approximately 146 kJ/mol), initiating decomposition. We need to identify which option can reduce this photodecomposition.

Urea (NH$$_2$$CONH$$_2$$) can stabilize H$$_2$$O$$_2$$ by forming hydrogen-bonded complexes, but it primarily works against thermal and catalytic decomposition rather than specifically blocking light.

Alkaline conditions actually accelerate the decomposition of H$$_2$$O$$_2$$ because the hydroperoxide ion (HO$$_2^-$$) formed in basic solution is more reactive.

Now, H$$_2$$O$$_2$$ is stored in dark-coloured (brown or amber) glass containers. The coloured glass absorbs UV and visible light, preventing it from reaching the H$$_2$$O$$_2$$, thereby directly reducing photodecomposition.

Dust particles can contain metal ions (such as Fe$$^{2+}$$, Mn$$^{2+}$$) that act as catalysts for H$$_2$$O$$_2$$ decomposition, so dust would increase decomposition rather than reduce it.

Hence, the correct answer is Option 3: Glass containers.