The correct order of melting points of hydrides of group 16 elements is

We need to find the correct order of melting points of the Group 16 hydrides $$H_2O$$, $$H_2S$$, $$H_2Se$$, and $$H_2Te$$.

Since water ($$H_2O$$) exhibits extensive intermolecular hydrogen bonding due to the high electronegativity of oxygen and its small size, it has an anomalously high melting point compared to the other hydrides. The melting point of $$H_2O$$ is 0°C, which corresponds to 273 K.

In contrast, $$H_2S$$, $$H_2Se$$, and $$H_2Te$$ lack significant hydrogen bonding and are held together primarily by van der Waals (London dispersion) forces, which become stronger with increasing molecular size and mass. Consequently, their melting points increase in the order $$H_2S < H_2Se < H_2Te$$.

The approximate melting points are $$H_2S$$: $$-85°C$$ (188 K), $$H_2Se$$: $$-66°C$$ (207 K), $$H_2Te$$: $$-49°C$$ (224 K), and $$H_2O$$: $$0°C$$ (273 K).

Combining these observations gives $$H_2S < H_2Se < H_2Te < H_2O$$. Therefore, the correct answer is Option A: $$H_2S < H_2Se < H_2Te < H_2O$$.