If the boiling point of $$H_2O$$ is 373 K, and the boiling point of $$H_2S$$ will be:

We are asked to compare the boiling point of $$H_2S$$ with that of $$H_2O$$, whose boiling point is already given as $$373\ \text{K}$$. The guiding idea is that the boiling point of a liquid rises or falls according to how strong the intermolecular attractions are inside that liquid. In simple words, the stronger the attractions, the more heat is required for the molecules to escape into the vapour phase, and therefore the higher the boiling point becomes.

Now, we recall the following qualitative relation:

$$\text{Boiling point}\; \propto \; \text{Strength of intermolecular forces}$$

The main intermolecular forces relevant here are:

1. $$\text{Hydrogen bonding}$$ - a particularly strong dipole-dipole attraction that occurs when hydrogen is bonded to a small, highly electronegative atom such as $$\text{F, O, N}$$.
2. $$\text{Dipole-dipole interactions}$$ - weaker than hydrogen bonding, present between any polar molecules.
3. $$\text{London dispersion (van der Waals) forces}$$ - present in all molecules, but they are the only attractions available when the molecule is neither highly polar nor capable of hydrogen bonding.

We examine the two specific molecules in the question one by one.

For $$H_2O$$:

• Oxygen is highly electronegative.
• Each $$H$$ is covalently bonded to $$O$$, so every $$H$$ is attached to a strongly electronegative atom.
• This meets the classical criterion for hydrogen bonding, therefore $$H_2O$$ engages in extensive hydrogen bonding in the liquid state.
• Because hydrogen bonding is very strong, a large amount of heat is required to make $$H_2O$$ boil, giving it a high boiling point of $$T_\text{b}(H_2O)=373\ \text{K}$$.

For $$H_2S$$:

• Sulphur is far less electronegative than oxygen.
• Although the $$H-S$$ bond is polar, the polarity is not large enough to enable hydrogen bonding.
• Consequently, $$H_2S$$ molecules can only interact through ordinary dipole-dipole forces and London dispersion forces, which are much weaker than hydrogen bonds.
• Weaker intermolecular attractions imply that much less heat is needed for the molecules to escape into the vapour phase.
• Therefore the boiling point of $$H_2S$$ must be markedly lower than that of $$H_2O$$.

Next, we match this qualitative deduction with the numerical ranges supplied in the options. The actual experimental boiling point of $$H_2S$$ is around $$212\ \text{K}$$, which is well below $$300\ \text{K}$$. Even without memorising the exact numeric value, realising that it is far below $$373\ \text{K}$$ tells us it lies in the range

$$T_\text{b}(H_2S) < 300\ \text{K}.$$

Scanning the alternatives:

A. less than 300 K
B. equal to 373 K
C. more than 373 K
D. greater than 300 K but less than 373 K

Only Option A precisely states that the boiling point is lower than $$300\ \text{K}$$, which aligns with our analysis.

Hence, the correct answer is Option A.