Identify the intermolecular forces for each compound.

• $\text{CH}_4$ (methane): This is a nonpolar molecule due to its symmetrical tetrahedral geometry. The only intermolecular forces present are weak London dispersion forces.
• $\text{H}_2\text{S}$ (hydrogen sulfide): This is a polar molecule due to its bent geometry. It has dipole-dipole forces and London dispersion forces. Sulfur is not electronegative enough to form hydrogen bonds.
• $\text{H}_2\text{O}$ (water): This is a highly polar molecule with a bent geometry. Oxygen is highly electronegative and bonded to hydrogen, allowing water molecules to form strong hydrogen bonds. It also has dipole-dipole forces and London dispersion forces.

• London dispersion forces are the weakest.
• Dipole-dipole forces are stronger than London dispersion forces.
• Hydrogen bonds are the strongest of these three types of intermolecular forces.

The order of increasing intermolecular force strength is:
$\text{CH}_4$ (London dispersion) < $\text{H}_2\text{S}$ (dipole-dipole + London dispersion) < $\text{H}_2\text{O}$ (hydrogen bonding + dipole-dipole + London dispersion).

Therefore, the order of increasing melting point is:
$\text{CH}_4 < \text{H}_2\text{S} < \text{H}_2\text{O}$

This order matches option A.

The correct option is A.