Elements of group- 15, 16, and 17 form compounds with hydrogen with general formula \(MH_3\), \(H_2M\), and \(HX\) (called hydrogen halides). The stability of these hydrides, along with \(M-H\) bond strength and basic nature decreases down the group. Boiling point of these hydrides mainly depends on the strength of their intermolecular forces.

Which of the following is most basic among the following is most basic among the following hydrides of group 16?

\(H_2O\)

The correct answer is option 4. \(H_2O\).

The basicity of hydrides generally increases down a group in the periodic table. This is because the size of the atoms increases, leading to a more diffuse electron cloud, making it easier for the hydride ion (H⁻) to accept a proton (H⁺). In Group 16, the hydrides are \(H_2O\), \(H_2S\), \(H_2Se\), and \(H_2Te\).

So, in this case, the most basic hydride among the options is: 4. \(H_2O\)