Question

Oxygen is more electronegative than sulfur, yet ${H_2}S$ is acidic while ${H_2}O$ in neutral. This is because:
A.Water is highly associated compound
B.$H - S$ bond is weaker than $H - O$ bond
C.${H_2}S$ is a gas while ${H_2}O$ is a liquid
D.The molecular weight of ${H_2}S$ is more than that of ${H_2}O$

Answer 1

Hint: We know that electronegativity is an extent of an attraction of atoms for electrons present in a bond. Electronegativity denotes us how much electrons are required for a specific atom. We know that the electronegativity of an atom has its value ranging from 0 to 4; the greater the value, the atom could be more electronegative, and the more could be attraction of the electrons present in a bond.

Complete step by step answer:
We know that electronegativity is the measure of an atom’s ability to attract a shared pair of electrons. Atomic number and distance at which its outermost electrons reside from the charged nucleus affects the electronegativity of an atom. In the modern periodic table, the electronegativity increases when we move across a period and it decreases as we go down the group. Generally, metal has lower value of electronegativity when compared to non-metals. Non-metals are electronegative and metals are electropositive.
We know that electronegativity of sulfur is less when compared to the electronegativity of oxygen. This causes less polar nature of bonds of sulfur when compared to bonds of oxygen. This less polar nature of $H - S$ bond which contributes to weaker hydrogen bonding when compared to the more polar nature of $H - O$ that contributes to stronger hydrogen bonding.
The acidic nature of ${H_2}S$ is because of the weak $H - S$ bond.
And hence option B is correct.

Note: We have to remember that the electronegativity of sulfur is 2.4 and the electronegativity of oxygen is 3.5. We can also say the $H - S$ bond is weaker than $H - O$ because of the larger size of sulfur atom. Also, $H - S$ due to less polar nature, it donates a proton easily. So, the acidity of ${H_2}S$ is more when compared to ${H_2}O$.