Question

Identify increasing order of acidity for following diprotic acids in aqueous solutions.

• A. \(\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}\)
• B. \(\text{H}_2\text{Se} < \text{H}_2\text{S} < \text{H}_2\text{Te}\)
• C. \(\text{H}_2\text{Te} < \text{H}_2\text{S} < \text{H}_2\text{Se}\)
• D. \(\text{H}_2\text{Se} < \text{H}_2\text{Te} < \text{H}_2\text{S}\)

Hint:

Down the group the H-E bond gets longer and weaker, so acidity rises.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Acidity of hydrides of one group depends on how easily the H-E bond breaks to release \(\text{H}^+\). A weaker bond means a stronger acid.

Step 2: Key Formula or Approach:
Going down group 16 from S to Se to Te, the atom gets bigger, so the E-H bond gets longer and its bond enthalpy falls.

Step 3: Detailed Explanation:
Bond enthalpy order: S-H > Se-H > Te-H.
The weaker the bond, the easier the release of \(\text{H}^+\) in water.
So acid strength rises as \(\text{H}_2\text{S} < \text{H}_2\text{Se} < \text{H}_2\text{Te}\).
Options (B), (C) and (D) place Te or Se at the wrong position and so reverse or scramble this bond-strength trend.

Step 4: Check:
Typical \(pK_{a1}\) values are about 7.0 for \(\text{H}_2\text{S}\), 3.9 for \(\text{H}_2\text{Se}\) and 2.6 for \(\text{H}_2\text{Te}\). A smaller \(pK_a\) means a stronger acid, which agrees with option (A).

Final Answer:
Acidity increases down the group from H2S to H2Te. \[ \boxed{\text{(A) }\text{H}_2\text{S}<\text{H}_2\text{Se}<\text{H}_2\text{Te}} \]