Question:
Which of the following hydrides of Group \(16\) elements has highest reducing property?
Which of the following hydrides of Group \(16\) elements has highest reducing property?
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Going down the group, the H-E bond gets weaker, so the hydride gives up hydrogen more easily.
Updated On: Oct 1, 2026
- \(\text{H}_2\text{O}\)
- \(\text{H}_2\text{S}\)
- \(\text{H}_2\text{Se}\)
- \(\text{H}_2\text{Te}\)
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The Correct Option is D
Solution and Explanation
Step 1: Recall the trend
The hydrides of group 16 are H\(_2\)O, H\(_2\)S, H\(_2\)Se and H\(_2\)Te. A reducing agent donates hydrogen or electrons.
Step 2: Link bond strength to reduction
Down the group, the size of the central atom grows and the E-H bond becomes longer and weaker.
A weaker bond breaks more easily, so the hydride reduces more strongly.
Step 3: Conclude
The order of reducing power is \(\text{H}_2\text{O}<\text{H}_2\text{S}<\text{H}_2\text{Se}<\text{H}_2\text{Te}\). So H\(_2\)Te is the strongest, option (D).
Final Answer:
H\(_2\)Te has the weakest bond and is the best reducing agent, option (D). \[ \boxed{\text{(D) H}_2\text{Te}} \]
The hydrides of group 16 are H\(_2\)O, H\(_2\)S, H\(_2\)Se and H\(_2\)Te. A reducing agent donates hydrogen or electrons.
Step 2: Link bond strength to reduction
Down the group, the size of the central atom grows and the E-H bond becomes longer and weaker.
A weaker bond breaks more easily, so the hydride reduces more strongly.
Step 3: Conclude
The order of reducing power is \(\text{H}_2\text{O}<\text{H}_2\text{S}<\text{H}_2\text{Se}<\text{H}_2\text{Te}\). So H\(_2\)Te is the strongest, option (D).
Final Answer:
H\(_2\)Te has the weakest bond and is the best reducing agent, option (D). \[ \boxed{\text{(D) H}_2\text{Te}} \]
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