The strongest reducing agent amont the following is:
- \( \text{NH}_3 \)
- \( \text{SbH}_3 \)
- \( \text{BiH}_3 \)
- \( \text{PH}_3 \)
The Correct Option is C
Solution and Explanation
The reducing strength of hydrides of Group 15 elements increases as we move down the group. This trend can be explained based on the following factors: - As we move down the group, the size of the central atom increases, which weakens the E − H bond (where E is the central atom.) The bond dissociation energy decreases. This makes it easier for the hydride to donate hydrogen (as H$^-$), thereby increasing its reducing power. Among the given options:
NH$_3$ (ammonia) has the strongest N--H bond and, therefore, is the weakest reducing agent.
PH$_3$, SbH$_3$, and BiH$_3$ follow the trend of increasing reducing strength due to weaker bonds as we move down the group.
BiH$_3$ (bismuthine) has the weakest Bi--H bond, making it the strongest reducing agent among the given hydrides.
Conclusion: The strongest reducing agent is BiH$_3$.
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A. The stability of the hydrides decreases in the order \(\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3 > \text{BiH}_3\)
B. The reducing ability of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
C. Among the hydrides, \(\text{NH}_3\) is a strong reducing agent while \(\text{BiH}_3\) is a mild reducing agent.
D. The basicity of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
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