Question:
Which of the following reactions give $\mathrm{H_2}$ as one of the products?
(Reactions are not balanced.)
- $\mathrm{NaBH_4 + I_2}$
- $\mathrm{B_2H_6 + N(CH_3)_3}$
- $\mathrm{Al + NaOH + H_2O}$
- $\mathrm{BF_3 + NaH}$
Which of the following reactions give $\mathrm{H_2}$ as one of the products?
(Reactions are not balanced.)
- $\mathrm{NaBH_4 + I_2}$
- $\mathrm{B_2H_6 + N(CH_3)_3}$
- $\mathrm{Al + NaOH + H_2O}$
- $\mathrm{BF_3 + NaH}$
Show Hint
Look for redox reactions or metal-water-type reactions to identify \(\mathrm{H_2}\) evolution.
Updated On: Jun 4, 2025
- I, II & III only
- II & IV only
- I & III only
- II, III & IV only
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The Correct Option is C
Solution and Explanation
Let’s evaluate each: - I. \(\mathrm{NaBH_4 + I_2 \rightarrow}\): Produces \(\mathrm{H_2}\) - II. \(\mathrm{B_2H_6 + N(CH_3)_3}\): Lewis acid-base reaction, no \(\mathrm{H_2}\) - III. \(\mathrm{Al + NaOH + H_2O \rightarrow NaAlO_2 + H_2}\) - IV. \(\mathrm{BF_3 + NaH}\): Forms \(\mathrm{Na[BF_4]}\), no \(\mathrm{H_2}\) Only I and III produce hydrogen gas.
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