Question:
Consider the following statements:
Statement-I: The products formed when diborane burns in air are \({B}_2{O}_3\), \({H}_2\), and \({O}_2\).
Statement-II: Hybridization of boron atom in orthoboric acid is \(sp^2\). The correct answer is:
Consider the following statements:
Statement-I: The products formed when diborane burns in air are \({B}_2{O}_3\), \({H}_2\), and \({O}_2\).
Statement-II: Hybridization of boron atom in orthoboric acid is \(sp^2\). The correct answer is:
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Remember combustion products of boranes and hybridization states of boron compounds.
Updated On: Jun 3, 2025
- Both statement-I and statement-II are correct
- Both statement-I and statement-II are not correct
- Statement-I is correct, but statement-II is not correct
- Statement-I is not correct, but statement-II is correct
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The Correct Option is D
Solution and Explanation
- When diborane burns in air, products are \(\mathrm{B}_2\mathrm{O}_3\) and \(\mathrm{H}_2\mathrm{O}\), not \(\mathrm{H}_2\) and \(\mathrm{O}_2\).
- Boron in orthoboric acid \(\mathrm{B(OH)}_3\) is trigonal planar with \(sp^2\) hybridization.
Therefore, Statement-I is incorrect, Statement-II is correct.
- Boron in orthoboric acid \(\mathrm{B(OH)}_3\) is trigonal planar with \(sp^2\) hybridization.
Therefore, Statement-I is incorrect, Statement-II is correct.
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