Q65 · CSIR-NET Chemistry, December 2015
Paper: CSIR-NET December 2015 · Subject: Inorganic Chemistry · Chapter: Main Group Elements · Topic: Hydrides · Marks: 2 · Difficulty: Hard
The order of increasing Bronsted acidity for boron hydrides is
(a)$\mathrm{B}_{5} \mathrm{H}_{9}<\mathrm{B}_{6} \mathrm{H}_{10}<\mathrm{B}_{10} \mathrm{H}_{14}$
(b)$\mathrm{B}_{10} \mathrm{H}_{14}<\mathrm{B}_{5} \mathrm{H}_{9}<\mathrm{B}_{6} \mathrm{H}_{10}$
(c)$\mathrm{B}_{6} \mathrm{H}_{10}<\mathrm{B}_{10} \mathrm{H}_{14}<\mathrm{B}_{5} \mathrm{H}$
(d)$\mathrm{B_{10}H_{14}<B_6H_{10}<B_5H_9}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Brønsted acidity of the bridging B–H–B hydrogens rises with cluster size (better charge delocalisation in the conjugate base): B₅H₉ < B₆H₁₀ < B₁₀H₁₄.
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