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JEE Advanced 2016 Paper 2 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

(s) about boron and aluminium compounds.

One or more options correct

Select the correct statement(s) about boron and aluminium compounds.
  1. A. In B2H6\mathrm{B_2H_6} the bridging B−H−B\mathrm{B{-}H{-}B} units are three-centre two-electron bonds, and the bridging B−H\mathrm{B{-}H} bonds are shorter than the terminal ones.
  2. B. BCl3\mathrm{BCl_3} is a stronger Lewis acid than BF3\mathrm{BF_3}.
  3. C. Boric acid, H3BO3\mathrm{H_3BO_3}, is a tribasic Brønsted acid.
  4. D. In the dimer Al2Cl6\mathrm{Al_2Cl_6} each aluminium atom is sp2sp^2 hybridised.
Show answer and solution

Answer: B

(A) Incorrect. The bridging B-H-B bonds are indeed 3c-2e bonds, but they are longer (about 134134 pm) than the terminal 2c-2e B-H bonds (about 119119 pm), so the statement is wrong in its second half. (B) Correct. In BF3\mathrm{BF_3} the lone pairs of the small fluorine atoms overlap efficiently with the empty 2p2p orbital of boron (pπp\pi-pπp\pi back bonding), reducing its electron deficiency; the 3p3p orbitals of chlorine overlap poorly, so BCl3\mathrm{BCl_3} accepts an electron pair more readily. (C) Incorrect. Boric acid is a monobasic Lewis acid: B(OH)3+H2O⇌[B(OH)4]−+H+\mathrm{B(OH)_3+H_2O\rightleftharpoons[B(OH)_4]^-+H^+}, accepting OH−\mathrm{OH^-} rather than donating a proton. (D) Incorrect. Each Al in Al2Cl6\mathrm{Al_2Cl_6} is surrounded tetrahedrally by four Cl (two bridging, two terminal), so it is sp3sp^3 hybridised. Only B is correct.
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Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake