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JEE Advanced 2023 Paper 1 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

Regarding the chemistry of boron and its compounds, the correct statement(s) is/are

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Regarding the chemistry of boron and its compounds, the correct statement(s) is/are
  1. A. In B2H6\mathrm{B_2H_6} the two bridging B ⁣− ⁣H ⁣− ⁣B\mathrm{B\!-\!H\!-\!B} units are three-centre two-electron bonds, whereas the four terminal B ⁣− ⁣H\mathrm{B\!-\!H} bonds are two-centre two-electron bonds.
  2. B. Towards NH3\mathrm{NH_3} the Lewis acidity follows BF3>BCl3>BBr3\mathrm{BF_3}>\mathrm{BCl_3}>\mathrm{BBr_3} because the electronegativity of the halogen decreases down the group.
  3. C. Borazine, B3N3H6\mathrm{B_3N_3H_6}, adds three molecules of HCl\mathrm{HCl} to give B3N3H9Cl3\mathrm{B_3N_3H_9Cl_3} in which each chlorine is attached to a boron atom.
  4. D. Heating borax, Na2B4O7⋅10H2O\mathrm{Na_2B_4O_7\cdot10H_2O}, gives a glassy mass of NaBO2\mathrm{NaBO_2} and B2O3\mathrm{B_2O_3}, which on fusion with a Cu2+\mathrm{Cu^{2+}} salt in the oxidising flame gives a blue bead of Cu(BO2)2\mathrm{Cu(BO_2)_2}.
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Answer: A, C, D

(A) True. Diborane has 12 valence electrons; four terminal B−H\mathrm{B-H} bonds use 8, leaving 4 electrons for two bridging three-centre two-electron bonds.
(B) False. The order is BF3<BCl3<BBr3\mathrm{BF_3}<\mathrm{BCl_3}<\mathrm{BBr_3}. Back-donation of the lone pair of 2p2p fluorine into the empty 2p2p orbital of boron is most efficient because of the good 2p2p-2p2p overlap, which reduces the electron deficiency of boron in BF3\mathrm{BF_3}; the electronegativity argument is overridden.
(C) True. Polar Bδ+−Nδ−\mathrm{B^{\delta+}-N^{\delta-}} bonds add H+\mathrm{H^{+}} to nitrogen and Cl−\mathrm{Cl^-} to boron, giving B3N3H9Cl3\mathrm{B_3N_3H_9Cl_3} with B−Cl\mathrm{B-Cl} and N−H\mathrm{N-H} bonds.
(D) True. Borax on heating loses water and gives Na2B4O7→2NaBO2+B2O3\mathrm{Na_2B_4O_7\to 2NaBO_2+B_2O_3}; B2O3\mathrm{B_2O_3} combines with CuO\mathrm{CuO} to give blue copper(II) metaborate (borax-bead test).
Correct: A, C, D.
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