(s) about boron and its compounds.

One or more options correct

Choose the correct statement(s) about boron and its compounds.

A. Heating borax with a cobalt(II) salt on a platinum loop in an oxidising flame gives a blue bead of $\mathrm{Co(BO_2)_2}$
B. Boric acid, $\mathrm{B(OH)_3}$ , is a monobasic acid in water because it accepts $\mathrm{OH^-}$ rather than donating a proton, giving $\mathrm{[B(OH)_4]^-}$
C. Diborane reacts with excess $\mathrm{NH_3}$ at low temperature by unsymmetrical cleavage to give $\mathrm{[BH_2(NH_3)_2]^+[BH_4]^-}$
D. In $\mathrm{B_2H_6}$ both boron atoms are $sp^2$ hybridised and all six hydrogen atoms lie in one plane with the two boron atoms

Show answer and solution

Answer: A, B, C

(A) Correct: borax gives

\mathrm{Na_2B_4O_7\rightarrow 2NaBO_2+B_2O_3}

on heating, and

\mathrm{B_2O_3+CoO\rightarrow Co(BO_2)_2}

is the blue bead (borax bead test). (B) Correct:

\mathrm{B(OH)_3+2H_2O\rightleftharpoons[B(OH)_4]^-+H_3O^+}

; it behaves as a weak Lewis, monobasic acid. (C) Correct: with a small excess/low temperature

\mathrm{NH_3}

attacks one boron unsymmetrically, giving the ionic salt

\mathrm{[BH_2(NH_3)_2]^+[BH_4]^-}

(bulkier amines such as

\mathrm{NMe_3}

instead cleave diborane symmetrically, giving

\mathrm{BH_3\cdot NMe_3}

). (D) Incorrect: each boron in diborane is

sp^3

hybridised; the two borons and four terminal hydrogens lie in one plane, but the two bridging

\mathrm{B-H-B}

(three-centre, two-electron) hydrogens lie above and below that plane. Hence (A), (B), (C).

Original question written in the JEE Advanced 2023 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