Question

What is the correct formula for boron tribromide?
A. ${{BaB}}{{{r}}_{{3}}}$
B. ${{BB}}{{{r}}_{{3}}}$
C. ${{BeBr}}$
D. ${{{B}}_{{3}}}{{Br}}$
E. ${{B}}{{{a}}_{{3}}}{{Br}}$

Answer 1

In boron tribromide the charge on boron is ${{ + 3}}$ and the charge on bromide is ${{ - 1}}$ and the nature of bonding in boron trihalides are covalent. Boron elements have an unpaired electron in their p -orbital and it belongs to the p – block element of the periodic table.

Complete step by step answer:
Boron tribromide has got a formula of ${{BB}}{{{r}}_{{3}}}$ where Boron has a charge of ${{ + 3}}$ and bromide has a charge of ${{ - 1}}$ and has got a trigonal planar structure
Boron tribromide can be formed by the reaction of amorphous boron with bromine at a lower temperature
${{2B + 3B}}{{{r}}_{{2}}}\xrightarrow{{}}{{2BB}}{{{r}}_{{3}}}$
Trihalides of boron are covalent compounds
i.e., ${{B}}{{{F}}_{{3}}}{{, BB}}{{{r}}_{{3}}}{{, BC}}{{{l}}_{{3}}}{{, B}}{{{I}}_{{3}}}$
The trihalides of Boron are monomeric, highly reactive and volatile
Boron tribromide is a colourless fuming liquid
A binary compound of boron and bromine, is a strong Lewis acid as it is an electron pair acceptor
Reactions:
${{BB}}{{{r}}_{{3}}}$ reacts with water forming a significant amount of gas as and boric acid
${{BB}}{{{r}}_{{3}}}{{ + }}{{{H}}_{{2}}}{{O}}\xrightarrow{{}}{{{H}}_{{3}}}{{B}}{{{O}}_{{3}}}{{ + HBr}}$
${{BB}}{{{r}}_{{3}}}$ reacts with ${{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}$ forming Bromine
${{2BB}}{{{r}}_{{3}}}{{ + 3}}{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}\xrightarrow{{}}{{2}}{{{H}}_{{3}}}{{B}}{{{O}}_{{3}}}{{ + 3B}}{{{r}}_{{2}}}{{ + 3S}}{{{O}}_{{2}}} \uparrow$

So, the correct answer is Option A.

Additional Information:
The commercial samples of Boron tribromide will be sometimes red/brown in colour because of weak bromine contamination. Boron tribromide has got a molar mass of ${{250}}{{.52 gmo}}{{{l}}^{{{ - 1}}}}$ and has got a sharp and irritating odour and it is soluble in ethanol, Carbon tetrachloride. It has a viscosity of ${{7}}{{.31 \times 1}}{{{0}}^{{{ - 4}}}}{{ Pas}}$ and refractive index of ${{1}}{{.00207}}$, vapour pressure of ${{7}}{{.2 kPa}}$, density of ${{2}}{{.643 g/c}}{{{m}}^{{3}}}$ and boiling point of ${{91}}{{.3}}{{{ }}^{{o}}}{{C}}$.
Boron tribromide acts as a demethylating or dealkylating agent for the cleavage of ethers.
The dealkylation mechanism of boron tribromide upon tertiary alkyl ethers proceeds via formation of a complex between the boron centre and ether oxygen which is followed by the elimination of alkyl bromide.
${{ROR + BB}}{{{r}}_{{3}}}\xrightarrow{{}}{{R}}{{{O}}^{{ + }}}\left( {^{{ - }}{{BB}}{{{r}}_{{3}}}} \right){{R}}\xrightarrow{{}}{{ROBB}}{{{r}}_{{2}}}{{ + RBr}}$

Note: Boron tribromide reacts violently with potassium, water, alcohols, sodium, attacks metal, rubber and wood.
Boron tribromide is used in pharmaceuticals, organic synthesis, semiconductor doping, semiconductor plasma etching, image processing and in photovoltaic manufacturing.