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Standard XII

Chemistry

Borax

A compound ...

Question

A compound $A$ of boron reacts with $N M e_{3}$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$ . Compound $C$ is an acid.

Compound $C$ is :

diborane

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boric acid

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borate salt

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none of these

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Solution

The correct option is B boric acid
$B_{2} H_{6} + 2 N M e_{3} ⟶ 2 B H_{3} . N M e_{3}$
Diborane (A) Adduct (B)

$B H_{3} . N M e_{3} + 3 H_{2} O ⟶ H_{3} B O_{3} + N M e_{3} + 3 H_{2}$
Boric acid $↓$
(C) (D) gas

Compound "C" is boric acid $- H_{3} B O_{3}$

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A compound A of boron reacts with NMe3 to give an adduct B which on hydrolysis gives a compound C and a gas D. Compound C is an acid.Compound C is :

The correct option is B boric acid B2H6+2NMe3⟶2BH3.NMe3Diborane (A) Adduct (B)BH3.NMe3+3H2O⟶H3BO3+NMe3+3H2 Boric acid ↓ (C) (D) gasCompound "C" is boric acid −H3BO3

A compound $A$ of boron reacts with $N M e_{3}$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$ . Compound $C$ is an acid.

Compound $C$ is :

The correct option is B boric acid
$B_{2} H_{6} + 2 N M e_{3} ⟶ 2 B H_{3} . N M e_{3}$
Diborane (A) Adduct (B)

$B H_{3} . N M e_{3} + 3 H_{2} O ⟶ H_{3} B O_{3} + N M e_{3} + 3 H_{2}$
Boric acid $↓$
(C) (D) gas

Compound "C" is boric acid $- H_{3} B O_{3}$