Borazine is sometimes called inorganic benzene. Which of the following reactions is expected to give this compound?

B_{2} H_{6} + N H_{3} l o w t e m - ------ \to e x c e s s {N H}_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B_{2} H_{6} + N H_{3} h i g h t e m - ------ \to e x c e s s {N H}_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B_{2} H_{6} + N H_{3} h i g h t e m - ----------- \to r a t i o 2 {N H}_{3} : 1 B_{2} H_{6}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A l l o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $B_{2} H_{6} + N H_{3} h i g h t e m - ----------- \to r a t i o 2 {N H}_{3} : 1 B_{2} H_{6}$
Borazine is an inorganic compound with the chemical formula ${(B H}_{3} {) (N H}_{3})$ In this cyclic compound, the three BH units and three NH units alternate. The compound is isoelectronic and isostructural with benzene. Like benzene, borazine is a colourless liquid. For this reason, borazine is sometimes referred to as "inorganic benzene"
Borazine is synthesized from diborane and ammonia in a 1:2 ratio at 250–300 °C with a conversion of 50%.
$3 B_{2} H_{6} + 6 N H_{3} \to 2 B_{3} H_{6} N_{3} + 12 H_{2}$

Suggest Corrections

Similar questions

Q. Diborane is a Lewis acid forming addition compound

B_{2} H_{6} \cdot 2 N H_{3}

with

N H_{3},

a Lewis base. This -

A compound of B (Boron), X reacts at $200^{\circ} C$ with $N H_{3}$ to give another compound Y which is called as inorganic benzene. The compound Y is colorless liquid and is highly light sensitive. The compound X with excess of $N H_{3}$ and at still higher temp gives $(B N)_{n}$ . The compound X and Y respectively.