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Diborane

B 2 H 6 react...

Question

$B_{2} H_{6}$ reacts with $O_{2}$ and $H_{2} O$ respectvely to form:

A

B_{2} O_{3}, H_{3} B O_{3}

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B

B_{2} O_{3}, B H_{4}

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C

H B O_{2}, H_{3} B O_{3}

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D

H_{2} B O_{3}, H B O_{2}

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Solution

The correct option is A $B_{2} O_{3}, H_{3} B O_{3}$
Diborane catches fire spontaneously upon exposure to air. When diborane $(B_{2} H_{6})$ reacts with oxygen (or air) burns spontaneously to produce a large amount of heat. Thus, the reaction is exothermic. The reaction which takes place is:
$B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O$
$Δ H = - 1976 K J m o l^{- 1}$

Diborane react with water to form boric acid.
$B_{2} H_{6} + 6 H_{2} O \to 2 H_{3} B O_{3} + 6 H_{2}$

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Similar questions

B_{2} H_{6}

) reacts independently with

O_{2}

H_{2} O

to produce, respectively:

B_{2} H_{6}

O_{2}

H_{2} O

respectvely to form:

B_{2} H_{6}

O_{2}

H_{2} O

respectvely to form:

Q. Orthoboric acid, on heating, decomposes in two ways:
(I)

H_{3} B O_{3} \to H B O_{2} + H_{2} O

H_{5} B O_{3} \to B_{2} O_{3} + H_{2} O

H_{3} B O_{3}

decomposes by (I) pathway and remaining by (II) pathway. If total 11 moles of water are formed, then the number of moles of

B_{2} O_{3}

(B_{2} H_{6})

reacts independently with

O_{2}

H_{2} O

to produce, respectively.

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