$B e H_{2}$ is prepared by the reaction of:

2 B e C l_{2} + L i A l H_{4} e t h e r \to 2 B e H_{2} + A l C l_{3} + L i C l

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

B e + H_{2} \to B e H_{2}

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

Both

(A)

and

(B)

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

None of these

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

Open in App

Solution

The correct option is D $2 B e C l_{2} + L i A l H_{4} e t h e r \to 2 B e H_{2} + A l C l_{3} + L i C l$
Reaction for preparation of $B e H_{2}$
$2 B e C l_{2} + L i A l H_{4} e t h e r \to 2 B e H_{2} + A l C l_{3} + L i C l$
It is difficult to prepare $B e H_{2}$ by direct reaction of $B e$ and hydrogen gas.
All other alkali metal hydrides can be prepare accept Be:
$M + H_{2} \to M H_{2}$

Suggest Corrections

Similar questions

$2 B e C l_{2} + L i A l H_{4} \to A - - + L i C l + A l C l_{3}$ . $A$ is

B e H_{2}

M g H_{2}

C a H_{2}

B e H_{2}

B a H_{2}

B C l_{3} + L i A l H_{4} \to A + L i C l + A l C l_{3}

A + H_{2} O \to B + H_{2}

B \to C

C a H_{2}

B e H_{2}

B a H_{2}

Join BYJU'S Learning Program