$B e H_{2}$ and $M g H_{2}$ are prepared using the reagent:

L i H

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A l H_{3}

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L i A l H_{4}

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H_{2}

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Solution

The correct options are
C $L i A l H_{4}$
D $H_{2}$
Magnesium directly combines with hydrogen to form magnesium hydride.
$M g + H_{2} H e a t - -- \to M g H_{2}$
Beryllium hydride is prepared by the reaction of beryllium dichloride with lithium aluminum hydride.
$2 B e C l_{2} + L i A l H_{4} \to 2 B e H_{2} + L i C l + A l C l_{3}$

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Similar questions

L i H + A l H_{3} ⟶ L i A l H_{4}

A l H_{3}

L i H

Look at the following reaction:
$L i H + A l H_{3} ⟶ L i A l H_{4}$

Here $A l H_{3}$ and $L i H$ act as:

L i H, M g H_{2}

C u H

Arrange the following

(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.

(ii) LiH, NaH and CsH in order of increasing ionic character.

(iii) H–H, D–D and F–F in order of increasing bond dissociation enthalpy.

(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.

Arrange the following.
$C a H_{2}, B e H_{2}$ and $T i H_{2}$ in order of increasing electrical conductance.

LiH, NaH and CsH in order of increasing ionic character.

H - H, D - D and F - F in order in order of increasing bond dissociation enthalpy.

$N a H, m g H_{2}$ and $H_{2} O$ in order of increasing reducing property.

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