Which of the following will liberate H $_{2}$ when reacting with NaOH?

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Solution

The correct options are
A
Be

C
Al

D
Zn

Na does not react with NaOH. We have already seen that Zn liberates H $_{2}$ when reacting with NaOH. Similar to Zn, Al also reacts with NaOH to liberate H $_{2}$ .

2Al + 2NaOH + 2H $_{2}$ O $⟶$ 2NaAlO $_{2}$ + 3H $_{2}$

Sodium Meta Aluminate

2Al + 6NaOH $f u s e ⟶$ 2NaAlO $_{2}$ + 3H $_{2}$

Sodium Aluminate

Be + 2NaOH $⟶$ Na $_{2}$ BeO $_{2}$ + H $_{2}$

Sodium Beryillate

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