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Question:
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In which of the following reactions, hydrogen is liberated?
(i) $\mathrm{Al}(s)+\mathrm{HCl}(a q) \longrightarrow$
(ii) $\mathrm{Al}(s)+\mathrm{NaOH}(a q) \longrightarrow$
(iii) $\mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow$
Options:
(i) $\mathrm{Al}(s)+\mathrm{HCl}(a q) \longrightarrow$
(ii) $\mathrm{Al}(s)+\mathrm{NaOH}(a q) \longrightarrow$
(iii) $\mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow$
Solution:
2255 Upvotes
Verified Answer
The correct answer is:
i, ii, iii
(i) $2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{AlCl}_3(a q)+3 \mathrm{H}_2$
(ii)
$$
\begin{aligned}
2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q) \longrightarrow & \\
& 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]+3 \mathrm{H}_2 \\
& \begin{array}{l}
\text { Sodium tetrahydroxo } \\
\text { aluminate (III) }
\end{array}
\end{aligned}
$$
(iii)
$$
\begin{aligned}
\mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow & \mathrm{BH}_3+\mathrm{NaI}+\mathrm{HI} \\
& \downarrow\left(\mathrm{NaBH}_4\right) \\
& 2 \mathrm{BH}_3+2 \mathrm{NaI}+\mathrm{H}_2
\end{aligned}
$$
(ii)
$$
\begin{aligned}
2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q) \longrightarrow & \\
& 2 \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_4\right]+3 \mathrm{H}_2 \\
& \begin{array}{l}
\text { Sodium tetrahydroxo } \\
\text { aluminate (III) }
\end{array}
\end{aligned}
$$
(iii)
$$
\begin{aligned}
\mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow & \mathrm{BH}_3+\mathrm{NaI}+\mathrm{HI} \\
& \downarrow\left(\mathrm{NaBH}_4\right) \\
& 2 \mathrm{BH}_3+2 \mathrm{NaI}+\mathrm{H}_2
\end{aligned}
$$
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