Q.Write balanced equations for:

1. $2\text{BF}_3 + 6\text{LiH} \rightarrow \text{B}_2\text{H}_6 + 6\text{LiF}$.
2. $\text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{BO}_3 + 6\text{H}_2$.
3. $2\text{NaH} + \text{B}_2\text{H}_6 \rightarrow 2\text{NaBH}_4$.
4. $2\text{H}_3\text{BO}_3 \xrightarrow{\Delta} \text{B}_2\text{O}_3 + 3\text{H}_2\text{O}$. …

1. (i) $\text{BF}_3 + \text{LiH}$. Lithium hydride acts as a reducing agent, reducing BF$_3$ to diborane:

2. (ii) $\text{B}_2\text{H}_6 + \text{H}_2\text{O}$. Diborane is readily hydrolysed by water to give boric acid and hydrogen gas:

3. (iii) $\text{NaH} + \text{B}_2\text{H}_6$. Sodium hydride adds across diborane to give sodium borohydride, an important reducing agent:

4. (iv) $\text{H}_3\text{BO}_3 \xrightarrow{\Delta}$. On heating, boric acid undergoes stepwise loss of water — first to metaboric acid, and ultimately (on strong/red heat) to boric oxide:

5. (v) $\text{Al} + \text{NaOH}$. Amphoteric aluminium dissolves in strong alkali, evolving hydrogen and forming sodium aluminate:

6. (vi) $\text{B}_2\text{H}_6 + \text{NH}_3$. At low temperature, diborane first forms a simple Lewis adduct with ammonia; on strong heating this adduct loses hydrogen and cyclises into borazine ("inorganic benzene"): …

Recall each named reaction from Group 13 chemistry — reduction of BF$_3$ by hydride reagents, hydrolysis of diborane, hydride-addition to give borohydrides, thermal dehydration of boric acid, amphoteric dissolution of Al, and diborane–ammonia adduct/borazine formation — a …

• Leaving BF$_3$/LiH or NaH/B$_2$H$_6$ equations unbalanced (getting the 1:3 or 1:1 stoichiometry wrong).
• Writing the B$_2$H$_6$ + NH$_3$ reaction only as the low-temperature adduct without noting that strong heating gives the distinct product borazine, B$_3$N$_3$H$_6$. …