Step 1: Understanding the given reactions
- Sodium borohydride (\( NaBH_4 \)) reacts with iodine to release \( H_2 \):
\[
NaBH_4 + I_2 \rightarrow NaI + B_2H_6 + H_2
\]
- Hydrolysis of boranes also releases \( H_2 \):
\[
B_2H_6 + H_2O \rightarrow B(OH)_3 + H_2
\]
- The reaction of ammonia with diborane forms an adduct and releases \( H_2 \):
\[
B_2H_6 + NH_3 \rightarrow (BH_3)_2 \cdot NH_3 + H_2
\]
- However, burning diborane in oxygen results in complete combustion, forming \( B_2O_3 \) and water vapor, without producing dihydrogen:
\[
B_2H_6 + O_2 \rightarrow B_2O_3 + H_2O
\]