$\text{LiAlH}_4$ has $\text{AlH}_4^-$ ions, in which aluminium is in oxidation state $+3$. $\text{LiAlH}_4$ reacts with water, as follows. $\text{LiAlH}_4 + 4\text{H}_2\text{O} \rightarrow 4\text{H}_2 + \text{LiOH} + \text{Al(OH)}_3$ In this reaction, each of the four water molecules forms one hydroxide ion. This happens by losing one $\text{H}^+$ ion, which then reacts with the $\text{LiAlH}_4$. Which changes in oxidation number take place in this reaction?
- AAl increases by $1$ and H decreases by $1$.
- BH decreases by $2$ and also increases by $1$.
- CH increases by $1$ and also decreases by $1$.
- DO decreases by $1$ and H increases by $1$.