Question

What is the oxidation number of $\mathrm{Li}$ in ${\mathrm{LiAlH}}_4$?

Answer 1

Oxidation number: Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element.

Calculating the oxidation number of Lithium(Li) in ${\mathrm{LiAlH}}_4$:

The oxidation state of Hydrogen(H) = -1

The oxidation state of Aluminium(Al) = +3

Consider the oxidation state of Lithium(Li) = x

Now,

$$\begin{gathered} x+3+4\left(-1\right)=0 \\ x+3-4=0 \\ x-1=0 \\ x=1 \end{gathered}$$

So, the oxidation number of Lithium $\left(\mathrm{Li}\right)$ in Lithium aluminium hydride$\left({\mathrm{LiAlH}}_4\right)$ is $+1$.