Question

The oxidation state of Al in $LiAl{H}_4$ is:

• A. zero
• B. $+1$
• C. $+2$
• D. $+3$

Answer 1

The correct option is D $+3$

${LiAlH}_4⟶{Li}^{+}\left[{AlH}_4^{-}\right]$

Let the oxidation state of $Al$ be $x$

Since hydrogen is more electronegtive than aluminium, thus the oxidation state of $H$ will be taken as $-1$.

The oxidation state of $H$ in ${AIH}_4^{-}$ is $-1$

$\therefore$ For ${AIH}_4^{-}\Rightarrow x+4(-1)=-1$

$x-4=-1$

$x=+3$

Therefore, the oxidation state of $AI$ in ${LiAIH}_4$ is $+3$.