Match the following compounds with their respective oxidation numbers of the central atom or specific element:

Compound	Oxidation number
$1$. $C_2H_6$	$a$. $+3$
$2$. $H_2O_2$	$b$. $+6$
$3$. $LiAlH_4$	$c$. $-3$
$4$. $HO_2^-$	$d$. $-1$
$5$. $PbSO_4$	$e$. $+1$

(A) The oxidation numbers are calculated as follows:
$1$. In $C_2H_6$,$2x + 6(+1) = 0 \implies x = -3$.
$2$. In $H_2O_2$,$2(+1) + 2x = 0 \implies x = -1$.
$3$. In $LiAlH_4$,$+1 + x + 4(-1) = 0 \implies x = +3$.
$4$. In $HO_2^-$,$(+1) + 2x = -1 \implies 2x = -2 \implies x = -1$. (Note: The provided option $e$ is $+1$,which is incorrect for $O$ in $HO_2^-$. However,based on standard matching,$4$ matches $d$ or $e$ depending on the intended element. Given the options,$4$ matches $d$ $(-1)$).
$5$. In $PbSO_4$,$Pb^{2+} + SO_4^{2-}$,$S$ in $SO_4^{2-}$ is $x + 4(-2) = -2 \implies x = +6$.
Correct matching: $1-c, 2-d, 3-a, 4-d, 5-b$.