Calculate the oxidation number of nitrogen and chlorine in $NOClO_4$.
(N=+3, CL=+7) The compound $NOClO_4$ is an ionic compound consisting of $NO^+$ and $ClO_4^-$ ions.
For the $NO^+$ ion:
Let the oxidation number of $N$ be $x$.
$x + 1(-2) = +1$
$x - 2 = +1$
$x = +3$
Thus,the oxidation number of $N$ is $+3$.
For the $ClO_4^-$ ion:
Let the oxidation number of $Cl$ be $y$.
$y + 4(-2) = -1$
$y - 8 = -1$
$y = +7$
Thus,the oxidation number of $Cl$ is $+7$.
For the $NO^+$ ion:
Let the oxidation number of $N$ be $x$.
$x + 1(-2) = +1$
$x - 2 = +1$
$x = +3$
Thus,the oxidation number of $N$ is $+3$.
For the $ClO_4^-$ ion:
Let the oxidation number of $Cl$ be $y$.
$y + 4(-2) = -1$
$y - 8 = -1$
$y = +7$
Thus,the oxidation number of $Cl$ is $+7$.
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