The oxidation number of N and $C l$ in $N O C l O_{4}$ respectively are :

+2 and +7

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+3 and +7

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-3 and +5

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+2 and -7

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Solution

The correct option is A +3 and +7
$N O C l O_{4}$ is actually ${N O}^{+} {C l O}_{4}^{-}$ .

Let the oxidation state of N in ${N O}^{+}$ is.

$x + (- 2) = + 1$

$x = + 1 + 2 = + 3$

Let the oxidation state of $C l$ in ${C l O}_{4}^{-}$ is y.

$y + (- 2) \times 4 = - 1$

$y - 8 = - 1; y = + 7$

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The oxidation number of N and Cl in NOClO4 respectively are :

The correct option is A +3 and +7NOClO4 is actually NO+ClO−4.Let the oxidation state of N in NO+ is.x+(−2)=+1x=+1+2=+3Let the oxidation state of Cl in ClO−4 is y.y+(−2)×4=−1y−8=−1;y=+7

The oxidation number of N and $C l$ in $N O C l O_{4}$ respectively are :

The correct option is A +3 and +7
$N O C l O_{4}$ is actually ${N O}^{+} {C l O}_{4}^{-}$ .

Let the oxidation state of N in ${N O}^{+}$ is.

$x + (- 2) = + 1$

$x = + 1 + 2 = + 3$

Let the oxidation state of $C l$ in ${C l O}_{4}^{-}$ is y.

$y + (- 2) \times 4 = - 1$

$y - 8 = - 1; y = + 7$