How do you find the oxidation number of nitrogen in these molecules?

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How do you find the oxidation number of nitrogen in these molecules?

$H N O_{3}$ $HNO_3$
$N_{2} H_{4}$ $N_2H_4$
$N O_{2}^{+}$ $NO_2^+$
$N_{2}$ $N_2$
$N O_{2}^{-}$ $NO_2^-$
$N O_{2}$ $NO_2$
$N_{2} O_{4}$ $N_2O_4$
$N_{2} O$ $N_2O$
$N O$ $NO$

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Answer 1 · 2016-10-27T15:35:53

$H N O_{3}$ $HNO_3$ $Nitrogen +V$ $"Nitrogen +V"$

Explanation:

$N_{2} H_{4},$ $N_2H_4,$ $Nitrogen -II$ $"Nitrogen -II"$

$N O_{2}^{+},$ $NO_2^+,$ $Nitrogen +V$ $"Nitrogen +V"$

$N_{2},$ $N_2,$ $Nitrogen 0$ $"Nitrogen 0"$

$N O_{2}^{-},$ $NO_2^-,$ $Nitrogen +III$ $"Nitrogen +III"$

$N O_{2},$ $NO_2,$ $Nitrogen +IV$ $"Nitrogen +IV"$

$N_{2} H_{4},$ $N_2H_4,$ $Nitrogen -II$ $"Nitrogen -II"$

$N_{2} O,$ $N_2O,$ $Nitrogen +I$ $"Nitrogen +I"$ , $N \equiv N^{+} - O^{-}$ $N-=N^(+)-O^(-)$ , the terminal nitrogen is zerovalent, and the central $N$ $N$ is $+ I I$ $+II$ , i.e. average $+ I$ $+I$ state.

$N_{2} O_{4},$ $N_2O_4,$ $Nitrogen +IV$ $"Nitrogen +IV"$

$N O,$ $NO,$ $Nitrogen +II$ $"Nitrogen +II"$

I think I have got these right. Oxygen is more electronegative than nitrogen so it gets the electric charge in determining oxidation state.

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How do you find the oxidation number of nitrogen in these molecules?

$H N O_{3}$ $HNO_3$
$N_{2} H_{4}$ $N_2H_4$
$N O_{2}^{+}$ $NO_2^+$
$N_{2}$ $N_2$
$N O_{2}^{-}$ $NO_2^-$
$N O_{2}$ $NO_2$
$N_{2} O_{4}$ $N_2O_4$
$N_{2} O$ $N_2O$
$N O$ $NO$

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How do you find the oxidation number of nitrogen in these molecules?

HNO3HNO_3 N2H4N_2H_4 NO+2NO_2^+ N2N_2 NO−2NO_2^- NO2NO_2 N2O4N_2O_4 N2ON_2O NONO

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$H N O_{3}$ $HNO_3$
$N_{2} H_{4}$ $N_2H_4$
$N O_{2}^{+}$ $NO_2^+$
$N_{2}$ $N_2$
$N O_{2}^{-}$ $NO_2^-$
$N O_{2}$ $NO_2$
$N_{2} O_{4}$ $N_2O_4$
$N_{2} O$ $N_2O$
$N O$ $NO$