CHEMISTRY PLEASE ANSWER ASAP?

O.N. = oxidation number

In NH₄⁺ ion :
In an ion, total oxidation number = charge of the ion
(O.N. of N) + (O.N. of H)×4 = charge of the ion
(O.N. of N) + (+1)×4 = -1
Hence, O.N. of N in NH₄⁺ ion = -3

In NO₂ molecule :
In a molecule, total oxidation number = 0
(O.N. of N) + (O.N. of O)×2 = 0
(O.N. of N) + (-2)×2 = 0
Hence, O.N. of N in NO₂ molecule = +4

In compound NaNO₃ :
In a neutral compound, total oxidation number = 0
(O.N. of Na) + (O.N. of N) + (O.N. of O)×3 = 0
(+1) + (O.N. of N) + (-2)×3 = 0
Hence, O.N. of N in NaNO₃= +5


The answer: A) -3, +4, +5

A) -3, +4, +5
Nitrogen has a higher electronegativity than hydrogen, so in NH4+, the N has a -3 oxidation state. In a nitrate ion, O has a higher electronegativity, and so the N has an oxidation state of +4.