Arrange the species according to the oxidation state of nitrogen in each. Note that highest refers to the most positive oxidation state of nitrogen, whereas lowest refers to the most negative oxidation state. A. NO3-
B. NO4+
C. N2
D. NH2OH
E. NO2-
1. Highest oxidation state
2. Lowest oxidation state

A. NO₃⁻ ⇒ +5

B. NO₄⁺ ⇒ +9

C. N₂ ⇒ 0

D. NH₂OH ⇒ -1

E. NO₂⁻ ⇒ +3

Explanation:

To calculate the oxidation number of an element in a compound, we have to know the oxidation number of the other elements. Then, we have to consider that the sum of the oxidation number of each atom multiplied by the subscripts is equal to the net charge of the compound.

A. NO₃⁻

This is the ion nitrate. Oxygen atoms (O) has an oxidation number of -2 because it derives from an oxide. In this case, the net charge of the ion is -1. Thus, we calculate the oxidation number of N as follows:

N + (3 x (-2)) = -1

N - 6 = -1 ⇒ N= -1 + 6 = +5

B. NO₄⁺

In this case, the sum of the oxidation numbers of O and N multiplied by the subscripts is equal to +1:

N + (4 x (-2)) = +1

N - 8 = +1 ⇒ N = +1 +8 = +9

C. N₂

The oxidation number of N is 0 because N₂ is an elemental substance.

D. NH₂OH

The oxidation number of H is +1 and -2 for O. The net charge of the molecule is 0.

N + (H x 3) + (O) = 0

N + (+1 x 3) + (-2) = 0

N + 3 -2 = 0 ⇒ N = 2 - 3 = -1

E. NO₂⁻

The net charge of the ion nitrite is -1.

N + (2 x O) = -1

N + (2 x (-2)) = -1 ⇒ N = -1 + 4 = +3

Therefore:

Highest oxidation number = B. NO₄⁺ (+9)

Lowest oxidation number = D. NH₂OH (-1)

i think that the answer is c