What are the oxidation numbers of the elements in the nitrite ion, NO2−?

Question

What are the oxidation states of the elements in the nitrite ion (NO2−)? Specifically, how do we determine the oxidation number of nitrogen and the oxidation numbers of the oxygen atoms in this ion?

Anonymous · Accepted Answer

In order to get the oxidation number of each element, refer to the rules of assigning oxidation number.

First, the rule says that "O" oxygen always have a (-2) charge(except if peroxide), this means the "x" is the unknown charge of nitrogen.

x= charge of nitrogen

-2 = charge of oxygen

Simply multiply its charge into how many atoms of an elements is present.

1x= x is multiplied by 1 since there is only 1 nitrogen

[2(-2)] = the oxygen's charge of -2 is multiplied to 2 since there are 2 oxygen atoms

And since (NO2-) is a polyatmic ion with a charge of -1 , the oxidation numbers should all add up to (-) or (-1)

Represent each in the equation:

1x + [2(-2)] = -1

1x -4 = -1

1x = -1 + 4

x= +3

This means that nitrogen has an oxidation of +3, and oxygen is -2

There is an easy way to assign oxidation by the way, I just chose to demonstrate it algebraically to show how it's being solved.

May Hayes III · Answer

Charge on the polyatomic anion is -1, so the sum of the charges must be -1
Oxygen's oxidation # is -2. Since there are 2 oxygen atoms in the compound,
the product is -4. 3 must be the oxidation # of N, since 3+ -4 = -1.
nitrate's oxidation #'s for N and O2, respectively, are 3 and -2

Anonymous · Answer

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Mrs. Demetris Swift Jr.

What are the oxidation numbers of the elements in the nitrite ion, NO2−?

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