IMAT 2013 Q50 [Nitrogen Compounds]

The following are some compounds of nitrogen:
NOCl, KNO_2, NO_2, NO_2Cl, Ca(NO_3)_2

What oxidation numbers are shown by nitrogen in these compounds?

A. 3, 4, 5, 6
B. 2, 3, 4, 5
C. 2, 3, 4, 5, 6
D. 4, 5, 6
E. 3, 4, 5

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We can first discuss how each compound has been formed to understand the oxidation number of nitrogen in each.

For NOCl- The formation of this compound is a redox reaction.

Cl^0 + e → Cl^{-I} (reduction)

N^0 - 3 e → N^{III} (oxidation)

O^0 + 2 e → O^{-II} (reduction)

We can see that for this compound, Nitrogen’s oxidation number is 3. We can eliminate option D.

For KNO_2- The formation of this compound is a redox reaction.

K^0 - 1 e → K^I (oxidation)

N^0 - 3 e → N^{III} (oxidation)

2 O^0 + 4 e → 2 O^{-II} (reduction)

We can see that for this compound, Nitrogen’s oxidation number is also 3.

For NO_2- The formation of this compound is a redox reaction.

N^0 - 4 e → N^{IV} (oxidation)

2 O^0 + 4 e → 2 O^{-II} (reduction)

We can see that for this compound, Nitrogen’s oxidation number is 4.

For NO_2Cl- The formation of this compound is a redox reaction.

Cl^0 + e → Cl^{-I} (reduction)

N^0 - 5 e → N^V (oxidation)

2 O^0 + 4 e → 2 O^{-II} (reduction)

We can see that for this compound, Nitrogen’s oxidation number is 5.

For Ca(NO_3)_2- The formation of this compound is a redox reaction.

2 N^0 - 10 e → 2 N^V (oxidation)

Ca^0 - 2 e → Ca^{II} (oxidation)

6 O^0 + 12 e → 6 O^{-II} (reduction)

We can see that for this compound, Nitrogen’s oxidation number is also 5.

As a result, our answer is 3-4-5, E.

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