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Question
What is the change in oxidation number of nitrogen from HNO2 to HNO3 in the following reaction? Ignore the other two products.
3HNO2→HNO3+2NO+H2O
What is the change in oxidation number of nitrogen from HNO2 to HNO3 in the following reaction? Ignore the other two products.
3HNO2→HNO3+2NO+H2O
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Solution
The correct option is A 2
Let us briefly recall the rules for allocating oxidation numbers:
1. The oxidation number of an element in its free state is zero — for example, H2,S8,P4 etc.
2. In the case monoatomic ions, it is the same as the charge on the ion, for example: Ca2+=+2 and N3−=−3
3. For a neutral compound, the sum of all oxidation numbers is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the overall charge on the ion.
4. The first two groups - alkali metal (IA family) and alkaline earth metal (IIA family) generally tend to have +1 and +2 oxidation states respectively.
5. For most compounds, the oxidation number of oxygen is generally –2. There is, however, an exception in the case of peroxides (H2O2,Na2O2, Benzoyl peroxide). In cases where an - O – O bond is involved, oxygen attains an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1.
6. Fluorine is the most electronegative element in the periodic table. It almost always has the oxidation number as –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they’re in combination with an oxygen, fluorine or between themselves (halogens).
7. Usually, the oxidation state of hydrogen in a compound is +1. As an exception they tend to have –1 oxidation states in binary metal hydrides like NaH LiH etc.
Let the oxidation state of the central Nitrogen atom be ‘y’. Let us apply the usual rules from above:
Hydrogen has an oxidation number of +1 while oxygen has -2 (for both compounds)
The sum of oxidation states of all the atoms in a neutral atom is 0.
Using the above two on both the compounds:
For HNO2: (+1) + y + 2(-2) = 0; Solving y = +3
Similarly, for HNO3:
(+1) + y + 3(-2) = 0 ; Solving y = +5
The difference between +5 and +3 is +2
Let us briefly recall the rules for allocating oxidation numbers:
1. The oxidation number of an element in its free state is zero — for example, H2,S8,P4 etc.
2. In the case monoatomic ions, it is the same as the charge on the ion, for example: Ca2+=+2 and N3−=−3
3. For a neutral compound, the sum of all oxidation numbers is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the overall charge on the ion.
4. The first two groups - alkali metal (IA family) and alkaline earth metal (IIA family) generally tend to have +1 and +2 oxidation states respectively.
5. For most compounds, the oxidation number of oxygen is generally –2. There is, however, an exception in the case of peroxides (H2O2,Na2O2, Benzoyl peroxide). In cases where an - O – O bond is involved, oxygen attains an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1.
6. Fluorine is the most electronegative element in the periodic table. It almost always has the oxidation number as –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they’re in combination with an oxygen, fluorine or between themselves (halogens).
7. Usually, the oxidation state of hydrogen in a compound is +1. As an exception they tend to have –1 oxidation states in binary metal hydrides like NaH LiH etc.
Let the oxidation state of the central Nitrogen atom be ‘y’. Let us apply the usual rules from above:
Hydrogen has an oxidation number of +1 while oxygen has -2 (for both compounds)
The sum of oxidation states of all the atoms in a neutral atom is 0.
Using the above two on both the compounds:
For HNO2: (+1) + y + 2(-2) = 0; Solving y = +3
Similarly, for HNO3:
(+1) + y + 3(-2) = 0 ; Solving y = +5
The difference between +5 and +3 is +2
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