Finding Oxidation Numbers of Nitrogen in Oxides

To match the nitrogen oxides in List I with the oxidation number of nitrogen in List II, we need to calculate the oxidation number of nitrogen in each compound. The oxidation number of oxygen is typically -2 in these compounds.

Let's calculate the oxidation number of nitrogen for each oxide:

Calculating Oxidation Number for Each Nitrogen Oxide

• A. NO (Nitrogen Monoxide):

  Let the oxidation number of nitrogen be \(x\).

  The sum of oxidation numbers in a neutral molecule is zero.

  \(x + (\text{oxidation number of O}) = 0\)

  \(x + (-2) = 0\)

  \(x = +2\)

  So, the oxidation number of nitrogen in NO is +2. This corresponds to List II option II.

• B. N\(_2\)O\(_4\) (Dinitrogen Tetroxide):

  Let the oxidation number of nitrogen be \(x\).

  The sum of oxidation numbers in a neutral molecule is zero.

  \(2 \times (\text{oxidation number of N}) + 4 \times (\text{oxidation number of O}) = 0\)

  \(2x + 4(-2) = 0\)

  \(2x - 8 = 0\)

  \(2x = 8\)

  \(x = \frac{8}{2} = +4\)

  So, the oxidation number of nitrogen in N\(_2\)O\(_4\) is +4. This corresponds to List II option I.

• C. N\(_2\)O\(_3\) (Dinitrogen Trioxide):

  Let the oxidation number of nitrogen be \(x\).

  The sum of oxidation numbers in a neutral molecule is zero.

  \(2 \times (\text{oxidation number of N}) + 3 \times (\text{oxidation number of O}) = 0\)

  \(2x + 3(-2) = 0\)

  \(2x - 6 = 0\)

  \(2x = 6\)

  \(x = \frac{6}{2} = +3\)

  So, the oxidation number of nitrogen in N\(_2\)O\(_3\) is +3. This corresponds to List II option IV.

• D. N\(_2\)O\(_5\) (Dinitrogen Pentoxide):

  Let the oxidation number of nitrogen be \(x\).

  The sum of oxidation numbers in a neutral molecule is zero.

  \(2 \times (\text{oxidation number of N}) + 5 \times (\text{oxidation number of O}) = 0\)

  \(2x + 5(-2) = 0\)

  \(2x - 10 = 0\)

  \(2x = 10\)

  \(x = \frac{10}{2} = +5\)

  So, the oxidation number of nitrogen in N\(_2\)O\(_5\) is +5. This corresponds to List II option III.

Matching List I with List II

Based on our calculations, the matches are:

• A. NO \(\rightarrow\) II. +2
• B. N\(_2\)O\(_4\) \(\rightarrow\) I. +4
• C. N\(_2\)O\(_3\) \(\rightarrow\) IV. +3
• D. N\(_2\)O\(_5\) \(\rightarrow\) III. +5

Let's compare this with the given options:

The matches A-II, B-I, C-IV, D-III correspond to Option 3.

Revision Table: Nitrogen Oxides Oxidation States

Additional Information: Understanding Oxidation Numbers

The oxidation number (also called oxidation state) is a number assigned to an element in a compound that represents the number of electrons gained, lost, or shared by an atom of that element compared to its neutral elemental state. It's a hypothetical charge that an atom would have if all bonds were ionic.

Here are some general rules for assigning oxidation numbers:

• The oxidation number of an atom in a free element (like N\(_2\), O\(_2\), Fe) is zero.
• For a monatomic ion, the oxidation number is equal to the charge of the ion (e.g., Na\(^+\) is +1, Cl\(^-\) is -1).
• The sum of oxidation numbers in a neutral compound is zero.
• The sum of oxidation numbers in a polyatomic ion is equal to the charge of the ion.
• Oxygen usually has an oxidation number of -2 in compounds, except in peroxides (like H\(_2\)O\(_2\)), where it is -1, and in compounds with fluorine (like OF\(_2\)), where it can be positive.
• Hydrogen usually has an oxidation number of +1 in compounds with nonmetals (like H\(_2\)O, HCl) and -1 in metal hydrides (like NaH).
• Group 1 elements (alkali metals) always have an oxidation number of +1 in compounds.
• Group 2 elements (alkaline earth metals) always have an oxidation number of +2 in compounds.
• Fluorine always has an oxidation number of -1 in compounds.

By applying these rules, we can calculate the oxidation number of an element in a compound if the oxidation numbers of the other elements are known.