Question

State the oxidation number of \(\mathrm{N}\) in (a) \(\mathrm{NO}_{2}^{-}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{HNO}_{3}\) (d) \(\mathrm{NH}_{4}{ }^{+}\)

Short answer

Answer: The oxidation numbers of nitrogen (N) in the compounds are: (a) \(\mathrm{NO}_{2}^{-}\): +3, (b) \(\mathrm{NO}_{2}\): +4, (c) \(\mathrm{HNO}_{3}\): +5, and (d) \(\mathrm{NH}_{4}{ }^{+}\): -3.

Step-by-step solution

Review the rules for assigning oxidation numbers

Here is a summary of the rules to assign oxidation numbers: 1. Elements in their free state have an oxidation number of 0. 2. The oxidation number of oxygen in compounds is usually -2. 3. The oxidation number of hydrogen in compounds is usually +1. 4. The sum of oxidation numbers of each element in a compound must equal the overall charge of the compound. Now, we'll apply these rules for each given molecule.

Find oxidation number of N in \(\mathrm{NO}_{2}^{-}\)

We know that oxygen has an oxidation number of -2. Since there are 2 oxygen atoms in the molecule: 2 x (-2) = -4 The compound has an overall charge of -1, so the oxidation number of nitrogen (N) must be: x + (-4) = -1 Solve for x: x = +3 Thus, the oxidation number of N in \(\mathrm{NO}_{2}^{-}\) is +3.

Find oxidation number of N in \(\mathrm{NO}_{2}\)

Using the same approach as Step 2 for this neutral molecule, oxygen's oxidation number is -2. With 2 oxygen atoms in the molecule: 2 x (-2) = -4 Since the molecule is neutral, the sum of oxidation numbers should be 0, so the oxidation number of nitrogen (N) must be: x + (-4) = 0 Solve for x: x = +4 Thus, the oxidation number of N in \(\mathrm{NO}_{2}\) is +4.

Find oxidation number of N in \(\mathrm{HNO}_{3}\)

For this molecule, hydrogen's oxidation number is +1, and we have 1 hydrogen atom: 1 x (+1) = +1 Oxygen's oxidation number is -2, and we have 3 oxygen atoms: 3 x (-2) = -6 Thus, the sum of hydrogen and oxygen oxidation numbers is: +1 + (-6) = -5 The molecule is neutral, so the sum of oxidation numbers should be 0. Thus, the oxidation number of nitrogen (N) must be: x + (-5) = 0 Solve for x: x = +5 Thus, the oxidation number of N in \(\mathrm{HNO}_{3}\) is +5.

Find oxidation number of N in \(\mathrm{NH}_{4}{ }^{+}\)

For this molecule, oxidation number of hydrogen is +1, and we have 4 hydrogen atoms: 4 x (+1) = +4 Since the molecule has an overall charge of +1, the oxidation number of nitrogen (N) must be: x + (+4) = +1 Solve for x: x = -3 Thus, the oxidation number of N in \(\mathrm{NH}_{4}{ }^{+}\) is -3. To summarize, the oxidation numbers of nitrogen (N) in each compound are: (a) \(\mathrm{NO}_{2}^{-}\): +3 (b) \(\mathrm{NO}_{2}\): +4 (c) \(\mathrm{HNO}_{3}\): +5 (d) \(\mathrm{NH}_{4}{ }^{+}\): -3