Match List I with List II:

Choose the correct answer from the  options given below:

A-II, B-I, C-IV, D-III

The correct answer is option 3. A-II, B-I, C-IV, D-III.

Let us determine the oxidation numbers of nitrogen in each of the nitrogen oxides listed:

1. Nitric oxide (\(NO\):

In \(NO\), the oxidation state of nitrogen can be found by setting up the equation based on the known oxidation state of oxygen (\(-2\)).

Let \(x\) be the oxidation state of nitrogen in \(NO\). The total oxidation state should equal the charge of the molecule, which is 0.

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

\(x = +2\)

2. Dinitrogen tetroxide (\(N_2O_4\)):

In \(N_2O_4\), let \(x\) be the oxidation state of nitrogen.

Each oxygen has an oxidation state of \(-2\). So, for \(N_2O_4\), the equation is

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

\(2x - 8 = 0\)

\(2x = +8\)

\(x = +4\)

3. Dinitrogen trioxide (\(N_2O_3\)):

In \(N_2O_3\), let \(x\) be the oxidation state of nitrogen. The equation is

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

\(2x - 6 = 0\)

\(2x = +6\)

\(x = +3\)

4. Dinitrogen pentoxide (\(N_2O_5\)):

In \(N_2O_5\), let \(x\) be the oxidation state of nitrogen. The equation is

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

\(2x - 10 = 0\)

\(2x = +10\)

\(x = +5\)

So, the correct matches are:

A → II (\(NO\) has oxidation state \(+2\))

B → I (\(N_2O_4\) has oxidation state \(+4\))

C → IV (\(N_2O_3\) has oxidation state \(+3\))

D → III (\(N_2O_5\) has oxidation state \(+5\))

The correct answer is: 3. A-II, B-I, C-IV, D-III