Answer the question on the basis of passage given below:
Nitrogen differs from rest of the members of group 15 due to its small size, high electronegativity, high ionization enthalpy and non-availability of d orbitals. Nitrogen has unique ability to form pл - рл multiple bonds with itself and with other elements having small size and high electronegativity. Heavier elements of this group do not form \(р\pi -р\pi \) bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping.

Match List I with List II

Choose the correct answer from the options given below:

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

The correct answer is option 2. A-III, B-I, C-IV, D-II.

Let us break down the matching of nitrogen oxides with their corresponding oxidation states in detail:

A. \(N_2O\) (Nitrous Oxide or Dinitrogen Monoxide)

\(N_2O\) consists of two nitrogen atoms and one oxygen atom. 

Let the oxidation state of nitrogen in \(N_2O\) be \(x\).

The total oxidation state of two nitrogen atoms is \(2x\).

Oxygen typically has an oxidation state of \(-2\).

The molecule is neutral, so the total charge is 0.

Therefore:

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

Solving for \(x\):

\(2x = +2 \quad \Rightarrow \quad x = +1\)

Hence, the oxidation state of nitrogen in \(N_2O\) is +1.

B. \(N_2O_3\) (Dinitrogen Trioxide)

\(N_2O_3\) consists of two nitrogen atoms and three oxygen atoms. It is an anhydride of nitrous acid (HNO₂).

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

There are two nitrogen atoms, so the total oxidation state from nitrogen is \(2x\).

The oxidation state of oxygen is \(-2\), and there are three oxygen atoms, so the total contribution from oxygen is \(3 \times (-2) = -6\).

Therefore:

\(2x + (-6) = 0\)

Solving for \(x\):

\(2x = +6 \quad \Rightarrow \quad x = +3\)

Hence, the oxidation state of nitrogen in \(N_2O_3\) is +3.

C. \(N_2O_4\) (Dinitrogen Tetroxide)

\(N_2O_4\) is a dimer of nitrogen dioxide (\(NO_2\)). It consists of two nitrogen atoms and four oxygen atoms.

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

There are two nitrogen atoms, so the total oxidation state from nitrogen is \(2x\).

The oxidation state of oxygen is \(-2\), and there are four oxygen atoms, so the total contribution from oxygen is \(4 \times (-2) = -8\).

Therefore:

\(2x + (-8) = 0\)

Solving for \(x\):

\(2x = +8 \quad \Rightarrow \quad x = +4\)

Hence, the oxidation state of nitrogen in \(N_2O_4\) is +4.

D. \(N_2O_5\) (Dinitrogen Pentoxide)

\(N_2O_5\) is the anhydride of nitric acid (\(HNO_3\)). It consists of two nitrogen atoms and five oxygen atoms.

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

There are two nitrogen atoms, so the total oxidation state from nitrogen is \(2x\).

The oxidation state of oxygen is \(-2\), and there are five oxygen atoms, so the total contribution from oxygen is \(5 \times (-2) = -10\).

Therefore:

\(2x + (-10) = 0\)

Solving for \(x\):

\(2x = +10 \quad \Rightarrow \quad x = +5\)

Hence, the oxidation state of nitrogen in \(N_2O_5\) is +5.

Thus, the correct matching is: 2. A-III, B-I, C-IV, D-II