Arrange the following in the increasing order of oxidation state of n…

The correct answer is option 4. D < A < E < B < C.

The oxidation state (or oxidation number) of an element in a compound represents the number of electrons lost or gained by an atom of that element when forming the compound. In general:

Oxygen (O) usually has an oxidation state of -2.

Nitrogen (N) can have various oxidation states ranging from \(-3\) to \(+5\), depending on the compound.

Calculation of Oxidation States in Each Compound

A. \(NO\) (Nitric oxide)

Composition: 1 nitrogen (N) and 1 oxygen (O).

Let us assume the oxidation state of nitrogen is \(x\).

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

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

Oxidation state of nitrogen in \(NO\): +2.

B. \(NO_2\) (Nitrogen dioxide)

Composition: 1 nitrogen (N) and 2 oxygens (O).

Let us assume the oxidation state of nitrogen is \(x\).

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

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

Oxidation state of nitrogen in \(NO_2\): +4.

C. \(N_2O_5\) (Dinitrogen pentoxide)

Composition: 2 nitrogens (N) and 5 oxygens (O).

Let us assume the oxidation state of nitrogen is \(x\).

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

\(2x + 5(-2) = 0 \quad \Rightarrow \quad 2x - 10 = 0 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = +5\)

Oxidation state of nitrogen in \(N_2O_5\): +5.

D. \(N_2O\) (Nitrous oxide)

Composition: 2 nitrogens (N) and 1 oxygen (O).

Let us assume the oxidation state of nitrogen is \(x\).

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

\(2x + (-2) = 0 \quad \Rightarrow \quad 2x - 2 = 0 \quad \Rightarrow \quad 2x = 2 \quad \Rightarrow \quad x = +1\)

Oxidation state of nitrogen in \(N_2O\): +1.

E. \(N_2O_3\) (Dinitrogen trioxide)

Composition: 2 nitrogens (N) and 3 oxygens (O).

Let us assume the oxidation state of nitrogen is \(x\).

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

\(2x + 3(-2) = 0 \quad \Rightarrow \quad 2x - 6 = 0 \quad \Rightarrow \quad 2x = 6 \quad \Rightarrow \quad x = +3\)

Oxidation state of nitrogen in \(N_2O_3\): +3

Here are the oxidation states of nitrogen in each compound:

\(N_2O\): +1

\(NO\): +2

\(N_2O_3\): +3

\(NO_2\): +4

\(N_2O_5\): +5

Now, we arrange the compounds in order of increasing oxidation states of nitrogen:

The lowest oxidation state is +1 in \(N_2O\).

The next is +2 in \(NO\).

Then, +3 in \(N_2O_3\).

Followed by +4 in \(NO_2\).

Finally, the highest oxidation state is +5 in \(N_2O_5\).

Thus, the increasing order of oxidation states is:

\(\textbf{D (N}_2\textbf{O)} < \textbf{A (NO)} < \textbf{E (N}_2\textbf{O}_3\textbf{)} < \textbf{B (NO}_2\textbf{)} < \textbf{C (N}_2\textbf{O}_5\textbf{)} \)

Conclusion

The correct answer is option 4: D < A < E < B < C.

This sequence correctly represents the increasing order of oxidation states of nitrogen in the given compounds.