JEE Chemistry 2023 Question with Solution

Given: The species are $\mathrm{NO}_2$, $\mathrm{ICl}_4^-$, $\mathrm{BrF}_3$, $\mathrm{ClO}_2$, $\mathrm{NO}_2^+$, and $\mathrm{NO}$.

Find: The number of species that do not have an odd number of electrons.

Odd-electron species have an odd total number of electrons. So, count total valence electrons for each species.

1. For $\mathrm{NO}_2$:

This is odd.

2. For $\mathrm{ICl}_4^-$:

This is even.

3. For $\mathrm{BrF}_3$:

This is even.

4. For $\mathrm{ClO}_2$:

This is odd.

5. For $\mathrm{NO}_2^+$:

This is even.

6. For $\mathrm{NO}$:

This is odd.

Species without odd electrons are $\mathrm{ICl}_4^-$, $\mathrm{BrF}_3$, and $\mathrm{NO}_2^+$.

Therefore, the required number is $3$.

Given: Determine which listed species have an even total number of electrons.

Find: Total count of species that are not odd-electron species.

Classify each one by total valence electron count:

• $\mathrm{NO}_2 \rightarrow 17$ electrons, so odd
• $\mathrm{ICl}_4^- \rightarrow 36$ electrons, so even
• $\mathrm{BrF}_3 \rightarrow 28$ electrons, so even
• $\mathrm{ClO}_2 \rightarrow 19$ electrons, so odd
• $\mathrm{NO}_2^+ \rightarrow 16$ electrons, so even
• $\mathrm{NO} \rightarrow 11$ electrons, so odd

Thus, exactly three species have an even number of electrons.

Therefore, the final answer is $3$.