JEE Chemistry 2024 Question with Solution

Given: Two statements about oxidation states of group $16$ elements.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

For Statement (I), oxygen does not exhibit only the oxidation state $-2$. Although $-2$ is the most common oxidation state of oxygen, it also shows other oxidation states in special compounds.

Examples given in the solution are:

• In $\text{H}_2\text{O}_2$, oxygen has oxidation state $-1$.
• In $\text{OF}_2$, oxygen has oxidation state $+2$.

Therefore, Statement (I) is incorrect.

For Statement (II), down group $16$ the stability trend stated is reversed. Due to the inert pair effect in heavier elements, the $+6$ oxidation state decreases in stability down the group, while the $+4$ oxidation state becomes more stable for heavier members such as selenium and tellurium.

Therefore, Statement (II) is incorrect.

Thus, both Statement I and Statement II are incorrect.

The correct option is C.

Given:

• Statement I says oxygen exhibits only $-2$ oxidation state.
• Statement II says down group $16$, stability of $+4$ decreases and $+6$ increases.

Find: The correct option.

Analyze Statement I: Oxygen is the first member of group $16$ and commonly shows oxidation state $-2$. However, it is not restricted to only this value.

The solution mentions:

• $\text{H}_2\text{O}_2$ where oxygen is $-1$
• $\text{OF}_2$ where oxygen is $+2$
• molecular oxygen $\text{O}_2$ where oxygen is $0$

Hence the word only makes Statement I false.

Analyze Statement II: For heavier chalcogens, the inert pair effect becomes more important. Because of this, the higher oxidation state $+6$ becomes less stable down the group, whereas the lower oxidation state $+4$ becomes relatively more stable.

So the actual trend is:

Therefore, Statement II is also false.

Conclusion: Since both statements are false, the correct answer is Both Statement I and Statement II are incorrect, that is, option C.