Question

Nitrosyl ligand binds to d-metal atoms in linear and bent fashion and behaves, respectively, as

• A. NO$^+$ and NO$^+$
• B. NO$^-$ and NO$^-$
• C. NO$^-$ and NO$^+$
• D. NO$^+$ and NO$^-$

Hint:

Remember: linear NO ≈ NO$^+$ (π-acceptor), bent NO ≈ NO$^-$.

Answer 1

The Correct Option is B

Step 1: Understanding NO coordination.
The nitrosyl ligand (NO) can coordinate either linearly or in a bent fashion. A linear M–NO bond corresponds to NO acting as NO$^+$ (a strong π-acceptor). A bent M–NO bond corresponds to NO acting as NO$^-$.
Step 2: Reason for behavior.
Linear NO has strong back bonding, consistent with NO$^+$. Bent NO has weaker back bonding and behaves more like NO$^-$.
Step 3: Conclusion.
Thus, in linear mode NO behaves as NO$^+$ and in bent mode as NO$^-$.