JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

For NO _2^+, the central nitrogen atom is sp hybridized with no lone pairs, resulting in a linear geometry and a bond angle of 180^ . For NO _2, the central nitrogen atom is sp^2 hybridized with one unpaired electron. The repulsion from a single electron is less than that from a bond pair, so the bond angle opens up to be greater than 120^ (approximately 134^ ). For NO _3^-, the central nitrogen atom is sp^2 hybridized with no lone pairs, resulting in a trigonal planar geometry and a bond angle of exactly 120^ . For NO _2^-, the central nitrogen atom is sp^2 hybridized with one lone pair. The lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, compressing the bond angle to less than 120^ (approximately 115^ ). Therefore, the correct order of O-N-O bond angles is NO _2^- Answer: NO _2^- < NO _3^- < NO _2 < NO _2^+