See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Solution: H-N-H Bond Angle Order **Step 1: Identify the structures and hybridization of nitrogen** - **I. Aniline** ($C_6H_5-NH_2$): N is $sp^2$ hybridized (aromatic ring conjugation) - **II. Ammonia** ($NH_3$): N is $sp^3$ hybridized - **III. Methylamine** ($CH_3-NH_2$): N is $sp^3$ hybridized - **IV. Urea** ($CH_3-C(=O)-NH_2$): N is $sp^2$ hybridized (resonance with C=O) **Step 2: Apply VSEPR theory and lone pair effects** Bond angles depend on hybridization and lone pair repulsion: - $sp^2$ hybridization → ideal angle = $120°$ - $sp^3$ hybridization → ideal angle = $109.5°$ **Step 3: Account for lone pair repulsion** Lone pair-lone pair repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion **IV (urea)**: $sp^2$, resonance stabilizes lone pair on N → **~120°** (closest to $sp^2$ ideal) **I (aniline)**: $sp^2$, but lone pair delocalized into aromatic ring (reduced repulsion) → **~118°** **II (ammonia)**: $sp^3$ with one lone pair, maximum lone pair-bonding pair compression → **~107°** (compressed from 109.5°) **III (methylamine)**: $sp^3$ with one lone pair, similar to ammonia but slightly higher → **~108°** (slightly higher due to less electronegative R-group effect) **Step 4: Order the bond angles** $$\boxed{IV > I > II > III}$$ **Answer: (A)** — Urea has the maximum angle due to $sp^2$ hybridization with resonance stabilization, followed by aniline, then ammonia and methylamine (both $sp^3$ with comparable compression, II slightly less than III).