Class 11 NEET Chemistry Chemical Bonding And Molecular Structure Part 2

Concept:Count sigma (σ) bonds, pi (π) bonds, and lone pairs in each molecule to match with the given descriptions.Chemical Equation / Formula:$\mathrm{C}_2\mathrm{H}_4$ (ethene), $\mathrm{C}_2\mathrm{H}_2$ (ethyne), $\mathrm{CH}_4$ (methane), $\mathrm{NH}_3$ (ammonia).Explanation:Step 1: $\mathrm{C}_2\mathrm{H}_4$ has a double bond between carbons.It contains 4 $\mathrm{C-H}$ σ bonds and 1 $\mathrm{C-C}$ σ bond = 5 σ bonds total, plus 1 π bond.So it matches IV ($5 \sigma$ bonds, $1 \pi$ bond).Step 2: $\mathrm{C}_2\mathrm{H}_2$ has a triple bond between carbons.It has 2 $\mathrm{C-H}$ σ bonds and 1 $\mathrm{C-C}$ σ bond = 3 σ bonds, plus 2 π bonds.So it matches I ($3 \sigma$ bonds, $2 \pi$ bonds).Step 3: $\mathrm{CH}_4$ has four single bonds.It has 4 $\mathrm{C-H}$ σ bonds and no π bonds.So it matches III ($4 \sigma$ bonds).Step 4: $\mathrm{NH}_3$ has three single bonds and one lone pair on nitrogen.It has 3 $\mathrm{N-H}$ σ bonds and one lone pair.So it matches II ($3 \sigma$ bonds, one lone pair).Thus correct pairing: A→IV, B→I, C→III, D→II.Answer:Option B: A-IV, B-I, C-III, D-II.