JEE Advanced 2026 Paper 1, Chemistry Section 1 Q3: Valence Shell Electron Pair Repulsion (VSEPR) Theory

Symmetry analysis.

• BF${}_3$ is trigonal planar ($D_{3h}$): three equal B–F bond dipoles at 120° cancel, ${\mu }_{\text{net}}=0$.
• NH${}_4^{+}$ is regular tetrahedral ($T_d$): four equal N–H bond dipoles cancel, ${\mu }_{\text{net}}=0$.
• NF${}_3$ is pyramidal with the lone pair on N. The N–F bond dipoles point from N toward F (since F is more electronegative), while the lone pair dipole points opposite to the resultant of the N–F bonds. The lone-pair contribution partially cancels the bond contribution, giving a small net dipole $\approx 0.23$ D.
• NH${}_3$ is also pyramidal, but the N–H bond dipoles point from H toward N (N is more electronegative), in the same direction as the lone-pair dipole. The contributions reinforce, giving a larger net dipole $\approx 1.47$ D.

Order: ${\text{BF}}_3={\text{NH}}_4^{+}<{\text{NF}}_3<{\text{NH}}_3$. Option (A).