NEET 2021 Chemistry Question 98 Solution

A molecule is non-polar if the vector sum of all individual bond dipoles and lone-pair moments cancels to zero ($\vec{\mu}_{\text{net}} = \sum \vec{\mu}_i = 0$). This occurs in regular, symmetrical geometries where all ligands attached to the central atom are identical and there are no stereochemically active lone pairs disrupting symmetry.

Antimony ($Sb$) is in Group 15 with 5 valence electrons. In $SbCl_5$, it forms five single covalent bonds with chlorine atoms and possesses zero lone pairs ($sp^3d$ hybridization). Its molecular geometry is a regular trigonal bipyramidal (TBP): - Equatorial Plane: Three $Sb-Cl$ bonds are directed towards the corners of an equilateral triangle at $120^\circ$ bond angles. The vector sum of three equal planar vectors separated by $120^\circ$ is identically zero: $$\vec{\mu}_{\text{eq1}} + \vec{\mu}_{\text{eq2}} + \vec{\mu}_{\text{eq3}} = 0$$ - Axial Axis: Two $Sb-Cl$ bonds lie along a straight line perpendicular to the equatorial plane at $180^\circ$ pointing in opposite directions: $$\vec{\mu}_{\text{ax1}} + \vec{\mu}_{\text{ax2}} = 0$$ Consequently, the overall net dipole moment is zero:

- $POCl_3$: Distorted tetrahedral geometry where the strong $P=O$ dipole cannot be cancelled by the three $P-Cl$ bond dipoles ($\mu = 2.4\text{ D} \neq 0$). - $CH_2O$: Planar geometry, but the strong polar $C=O$ dipole points in the opposite direction of the two weakly polar $C-H$ bonds ($\mu = 2.3\text{ D} \neq 0$). - $NO_2$: Angular/bent molecule (bond angle $\approx 134^\circ$) with an odd electron on nitrogen, giving a permanent dipole moment ($\mu = 0.32\text{ D} \neq 0$).