Is the molecule BF3 polar or nonpolar? (b) If you react BF3 to make the ion BF4-, is this ion planar? (c) Does the molecule BF2Cl have a dipole moment? 9.51 Indicate the hybridization of the central atom in (a) H2S, (b) SeF4, (c) POH3, (d) AlCl3.
The Correct Answer and Explanation is:
9.50 (a) BF₃ is nonpolar.
(b) The ion BF₄⁻ is not planar.
(c) The molecule BF₂Cl has a dipole moment.
9.51 Hybridization of the central atom:
(a) H₂S → sp³
(b) SeF₄ → sp³d
(c) POH₃ → sp³
(d) AlCl₃ → sp²
Conceptual Explanation
Molecular Polarity and Geometry
The polarity of a molecule depends on both the polarity of its individual bonds and its molecular geometry. BF₃ (boron trifluoride) has three identical B–F bonds, each polar due to the electronegativity difference between B and F. However, the molecule adopts a trigonal planar shape with 120° bond angles, and the three bond dipoles cancel each other out. Therefore, BF₃ is nonpolar overall despite having polar bonds.
When BF₃ reacts with a fluoride ion to form BF₄⁻, the molecule becomes tetrahedral, with four equivalent B–F bonds at angles of 109.5°. This shape is not planar, and the added electron pair is delocalized over the structure. The ion is symmetrical and nonpolar.
In BF₂Cl, however, the situation is different. It has a trigonal planar geometry like BF₃, but one of the fluorine atoms is replaced by a chlorine atom. Since Cl and F have different electronegativities, the dipoles do not cancel out. As a result, the molecule has a net dipole moment and is polar.
Hybridization of Central Atoms
In H₂S, sulfur forms two sigma bonds and has two lone pairs, giving it a bent shape. This requires sp³ hybridization.
In SeF₄, selenium forms four bonds and has one lone pair, leading to a see-saw geometry, requiring sp³d hybridization.
In POH₃, phosphorus forms three single bonds with hydrogen and one with oxygen (which has a lone pair on O), requiring sp³ hybridization.
In AlCl₃, aluminum forms three sigma bonds with chlorine atoms and has no lone pairs, adopting a trigonal planar geometry, so it is sp² hybridized.
These hybridizations explain the geometries and bonding observed in each compound.
