The molecular geometry of the CHBr3 molecule is Explain and the molecule is A trigonal pyramidal, polar B) tetrahedral, nonpolar seesaw nonpolar D) tetrahedral, polar E) seesaw polar
The Correct Answer and Explanation is:
Correct choice: D) tetrahedral, polar
Explanation
Bromoform, CHBr₃, contains a central carbon atom that forms four σ-bonds: one to hydrogen and one to each of three bromine atoms. Carbon supplies four valence electrons, and each X–C bond uses one electron pair, leaving no lone pairs on the central atom. Under Valence-Shell Electron-Pair Repulsion (VSEPR) theory, four regions of electron density arrange themselves as far apart as possible, giving an electron-pair geometry that is tetrahedral with ideal 109.5 ° angles. Because there are no lone pairs to distort the shape, the molecular geometry is also tetrahedral.
Whether a tetrahedral molecule is polar hinges on how its individual bond dipoles combine. Each C–Br bond is polar (bromine electronegativity χ ≈ 2.96 > carbon χ ≈ 2.55), with its dipole pointing from carbon toward bromine. By contrast, the C–H bond is weakly polar in the opposite direction (carbon χ > hydrogen χ ≈ 2.20), so its dipole points from hydrogen toward carbon.
If all four substituents were identical, the tetrahedron’s dipoles would cancel perfectly and the molecule would be non-polar, as in CCl₄. In CHBr₃, however, three identical C–Br vectors lie at the corners of a trigonal pyramid, while the single C–H vector occupies the fourth vertex. Adding the three equal C–Br dipoles yields a resultant vector pointing roughly opposite the C–H dipole, but the magnitude of each C–Br dipole is larger than that of C–H. Consequently, the vectors do not cancel completely, leaving a measurable molecular dipole (experimentally ≈ 1.25 D).
This residual charge separation classifies CHBr₃ as a polar molecule. Its polarity helps explain several physical properties: a higher boiling point (150 °C) than non-polar hydrocarbons of similar molar mass, appreciable solubility in moderately polar organic solvents, and immiscibility with water. All observations are fully consistent with option D) tetrahedral, polar.
