Intermolecular forces (IMFs) are the attractive forces that exist between neighboring molecules, influencing a substance’s physical properties, such as its boiling and melting points. These forces are significantly weaker than the covalent bonds that hold atoms together within a single molecule, such as Hydrogen Bromide (HBr). Understanding the specific IMFs at play is necessary to predict how HBr will behave. The strongest of these attractive forces is hydrogen bonding, and we will investigate the strict chemical requirements for this special type of bonding to determine the verdict for Hydrogen Bromide.
Defining Hydrogen Bonding
Hydrogen bonding is a strong type of dipole-dipole interaction, but it requires very specific conditions. The fundamental requirement is that a hydrogen atom must be covalently bonded to one of three highly electronegative atoms: Nitrogen (N), Oxygen (O), or Fluorine (F). These three elements possess high electronegativity values, which is a measure of an atom’s ability to attract electrons toward itself.
When hydrogen is bonded to one of these small, highly electronegative atoms, the shared electrons are pulled strongly toward the N, O, or F atom. This unequal sharing creates a highly polar covalent bond, causing the hydrogen atom to acquire a substantial partial positive charge (\(\delta^+\)). This proton is then strongly attracted to a lone pair of electrons on a nearby highly electronegative atom on an adjacent molecule, forming the hydrogen bond. The small size of the hydrogen atom allows it to get very close to the electron lone pair, contributing to the exceptional strength of this IMF.
The Verdict on HBr
Hydrogen Bromide (HBr) does not exhibit classic hydrogen bonding. It fails to meet the strict requirement of the hydrogen being bonded to Nitrogen, Oxygen, or Fluorine. Bromine (Br) is positioned below Fluorine on the periodic table. Although Bromine is electronegative (2.96 on the Pauling scale), its value is lower than the required elements (N=3.04, O=3.44, F=3.98). This lower electronegativity means that the H-Br bond, while polar, is not polar enough to create the necessary large partial positive charge on the hydrogen atom.
The difference in atomic size also plays a significant role. Bromine has a much larger atomic radius than the small, second-period elements (N, O, F). This larger size means that the negative charge on the Bromine atom is spread out over a greater volume, resulting in a lower charge density. The partial positive charge on the hydrogen atom is therefore not attracted strongly enough to the diffuse negative charge on a neighboring HBr molecule to form a true hydrogen bond.
Intermolecular Forces in HBr
Since HBr does not qualify for hydrogen bonding, its physical properties are determined by the other available intermolecular forces. Hydrogen Bromide is a polar molecule because Bromine is significantly more electronegative than Hydrogen, creating an uneven distribution of electron density. This polarity results in strong Dipole-Dipole interactions between adjacent HBr molecules. The partially positive end of one HBr molecule is attracted to the partially negative end of a neighboring HBr molecule, causing them to align.
All molecules, including HBr, also experience London Dispersion Forces (LDF), which arise from temporary, fluctuating dipoles. In HBr, LDF are significant because Bromine is a large atom with a high number of electrons, making the electron cloud highly polarizable. The ease with which the electron cloud can be distorted leads to strong LDF, which can become the dominant force among the heavier hydrogen halides. The cumulative effect of these dipole-dipole forces and strong LDF dictates HBr’s physical behavior, such as its boiling point.