the intermolecular forces responsible are van der vaals/london dispersion. Read here, but essentially at any moment in time there is a likelihood that there is more electrons on one "side" of the atom, so it will have a slight negative charge. This will repel electrons in neighboring molecules, and attract the molecule towards it.
the intermolecular forces responsible are van der vaals/london dispersion. Read here, but essentially at any moment in time there is a likelihood that there is more electrons on one "side" of the atom, so it will have a slight negative charge. This will repel electrons in neighboring molecules, and attract the molecule towards it.
the intermolecular forces responsible are van der vaals/london dispersion. Read here, but essentially at any moment in time there is a likelihood that there is more electrons on one "side" of the atom, so it will have a slight negative charge. This will repel electrons in neighboring molecules, and attract the molecule towards it.
Yes as said above it is due to Van-der-Waal forces. The bromine atoms are more 'polarisable', the term which appears in the formula predicting the strength of a VDW interaction. Br is more polarisable than the lighter halogens as the electrons are less tightly bound, and are more easily moved by external forces. The analogy of the atom being soft and squishy as opposed to a hard sphere is often used to visualise this.