Page 1 of 1

Dissociation energy

Posted: Mon Nov 04, 2019 9:15 am
by asannajust_1J
Explain how covalent bond dissociation energy is related to covalent bond multiplicity, atomic radius, and the presence of unpaired electrons. Am mostly confused on the reference to the presence of unpaired electrons.

Re: Dissociation energy

Posted: Mon Nov 04, 2019 12:00 pm
by Brian_Ho_2B
asannajust wrote:Explain how covalent bond dissociation energy is related to covalent bond multiplicity, atomic radius, and the presence of unpaired electrons. Am mostly confused on the reference to the presence of unpaired electrons.

Covalent Bond multiplicity: more bonds between two atoms means higher dissociation energy (eg. C=C is stronger than C-C) and vise versa
Atomic radius: larger atomic radii --> longer distance between atoms and thus longer bond --> weaker bond with less Dissociation energy
Presence of unpaired electrons: Unpaired electrons have very strong repulsion power (as we will see when we cover VSEPR); a molecule with a single bond has many unpaired electrons which repel the two atoms in the bond away from each other, weakening the bond; a molecule with a double/triple bond has less lone pairs repelling each other and thus stronger bonds.

Re: Dissociation energy

Posted: Mon Nov 04, 2019 12:02 pm
by McKenna_4A
The presence of unpaired electrons on atoms within a compound result in them repelling from other because of the negative charges working on each other. Consider F2, where each F atom has 6 unpaired electrons. They exert repulsive forces on one another because of the unpaired electrons in their valence shell. Therefore, their bon/dissociation energy is "unusually low," as Dr. Lavelle stated.

Re: Dissociation energy

Posted: Fri Nov 22, 2019 4:42 pm
by Angela Prince 1J
Just in terms of the number of bonds, a molecule with two double bonds would have a higher dissociation energy than one with two single bonds, granted there are no lone pairs on either molecule?