Outline 3 - Covalent Bond Dissociation Energy
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Outline 3 - Covalent Bond Dissociation Energy
Hello! I was looking over Outline 3 and couldn't understand what it meant with the following sentence "Explain how covalent bond dissociation energy is related to covalent bond multiplicity, atomic radius, and the presence of unpaired electrons." Is this simply referring to how the strength of a bond changes with whether it's a single/double/triple bond, how far away the atoms are, and whether there are unpaired electrons? I'm not sure if I'm understanding that right, so if there's anything I'm missing, it would be great if you could let me know! Also, what effect does unpaired electrons have on the strength of covalent bonds?
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Re: Outline 3 - Covalent Bond Dissociation Energy
Yes, you're right. Covalent bond multiplicity refers to bond order (a single bond is weaker than a triple bond) , atomic radius refers to the distance between the atoms (the larger the radius, the farther apart they are and the weaker the bond), and unpaired electrons refers to the lone pairs on neighboring atoms. The more unpaired electrons there are, the weaker the bond because the lone pairs repel each other and cause electron repulsion.
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Re: Outline 3 - Covalent Bond Dissociation Energy
Also, bond dissociation energy is related to the other periodic trends of atomic radius and ionization energy which contribute to how electrons shield and have electrostatic attractions.
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Re: Outline 3 - Covalent Bond Dissociation Energy
You almost have it! Like it has already been said, covalent bond multiplicity has to do with bond order, atomic radius is the distance between atoms, and unpaired e- refers to the lone pairs of neighboring atoms. More unpaired electrons means that it is more likely that you are dealing with a weaker bond. This comes back again in other topics mentioned such as bond dissociation energy.
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