littleboy Posted December 30, 2010 Posted December 30, 2010 What determines the internuclear distance of a covalent compound? What i guess is that the main factor determining the internuclear distance of a covalent compound is the original distance between valence electrons and the nucleus of atoms that the compound derived from(i.e the determinating factor does not include the bond enthalpy) Am i right?
mississippichem Posted December 30, 2010 Posted December 30, 2010 What determines the internuclear distance of a covalent compound? What i guess is that the main factor determining the internuclear distance of a covalent compound is the original distance between valence electrons and the nucleus of atoms that the compound derived from(i.e the determinating factor does not include the bond enthalpy) Am i right? You're on the right track. The amount of orbital overlap between the corresponding atomic orbitals to form a set of molecular orbitals is also a factor. Orbitals of similar shape, size, and energy overlap well and form strong short bonds. There are also symmetry considerations, but they don't matter much unless you are looking for an advanced treatment of the system.
littleboy Posted December 31, 2010 Author Posted December 31, 2010 then how can we explain that bond length of C-C is greater than that of C=C ?
mississippichem Posted December 31, 2010 Posted December 31, 2010 then how can we explain that bond length of C-C is greater than that of C=C ? C-C is comprised of one [math] \sigma [/math] bond. C=C is comprised of one [math] \sigma [/math] and one [math]\pi[/math] bond. C=C is more energetic and therefore a stronger bond and shorter. Remember, [math] \pi [/math] bonds are made by side on overlap between p or d orbitals, while [math] \sigma [/math] bonds are made by overlap between two s-orbitals, two p-orbitals (head on overlap), or an (s and p)/(s and d) (head on overlap). C=C is also a four electron bond while C-C is a two electron bond.
littleboy Posted December 31, 2010 Author Posted December 31, 2010 I understand what you mean above. But the answer above meanwhile implies that stronger bond leads to a shorter bond length. What I want to ask is that how can we explain the fact that F-F bond has a short bond length but a weak attraction. If F-F bond has a weak attraction,why cant its bond lengthen?
mississippichem Posted December 31, 2010 Posted December 31, 2010 I understand what you mean above. But the answer above meanwhile implies that stronger bond leads to a shorter bond length. What I want to ask is that how can we explain the fact that F-F bond has a short bond length but a weak attraction. If F-F bond has a weak attraction,why cant its bond lengthen? Difluorine has a fully occupied [math] 2p \pi [/math] orbital which makes the bond short like a double bond. However, difluorine also has a fully occupied [math]2p \pi^{*} [/math] antibonding orbital. So the next effect is a bond that is about as energetic as a typical single bond but about as long as a typical double bond. Another, more classical, way to look at this is there is a lot of electron-electron repulsion going on between the six sets of lone pairs between the two fluorine atoms. Each fluorine center is also highly electronegative, equally electronegative. However, this might be only trivially correct; the molecular orbital description above is more acceptable.
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