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How do I arrange these chemical bonds in an increasing/decreasing order based upon their lengths?
I understand that the bond (a) will be longer than (b) because (a) is a bond between sp² hybrid and sp hybrid carbon atoms, as opposed to (b) which is a bond between two sp hybrid carbon atoms. But how do we compare bond (c)?
Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).
At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).
We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).
You can draw Hyper conjugating Structures to compare bond length. More the α Hydrogen of alkene, more are the Hyper conjugating structures, more is the delocalization of π-bond and more is the bond length (since double bond character decreases, bond length increases).
Thus bond length C would be greater than A.
As there are 4 α hydrogens around C, whereas '0' α hydrogens around A.
Therefore the order of bond length would be C>A>B
