What is the decreasing order of carbon-carbon | vedclass

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(C) The bond length of a $C-C$ single bond depends on the hybridization of the carbon atoms involved.As the percentage of $s$-character increases,the

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$III > IV > I > II$

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(C) The bond length of a $C-C$ single bond depends on the hybridization of the carbon atoms involved.As the percentage of $s$-character increases,the electronegativity of the carbon atom increases,and the bond length decreases.Hybridization states for the $C-C$ single bond in each compound:$(I) CH_2 = CH - C \equiv CH$: The $C-C$ single bond is between $sp^2$ and $sp$ hybridized carbons.$(II) CH \equiv C - C \equiv CH$: The $C-C$ single bond is between $sp$ and $sp$ hybridized carbons.$(III) CH_3 - CH = CH_2$: The $C-C$ single bond is between $sp^3$ and $sp^2$ hybridized carbons.$(IV) CH_2 = CH - CH = CH_2$: The $C-C$ single bond is between $sp^2$ and $sp^2$ hybridized carbons.Comparing the $s$-character:$sp^3-sp^2$ (lowest $s$-character) > $sp^2-sp^2$ > $sp^2-sp$ > $sp-sp$ (highest $s$-character).Therefore,the order of bond length is: $(III) > (IV) > (I) > (II)$.

What is the decreasing order of carbon-carbon single bond length in the following compounds?
$(I) CH_2 = CH - C \equiv CH$
$(II) CH \equiv C - C \equiv CH$
$(III) CH_3 - CH = CH_2$
$(IV) CH_2 = CH - CH = CH_2$

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What is the decreasing order of carbon-carbon single bond length in the following compounds?$(I) CH_2 = CH - C \equiv CH$$(II) CH \equiv C - C \equiv CH$$(III) CH_3 - CH = CH_2$$(IV) CH_2 = CH - CH = CH_2$