Question

Which of the following has the shortest carbon-carbon bond length?
A.${C_2}{H_2}$
B.${C_2}{H_4}$
C.${C_2}{H_6}$
D.All have the same bond length

Answer 1

Hint:The relationship between bond length and bond order will be taken into consideration. Bond order is inversely proportional to bond length.

Complete step by step answer:

Bond order and bond length is inversely proportional to each other, which defines that the carbon-carbon bond which has a larger bond order will have the shortest carbon-carbon bond length.
$B.O \propto \dfrac{1}{{B.L}}$
Where, B.O is bond order and B.L is bond length.

Bond order is always based upon \[C - C\] type of bond, \[C - C\] triple bond will always have a higher bond order and shortest bond length.

\[C - C\] Single bond will always have a lesser bond order and longest bond length.

As to compare bond order, we need to find out bond order.
For${C_2}{H_2}$, (Ethyne)


The carbon-carbon bond in ethyne is a triple bond, six electrons are shared in the triple bond.
Bond order will be \[3\].

For${C_2}{H_4}$ (Ethene)


The carbon-carbon bond in ethene is double bond, two electrons are shared in the double bond.
Bond order will be \[2\].

For${C_2}{H_6}$ (ethane)


The carbon-carbon bond in ethane is a single bond, two electrons are shared in the single bond.
Bond order will be \[1\].
By comparing the above three compounds we can conclude that the B.O of carbon-carbon bond ${C_2}{H_2}$ is highest.

So, according to the relation between these two ${C_2}{H_2}$ will have the shortest carbon-carbon bond length.
Therefore, option A is correct.


Note:
Bond order tells us about the number of bonds between two atoms in a molecule. Atoms and molecules may form bonds in which they share more than two electrons. Bond length or bond distance can be defined as the average distance between nuclei of two bonded atoms in a molecule.