Question

Bond length of Ethane (I), Ethene(II), Acetylene (III) and Benzene (IV) following the order:
A) I >II>III>IV
B) I>II>IV>III
C) I>IV>II>III
D) III>IV>II>I

Answer 1

The distance between the centres of two covalently bound atoms is known as bond length. The number of bound electrons determines the length of the bond (the bond order). The heavier the pull between the two atoms and the shorter the bond length, the higher the bond order.

Complete answer: Bond order is trending in the opposite direction to bond time. First, let's look at the bond length pattern.
Consider the following two factors:
A single bond (alkane) has a longer bond length than a double bond (alkene).
One sigma and one pi bond are found in alkene bonds. One sigma and one pi bond are both found in benzene. In the case of benzene, however, the pi bonds are delocalized all over the ring, so the pi electron density is scattered all around the ring. Since the pi electrons are not scattered over many atoms, the C-C in Alkenes is shorter.
As a result, the bond lengths are listed in ascending order: Ethane < Benzene < Ethene < Acetylene
Finally, the bond lengths will be, in ascending order:
acetylene(120pm) < Ethene (133pm) < Benzene (139pm) < Ethane (154pm)
As a result, Option C is the right answer. That is, C) I>IV>II>III

Note:
If the molecule contains more than two atoms, take these steps to establish the bond order:
1. Make an understanding of the Lewis structure.
2. Count how many bonds there are in all.
3. Count the number of bond groups that exist between each atom.
4. Divide the total number of bond groups in the molecule by the number of bonds between atoms.