Question

Calculate the I−I distance in the given compound \[{H_2}C = C{I_2}\] , if \[C - I\] bond length is \[2.10{A^0}\] . Give answer in nearest integer value in \[{A^0}\].

Answer 1

Hint: We need to remember that the distances between the centers of bonded atoms are called bond lengths. Bond order affects the bond length greater the bond order shorter is the bond length.
Periodic trend:
As we know that across a period the bond length decreases and increases down a group. Because down the group, the atoms of the elements are increasing in size due to more number of electron shells. Since the distance of the nucleus to the outer electron of the other atom is longer, bond length will be longer.

Complete step by step answer:
It is given that bond length of \[C - I = 2.10{A^o}\]
The distance between I-I is given by the expression,
$2 \times \left( {{\text{Bond length}}{{\text{h}}_{C - I}} \times \sin {{60}^ \circ }} \right)$
Now we can substitute the known values we get,
\[ = 2 \times 2.10 \times 0.866 = 3.65 \simeq 4{A^0}\]
The distance between I-I in the given compound is $4{A^o}$.

Note: Now we can discuss about the concept of bond energy and bond order as,
Bond energy:
We need to know that the bond order is the number of electrons pairs shared between two atoms in the formation of the bond. The bond dissociation energy is the amount of energy needed to cleave the bond. Thus greater the bond order greater is the bond energy.
Bond order:
Bond order$ = \dfrac{{\left( {{N_b}} \right)\, - \left( {{N_a}} \right)}}{2}$
Where, ${{\text{N}}_{\text{b}}}$ is the number of electrons in a bonding orbital.
${{\text{N}}_{\text{a}}}$ is the number of electrons in an antibonding orbital.
${\text{Bond order}} \propto {\text{Bond energy}} \propto \dfrac{1}{{{\text{Bond length}}}}$
As the bond order decreases, the bond energy decreases and the bond length increases. The triple bonds between atoms are shorter than double bonds because it requires more energy to completely break all three bonds than to break two. Triple bonds are stronger than double bonds. A large amount of energy has to use large energy to triple bond than double bond. That is the reason why triple bonds are shorter than single and double bonds.