Question

${{\text{X}}_{\text{2}}}$ represents a halogen molecule. The bond energy of different halogen molecules will lie in the following sequences:
A)\[{{\text{F}}_{\text{2}}}\text{ }\rangle \text{ C}{{\text{l}}_{\text{2}}}\text{ }\rangle \text{ B}{{\text{r}}_{\text{2}}}\text{ }\rangle \text{ }{{\text{I}}_{\text{2}}}\]
B)\[\text{C}{{\text{l}}_{\text{2}}}\text{ }\rangle \text{ B}{{\text{r}}_{\text{2}}}\text{ }\rangle \text{ }{{\text{F}}_{\text{2}}}\text{ }\rangle \text{ }{{\text{I}}_{\text{2}}}\]
C) ${{\text{I}}_{\text{2}}}\rangle \text{ C}{{\text{l}}_{\text{2}}}\rangle \text{ B}{{\text{r}}_{\text{2}}}\rangle \text{ }{{\text{F}}_{\text{2}}}$
D) \[\text{B}{{\text{r}}_{\text{2}}}\text{ }\rangle \text{ }{{\text{F}}_{\text{2}}}\text{ }\rangle \text{ }{{\text{I}}_{\text{2}}}\rangle \text{ C}{{\text{l}}_{\text{2}}}\]

Answer 1

Hint: Bond energy or enthalpy is the amount of heat required to break one mole of a covalent bond between the atoms in a molecule into its corresponding atoms in the gas state. For a diatomic molecule, the bond energy is inversely proportional to the bond length.

Complete step by step answer:
-In the periodic table, halogen is the group VII A group element. There are five elements in which are; Fluorine \[\text{(F)}\], chlorine $\text{(Cl)}$, Bromine $\text{(Br)}$ , Iodine $\text{(I)}$ , and Astatine $\text{(As)}$.
-The general electronic configuration for halogen is$\text{ (n}{{\text{s}}^{\text{2}}}\text{n}{{\text{p}}^{\text{5}}}\text{)}$. The last electron enters into the p-orbital.
-We know that all the halogens exist as diatomic molecules. This is because of the overlapping of p-orbitals.
-The atomic size of the atoms goes on the increase as we move down in a group from fluorine to the iodine. As the size of the atom goes on increasing from $\text{Cl}$ $\text{I}$ the bond length between the atoms increases. We know that bond length is inversely related to bond energy. Therefore, the bond dissociation or bond energy goes on decreasing from $\text{Cl}$to$\text{I}$.
-Conventionally, the bond energies should be greater for fluorine and decrease as we move down in a group. However, halogen does not follow this trend.

-Because \[\text{F}\] have anomalous behavior.
The electronic configuration for \[\text{F}\] is given as follows,
\[\text{1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{5}}}\]
Here \[\text{2p}\] subshell is very compact and close to the nucleus. Therefore the number of electrons on the fluorine is held tightly in a compact volume. There exists a strong repulsive force between the non-bonded or lone pairs on the two fluorine atoms.

-The bond $\text{F-F}$ becomes weak because of the repulsive forces. Therefore the bond cannot hold the two fluorine atoms together to form a stable molecule. The amount of energy required to break $\text{F-F}$ is less than the other halogens.

-Bond energies for $\text{C}{{\text{l}}_{\text{2}}}$ is highest. Bond energy ${{\text{F}}_{\text{2}}}\text{,C}{{\text{l}}_{\text{2}}}\text{,B}{{\text{r}}_{\text{2}}}\text{ and }{{\text{I}}_{\text{2}}}$ is 37, 58, 46, and 36 $\text{Kcal/mol}$respectively.

-Therefore the bond energy trend in a halogen is,
\[\text{C}{{\text{l}}_{\text{2}}}\rangle \text{ B}{{\text{r}}_{\text{2}}}\rangle \text{ }{{\text{F}}_{\text{2}}}\rangle \text{ }{{\text{I}}_{\text{2}}}\]
So, the correct answer is “Option B”.

Additional Information:
In redox chemistry, the stronger is the reducing agent, the weaker is the oxidizing agent, for example, the fluorine gas is the strong oxidizing agent but this is the weak reducing agent as it cannot donate its electrons easily. Similarly, we can say that the weaker is the oxidizing agent, stronger is its corresponding reducing agent. For example,
\[\begin{align}
  & \begin{matrix}
   {{\text{F}}_{\text{2}}} & +\text{2}{{\text{e}}^{\text{-}}} & \to & \text{2}{{\text{F}}^{\text{-}}} \\
   (strong\text{ oxidising agent)} & {} & {} & \text{(weakest reducing agent) } \\
\end{matrix} \\
 & \text{ } \\
\end{align}\]
\[\begin{align}
  & \begin{matrix}
   \underset{{}}{\mathop{\text{N}{{\text{a}}^{\text{+}}}}}\, & \text{+ }{{\text{e}}^{\text{-}}} & \to & \underset{{}}{\mathop{\text{Na}}}\, \\
   \text{(weakest oxidising agent)} & {} & {} & \text{( Strongest reducing agent)} \\
\end{matrix} \\
 & \text{ } \\
\end{align}\]

Note: Remember that bond enthalpies are only applicable to the substances in a gaseous state. \[\text{B}{{\text{r}}_{\text{2}}}\] And \[{{\text{I}}_{\text{2}}}\] exists as a liquid and solid respectively. To study the bond energies \[\text{B}{{\text{r}}_{\text{2}}}\] and \[{{\text{I}}_{\text{2}}}\] they are required to be converted into gaseous form. The energy associated with the conversion of liquid and solid to gas form is calculated as the energy of atomization.