Question

Which has the highest bond energy and stronger bond?
A. ${{\text{F}}_{\text{2}}}$
B. ${\text{C}}{{\text{l}}_{\text{2}}}$
C. ${\text{B}}{{\text{r}}_{\text{2}}}$
D. ${{\text{I}}_{\text{2}}}$

Answer 1

To determine the answer we should know the general trends of periodic table. The distance between the nucleuses of two bonded atoms is known as bond length. Bond strength tells weakest and strongest bond. Bond energy and strength depends upon the size of atoms. So, we will check the size of given molecules.

Complete step-by-step answer:
According to the general trend of periodic table, on going left to right in the period, the size of the atom decreases and on going up to down in the group, the size of the atom increases.Down in the group, the electrons get added in the next higher shell, so the distance between the nucleus and outermost shell increases, so the effective nuclear charge of the nucleus decreases and thus size increases.

Due to increase in size of atoms, the atoms cannot come too close to each other, so the distance between nucleuses of bonded atoms increases and thus the bond becomes weak and the bond has small energy.

Fluorine, chlorine, bromine and iodine all are of group-seventeen molecules. Fluorine due to its smallest size should form the strongest bond and iodine should form the weakest bond. Strongest bonds require high energy to break, so the bond energy will be high for the strongest bond.

So, the order of bond strength or bond energy should be,
${{\text{F}}_{\text{2}}}\,{\text{ > }}\,{\text{C}}{{\text{l}}_{\text{2}}}\,{\text{ > }}\,{\text{B}}{{\text{r}}_{\text{2}}}\,{\text{ > }}\,{{\text{I}}_{\text{2}}}$
But the fluorine is of very small size, so F-F distance is very less thus, the repulsion in between the electron pairs is very high, so fluorine molecule is less stable than chlorine molecule.
So, the correct order of bond strength or bond energy is,
${\text{C}}{{\text{l}}_{\text{2}}}\,{\text{ > }}\,{{\text{F}}_{\text{2}}}\,{\text{ > }}\,\,{\text{B}}{{\text{r}}_{\text{2}}}\,{\text{ > }}\,{{\text{I}}_{\text{2}}}$
So, ${\text{C}}{{\text{l}}_{\text{2}}}$ has the highest bond energy and stronger bond.

Therefore, option (B) ${\text{C}}{{\text{l}}_{\text{2}}}$is correct.

Note: The bond energy, bond strength and bond stability, all are directly proportional to the bond order. All these, bond energy, bond strength, bond stability and bond order all inversely proportional to the bond length. Bond order for homo-diatomic molecules is determined by molecular orbital theory. With the size of the atom, electronegativity also affects the bond strength. Higher the electronegativity higher the bond stability.