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Question: Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules...

Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
A) I2>Br2>Cl2>F2{I_2} > B{r_2} > C{l_2} > {F_2}
B) Cl2>Br2>F2>I2C{l_2} > B{r_2} > {F_2} > {I_2}
C) Br2>I2>F2>Cl2B{r_2} > {I_2} > {F_2} > C{l_2}
D) F2>Cl2>Br2>I2{F_2} > C{l_2} > B{r_2} > {I_2}

Explanation

Solution

The bond dissociation enthalpy is the energy needed to break a bond between two atoms. The halogens are group 17{\text{17}} elements and their size increases down the group. One can think about these factors which affect the bond dissociation enthalpy and decide the order.

Complete step by step answer:

  1. First of all we will learn about the concept of bond dissociation enthalpy. Every atom has a specific size and while making the molecule the size of the individual atoms contributes to molecule formation.
  2. The more the size of an atom the more will be its bond length in the molecule. The more is the bond length in the molecule the less energy will be required to break the bond between the two atoms in a molecule. Therefore, there will be less bond dissociation enthalpy will be required as the bond will be readily breaking.
  3. This happens in the case of small size atoms exactly opposite. This means the smaller is the size of the atom the more will be bond dissociation enthalpy of that molecule.
  4. Now in the case of halogen molecules the iodine molecule is largest hence it will have the lowest bond dissociation enthalpy. The chlorine atom has the lowest size among all hence it will have the highest bond dissociation enthalpy.
  5. The fluorine molecule shows an exception in this order hence it is placed after the bromine molecule.
    Therefore, the order for the bond dissociation enthalpy of halogen molecules is Cl2>Br2>F2>I2C{l_2} > B{r_2} > {F_2} > {I_2} which shows option B as the correct choice.

Note: In the case of fluorine which is an exception for this order, due to the high electronegativity of fluorine atom it tends to break the bond present between the fluorine molecule hence requiring less energy for the breaking of the bond which results in requiring low bond dissociation enthalpy.