What is the order of bond dissociation enthalpy in halogens?
Correct order of bond dissociation enthalpy of halogens is Cl2 > Br2 > F2 > I2. Due to inter electronic repulsions F–F bond becomes weak and easily broken.
Which has more dissociation energy F2 or Cl2?
Therefore, the repulsion between electrons in the outer most shell of the two atoms in a fluorine molecule is much greater than that in a chlorine molecule. Hence, it requires less energy to break up the fluorine molecule, making its bond dissociation energy lesser than that of chlorine molecule.
Why do halogens have low bond energy?
Due to small atomic size number of electrons are held in a compact volume and there is strong repulsion amongst non bonded electrons. Hence bond becomes weak though bond is short. Due to this reason bond dissociation energy of fluorine less than other halogens.
What is enthalpy of bond dissociation?
The bond dissociation enthalpy is the energy needed to break one mole of the bond to give separated atoms – everything being in the gas state. As an example of bond dissociation enthalpy, to break up 1 mole of gaseous hydrogen chloride molecules into separate gaseous hydrogen and chlorine atoms takes 432 kJ.
Which of the following has highest bond enthalpy F2 Cl2 Br2 I2?
F2 is expected to have highest bond dissociation enthalpy but it is Cl2…. the correct decreasing order isCl2>Br2>F2>I2 is is so because fluorine atom has very small size due to which there is a high inter electronic repulsion between two fluorine.
Why is Cl2 bond energy higher than F2?
The size of a fluorine atom is very small as compared to a chlorine atom. Therefore, the repulsion between electrons in the outer most shell of the two atoms in a fluorine molecule is much greater than that in a chlorine molecule.
Which halogen has the lowest bond dissociation energy?
4) Now in the case of halogen molecules the iodine molecule is largest hence it will have the lowest bond dissociation enthalpy.