Question

Which one of the following orders is correct for the bond
dissociation enthalpy of halogen molecules? [2016]
(a) I₂ > Br₂ > Cl₂ > F₂ (b) Cl₂ > Br₂ > F₂ > I₂
(c) Br₂ > I₂ > F₂ > Cl₂ (d) F₂ > Cl₂ > Br₂ > I₂

Answer 1

Answer:

(b) option is correct one!!!!!✔✔✔✔

Answer 2

Cl₂ > Br₂ > F₂ > I₂ is the correct trend.

Explanation:

• Bond energy or the bond dissociation energy is defined as the amount of energy needed to break the bonds of a molecule.

• Halogens are the members of group 17 of periodic table.

• Members of halogen group are fluorine, chlorine, bromine, and iodine.

• As we go down the group, the size of the atoms increase due to addition of new shells.

• So the bond length increase and the bond strength decrease.

• So less amount of energy is needed to break a bond.

• So the bond dissociation enthalpy decrease down a group.

• But in case of fluorine, being very small shows some exception.

• Fluorine is very small in size and the inter electronic expulsions make the bond length longer than expected.

• So less energy is needed to break the bond.

• Thus the trend comes as Cl₂ > Br₂ > F₂ > I₂.

For more information about bond dissociation energy of halogens,

https://brainly.in/question/3378097

increasing order of bond dissociation of halogen - Brainly.in

https://brainly.in/question/8265663

Which halogen have highest bond dissociation energy among hf hcl hbr hi?