The breaking of which of the following bond is most endothermic?

F - F

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C l - C l

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B r - B r

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I - I

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Solution

The correct option is B $F - F$
The bond dissociation energy is directly related to the bond enthalpy. The higher the bond dissociation energy, the more endothermic the reaction will be. The order of bond dissociation energies is:
$F - F > C l - C l > B r - B r > I - I$
Therefore, $F - F$ is the most endothermic.

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Similar questions

Q. The correct order of polarizability is

I^{-}, B r^{-}, C l^{-}, F^{-}

Q. Arrange the following bonds according to their average bond energies in descending order:

C - C l, C - B r, C - F, C - I

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy ) is
(a) F > Cl > Br > I (b) F < Cl < Br < I (c) F < Cl > Br > I (d) F < Cl < Br < I

Q. Which of the following options does not give to the correct trends indicated against it?
(I)

F_{2} > C l_{2} > B r_{2} > I_{2}

: Bond dissociation energy
(II)

F_{2} > C l_{2} > B r_{2} > I_{2}

: Oxidizing nature
(III)

C l > F > B r > I

: Electron gain enthalpy
(IV)

F > C l > B r > I

: Electronegativity

Which of the following displaces oxygen from water?

F₂, Cl₂, Br₂, I₂

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The breaking of which of the following bond is most endothermic?

The correct option is B F−FThe bond dissociation energy is directly related to the bond enthalpy. The higher the bond dissociation energy, the more endothermic the reaction will be. The order of bond dissociation energies is:F−F>Cl−Cl>Br−Br>I−ITherefore, F−F is the most endothermic.