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Enthalpy

Bond energies...

Question

Bond energies can be obtained by using the following relation:
$Δ H (r e a c t i o n) = \sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products
Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.Bond energy of different halogen molecules will lie in the sequences of:

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

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

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

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

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Solution

The correct option is B $C l_{2} > B r_{2} > F_{2} > I_{2}$
Bond energy of $F_{2}$ is low due to lower size of F atoms and strong repulsion between the lone pairs of two fluorine atoms.
Among other halides as size increases bond energy decreases so order is
$C l_{2} > B r_{2} > F_{2} > I_{2}$

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Similar questions

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

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

Δ H

= \sum

- \sum

N - H

O - H

F - H

Δ H (r e a c t i o n) = \sum

- \sum

N O

+ 90 k J m o l^{- 1}

N O

B E_{N \equiv N} = 941 k J m o l^{- 1}

B E_{O = O} = 499 k J m o l^{- 1}

{C l}_{2}, {B r}_{2}

I_{2}

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