F2 cl2 br2 i2 increasing bond dissociation enthalpy
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Answered by
9
Generally,,,
It will be like,
I2<Br2<Cl2<F2
The above series is due to the fact that the bond having short bond length are difficult to dissociate and more energy is required to break them whereas the bond with larger size or bond length are easy to dissociate and thus have small bond dissociation Enthalpy.
Hope you get your answer.
It will be like,
I2<Br2<Cl2<F2
The above series is due to the fact that the bond having short bond length are difficult to dissociate and more energy is required to break them whereas the bond with larger size or bond length are easy to dissociate and thus have small bond dissociation Enthalpy.
Hope you get your answer.
Answered by
7
I2 Br2 F2 Cl2
since fluorine is small in size and also highly electronegative so it's due to it's repulsion
its bond dissociation enthalpy is lower than cl2
thus
increasing order is
I2 Br2 F2 Cl2
since fluorine is small in size and also highly electronegative so it's due to it's repulsion
its bond dissociation enthalpy is lower than cl2
thus
increasing order is
I2 Br2 F2 Cl2
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