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Periodic Properties

Bond energy o...

Question

Bond energy of different halogen molecules will lie in the sequence
$C l_{2} > B r_{2} > F_{2} > I_{2}$

True

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False

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Solution

The correct option is A True
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

X_{2}

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

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

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

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

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

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

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

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

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

(x^{2} + y^{2})

x

y

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

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