Order of reactivity towards halogenation is $F_{2} > C l_{2} > B r_{2} > I_{2}$ .
State True or False.

True

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False

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Solution

The correct option is A True
Bond length is inversely proportional to reactivity. It is the bond energy decides the reactivity order. Lower the bond energy more reactive will be the halogen.
The bond dissociation energy of Fluorine,lower than those of chlorine and bromine because of inter electronic repulsions present in the small atom of fluorine.Hence bond energy decreases in the order $C l_{2} > B r_{2} > F_{2} > I_{2}$
Due to small size,high electronic density,greater inter electronic repulsion(lower $F - F$ bond energy) $F_{2}$ is highly reactive among halogens. Hence the order,
$F_{2} > C l_{2} > B r_{2} > I_{2}$

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Similar questions

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

I_{2} < {B r}_{2} < {C l}_{2} < F_{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

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

I_{2} < {B r}_{2} < {C l}_{2} < F_{2}

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

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