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The van der Waals forces in halogens increase in the order :
$F_{2} < {C l}_{2} < B r_{2} < I_{2}$
$I_{2} > C l_{2} > {B r}_{2} > F_{2}$
$F_{2} < B r_{2} < I_{2} < {C l}_{2}$
$B r_{2} < F_{2} < {C l}_{2} < I_{2}$

A

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

B

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

C

I_{2} > C l_{2} > {B r}_{2} > F_{2}

D

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

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Solution

Verified by Toppr

The van der Waals forces in halogens increase in the order $F_{2} < {C l}_{2} < B r_{2} < I_{2}$ . This is the same order as that for the increase in the atomic size. Higher is the atomic size, higher will be the van der Waal's force.

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Similar Questions

Q1

The van der Waals forces in halogens increase in the order :

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Q2

Arrange the halogens

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

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Q3

Consider the following orders –
(1)

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

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(2)

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

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

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(4)

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

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

EA
(6)

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

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(7)

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

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(8)

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

Oxidizing nature
Then calculate

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

when

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Q4

Order of reactivity towards halogenation is

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

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State True or False.

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Q5

Bond energy of different halogen molecules will lie in the sequence

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

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