Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:

F > C l > B r > I

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F < C l > B r > I

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F > B r > C l > I

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I > C l > B r > F

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Solution

The correct option is B $F < C l > B r > I$
Electron gain enthalpy becomes less negative on moving down a group due to increase in atomic size. However, the electron gain enthalpy of $F$ is less negative than that of $C I$ due to electron-electron repulsion in small-sized $F$ atom. Thus, the correct order is $F < C l > B r > I$ .
$F, C l, B r$ and $I$ have $Δ_{e g} H$ values(in $k J m o l^{- 1}$ ) -328, -349, -325 and -295 respectively.

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Similar questions

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy ) is
(a) F > Cl > Br > I (b) F < Cl < Br < I (c) F < Cl > Br > I (d) F < Cl < Br < I

Q. The correct order of polarizability is

I^{-}, B r^{-}, C l^{-}, F^{-}

Q. Which of the following options does not give to the correct trends indicated against it?
(I)

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

: Bond dissociation energy
(II)

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

: Oxidizing nature
(III)

C l > F > B r > I

: Electron gain enthalpy
(IV)

F > C l > B r > I

: Electronegativity

Q. Which of the following options does not give to the correct trends indicated against it?
(I)

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

: Bond dissociation energy
(II)

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

: Oxidizing nature
(III)

C l > F > B r > I

: Electron gain enthalpy
(IV)

F > C l > B r > I

: Electronegativity

Q. The electron gain enthalpies of halogens in kJ mol

^{- 1}

are as given as F = -332, Cl = -349, Br = -324, I = -295. The less negative value for F as compared to that of Cl is due to :

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