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Periodic Properties

Which of the ...

Question

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

I only

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I I and I I I

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I I I and I V

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I I, I I I and I V

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Solution

The correct option is A $I only$
(I) Because of small size of fluorine atom there is a high inter electronic repulsion between two fluorine atoms.
The correct decreasing order is: $C l_{2} > B r_{2} > F_{2} > I_{2}$
(II) Oxidation ability decreases on moving down the group because of electronegativity and hydration energy decreases.
The correct decreasing order is $F_{2} > C l_{2} > B r_{2} > I_{2}$
(III) Electron gain enthalpy becomes less negative on moving down a group due to an increase in the atomic size. However, the electron gain enthalpy of F is less negative than that of Cl due to electron-electron repulsion in the small-sized F atom. Thus, the correct order is $F < C l > B r > I .$
(IV) Electronegativity decreases as we move doen the group, so the correct order is $F > C l > B r > I$

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Similar questions

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

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

C l > F > B r > I

F > C l > B r > I

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

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

C l > F > B r > I

F > C l > B r > I

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}

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

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