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Which one of the following arrangements does not give the correct picture of the trends indicated against it ?
A
$F_2 \gt Cl_2 \gt Br_2 \gt I_2$ : Bond dissociation energy
B
$F_2 \gt Cl_2 \gt Br_2 \gt I_2$ : Electronegativity
C
$F_2 \gt Cl_2 \gt Br_2 \gt I_2$ : Oxidizing power
D
$F_2 \gt Cl_2 \gt Br_2 \gt I_2$ : Electron gain enthalpy
Detailed Solution
In the case of the diatomic molecules ($X_2$) of the halogens, the bond dissociation energy decreases in the order: $Cl_2 \gt Br_2 \gt F_2 \gt I_2$
Approximate values: $Cl_2$ = 242.6, $Br_2$ = 192.8, $F_2$ = 158.8 and $I_2$ = 151.1 kJ mol$^{-1}$.
The F-F bond is unexpectedly weak because the F atom is very small, so the lone pairs on the two F atoms repel each other strongly.
So the arrangement $F_2 \gt Cl_2 \gt Br_2 \gt I_2$ for bond dissociation energy is not correct.
The oxidizing power, electronegativity and reactivity decrease in the order: $F_2 \gt Cl_2 \gt Br_2 \gt I_2$, so those two arrangements are correct.
The electron gain enthalpy of the halogens follows the order $Cl \gt F \gt Br \gt I$; the low value for fluorine is again due to the small size of the fluorine atom. Strictly, that arrangement is also not exact, but the answer intended by the paper is the bond dissociation energy arrangement.
Approximate values: $Cl_2$ = 242.6, $Br_2$ = 192.8, $F_2$ = 158.8 and $I_2$ = 151.1 kJ mol$^{-1}$.
The F-F bond is unexpectedly weak because the F atom is very small, so the lone pairs on the two F atoms repel each other strongly.
So the arrangement $F_2 \gt Cl_2 \gt Br_2 \gt I_2$ for bond dissociation energy is not correct.
The oxidizing power, electronegativity and reactivity decrease in the order: $F_2 \gt Cl_2 \gt Br_2 \gt I_2$, so those two arrangements are correct.
The electron gain enthalpy of the halogens follows the order $Cl \gt F \gt Br \gt I$; the low value for fluorine is again due to the small size of the fluorine atom. Strictly, that arrangement is also not exact, but the answer intended by the paper is the bond dissociation energy arrangement.