Which element in each of the following pairs has a more negative electron gain enthalpy?
$(i)$ $O$ or $F$ $(ii)$ $F$ or $Cl$
$(i)$ $O$ or $F$ $(ii)$ $F$ or $Cl$
(A) $(i)$ $F$ $(ii)$ $Cl$
$O$ has $-141 \ kJ \ mol^{-1}$,$F$ has $-328 \ kJ \ mol^{-1}$,and $Cl$ has $-349 \ kJ \ mol^{-1}$ as their electron gain enthalpy $(\Delta_{eg}H)$.
$O$ has $-141 \ kJ \ mol^{-1}$,$F$ has $-328 \ kJ \ mol^{-1}$,and $Cl$ has $-349 \ kJ \ mol^{-1}$ as their electron gain enthalpy $(\Delta_{eg}H)$.
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Among the following options, the correct trend in the electron gain enthalpy is
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View SolutionThe formation of the oxide ion $O^{2-}_{(g)}$ requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:
The amount of energy released for the process $X_{(g)} + e^- \to X^-_{(g)}$ is minimum and maximum respectively for :-
$(a) \ F$
$(b) \ Cl$
$(c) \ N$
$(d) \ B$
Correct answer is :-
$(a) \ F$
$(b) \ Cl$
$(c) \ N$
$(d) \ B$
Correct answer is :-
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