In which of the following processes is the hi | vedclass

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(D) The processes $Cu 	o Cu^{+}$ and $Li 	o Li^{+}$ represent ionization,which involves the removal of an electron. These elements are metals and ha

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$I 	o I^{-}$

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(D) The processes $Cu 	o Cu^{+}$ and $Li 	o Li^{+}$ represent ionization,which involves the removal of an electron. These elements are metals and have relatively low ionization energies.In contrast,$Br 	o Br^{-}$ and $I 	o I^{-}$ represent electron gain processes. The energy change associated with electron gain is known as electron gain enthalpy.For halogens,as we move down the group from $Br$ to $I$,the atomic size increases,which decreases the effective nuclear attraction for the incoming electron.However,the question asks for the process where the highest energy is absorbed. In the case of $I 	o I^{-}$,the electron gain enthalpy is less negative (or more positive/endothermic in some contexts compared to others) due to inter-electronic repulsion in the larger $I$ atom compared to $Br$.Comparing all options,the formation of $I^{-}$ is the least favorable process among the electron gain options,requiring the most energy input to overcome repulsion,making it the process where the highest energy is absorbed.

In which of the following processes is the highest energy absorbed?

Similar Questions

The formation of the oxide ion $O_{(g)}^{2-}$ requires first an exothermic and then an endothermic step as shown below. This is because
$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$
$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$

Which of the following elements will have the lowest electron affinity?

Which electron affinity process would liberate the most energy?

The correct order of the electron gain enthalpy of the given elements is

Electron affinity is the

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