Chemistry - PERIODIC TABLE Question with Solution | TestHub

A. Amongst noble gases, addition of electron is toughest for Ar.

Noble gases have stable, filled electron shells, making electron addition highly unfavorable (endothermic). Electron addition toughest for Ne, not Ar.

B. Ionisation Energy of ${\mathrm{F}}^{+}$ is greater than that of ${\mathrm{O}}^{+}$.

${\mathrm{F}}^{+}$ has an electron configuration of $1s^{2}2s^{2}2p^4$. ${\mathrm{O}}^{+}$ has an electron configuration of $1s^{2}2s^{2}2p^3$. Removing an electron from ${\mathrm{F}}^{+}$ means removing an electron from a $2p$ orbital, which is more tightly held than in ${\mathrm{O}}^{+}$ due to higher nuclear charge. Also, ${\mathrm{O}}^{+}$ has a half-filled $2p$ subshell ($2p^3$), which is relatively stable, making it harder to remove an electron from ${\mathrm{O}}^{+}$ than from ${\mathrm{F}}^{+}$. Therefore, $\mathrm{IE}({\mathrm{O}}^{+})>\mathrm{IE}({\mathrm{F}}^{+})$. This statement is incorrect.

C. ${\mathrm{A}}^{+}\to {\mathrm{A}}^{-}$ is always exothermic process.

This statement is generally true. The conversion of a cation ${\mathrm{A}}^{+}$ to an anion ${\mathrm{A}}^{-}$ involves the gain of two electrons. The first electron gain (${\mathrm{A}}^{+}\to \mathrm{A}$) is always highly exothermic (equal to the negative of the first ionization energy, $-{\mathrm{IE}}_1$). The second electron gain ($\mathrm{A}\to {\mathrm{A}}^{-}$) is the electron affinity (${\mathrm{EA}}_1$). While ${\mathrm{EA}}_1$ can be endothermic for some elements, the large exothermic energy from the first step usually dominates, making the overall process exothermic. For example, for chlorine: ${\mathrm{Cl}}^{+}\to \mathrm{Cl}$ is highly exothermic, and $\mathrm{Cl}\to {\mathrm{Cl}}^{-}$ is also exothermic.

D. Electron Affinity of ${\mathrm{Na}}^{+}$ is greater than that of ${\mathrm{Li}}^{+}$.

Electron affinity of a cation refers to the energy change when it gains an electron. ${\mathrm{Na}}^{+}$ has an electron configuration of $1s^{2}2s^{2}2p^6$. ${\mathrm{Li}}^{+}$ has an electron configuration of $1s^2$. Both are stable noble gas configurations. Gaining an electron for ${\mathrm{Na}}^{+}$ would lead to $\mathrm{Na}$ ($1s^{2}2s^{2}2p^{6}3s^1$), which is a stable atom. Gaining an electron for ${\mathrm{Li}}^{+}$ would lead to $\mathrm{Li}$ ($1s^{2}2s^1$). Due to larger size and lower effective nuclear charge, ${\mathrm{Na}}^{+}$ would have a lower attraction for an incoming electron compared to ${\mathrm{Li}}^{+}$. Therefore, $\mathrm{EA}({\mathrm{Li}}^{+})>\mathrm{EA}({\mathrm{Na}}^{+})$. This statement is incorrect.

Final Answer: The final answer is $\boxed{C}$