QJEEM24Chb891: JEE Main 2024 Chemistry - Class 11 JEEM24Chb891 Solution: The electron affinity value are negative for A. B. C. D. E.…

To determine which elements have negative electron affinity values, we'll analyze each option individually.

Understanding Electron Affinity

Electron Affinity (EA): The energy change when an electron is added to a neutral atom in the gas phase to form a negative ion.

Convention:

If energy is released (exothermic process), the electron affinity is considered positive.

If energy is absorbed (endothermic process), the electron affinity is considered negative.

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Option A: $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$

Electronic Configuration of Be: $1s^2\,2s^2$

Analysis:

Adding an electron to beryllium means placing it into the higher energy $2p$ orbital.

Beryllium has a filled $2s$ subshell, so the added electron experiences higher energy and electron-electron repulsion.

Result: Energy is absorbed; the process is endothermic.

Conclusion: Electron affinity of Be is negative.

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Option B: $\mathrm{N} \rightarrow \mathrm{N}^{-}$

Electronic Configuration of N: $1s^2\,2s^2\,2p^3$

Analysis:

Nitrogen has a half-filled $2p$ subshell.

Adding an electron introduces repulsion in the half-filled orbital.

Result: Energy is absorbed; the process is endothermic.

Conclusion: Electron affinity of N is negative.

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Option C: $\mathrm{O} \rightarrow \mathrm{O}^{2-}$

Electronic Configuration of O: $1s^2\,2s^2\,2p^4$

Analysis:

First Electron Affinity (O to O⁻): Exothermic (energy released).

Second Electron Affinity (O⁻ to O²⁻): Endothermic because adding an electron to a negative ion requires energy due to electron-electron repulsion.

Overall Process (O to O²⁻): The second step dominates, making the overall process endothermic.

Conclusion: Electron affinity for forming $\mathrm{O}^{2-}$ is negative.

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Option D: $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$

Electronic Configuration of Na: $1s^2\,2s^2\,2p^6\,3s^1$

Analysis:

Adding an electron fills the $3s$ orbital.

Sodium tends to lose an electron to form $\mathrm{Na}^+$ , not gain one.

However, adding an electron is still an exothermic process due to the low energy of the $3s$ orbital.

Result: Energy is released; the process is exothermic.

Conclusion: Electron affinity of Na is positive.

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Option E: $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$

Electronic Configuration of Al: $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$

Analysis:

Adding an electron to the $3p$ orbital.

The process releases energy due to the addition of an electron to a partially filled orbital.

Result: Energy is released; the process is exothermic.

Conclusion: Electron affinity of Al is positive.

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Final Conclusion

Negative Electron Affinity Values (Endothermic Processes):

Option A: Beryllium ( $\mathrm{Be}$ )

Option B: Nitrogen ( $\mathrm{N}$ )

Option C: Oxygen forming $\mathrm{O}^{2-}$

Therefore, the correct answer is:

Option D: A, B, and C only

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Answer: Option D