The electron affinity value are negative for

A. $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$

B. $\mathrm{N} \rightarrow \mathrm{N}^{-}$

C. $\mathrm{O} \rightarrow \mathrm{O}^{2-}$

D. $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$

E. $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$

Choose the most appropriate answer from the options given below :

<p>To determine which elements have <strong>negative electron affinity values</strong>, we'll analyze each option individually.</p> <h3>Understanding Electron Affinity</h3> <p><p><strong>Electron Affinity (EA):</strong> The energy change when an electron is added to a neutral atom in the gas phase to form a negative ion.</p></p> <p><p><strong>Convention:</strong></p></p> <p><p>If energy is <strong>released</strong> (exothermic process), the electron affinity is considered <strong>positive</strong>.</p></p> <p><p>If energy is <strong>absorbed</strong> (endothermic process), the electron affinity is considered <strong>negative</strong>.</p></p> <hr /> <h3>Option A: $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$</h3> <p><p><strong>Electronic Configuration of Be:</strong> $1s^2\,2s^2$</p></p> <p><p><strong>Analysis:</strong></p></p> <p><p>Adding an electron to beryllium means placing it into the higher energy $2p$ orbital.</p></p> <p><p>Beryllium has a filled $2s$ subshell, so the added electron experiences higher energy and electron-electron repulsion.</p></p> <p><p><strong>Result:</strong> Energy is <strong>absorbed</strong>; the process is <strong>endothermic</strong>.</p></p> <p><p><strong>Conclusion:</strong> Electron affinity of Be is <strong>negative</strong>.</p></p> <hr /> <h3>Option B: $\mathrm{N} \rightarrow \mathrm{N}^{-}$</h3> <p><p><strong>Electronic Configuration of N:</strong> $1s^2\,2s^2\,2p^3$</p></p> <p><p><strong>Analysis:</strong></p></p> <p><p>Nitrogen has a half-filled $2p$ subshell.</p></p> <p><p>Adding an electron introduces repulsion in the half-filled orbital.</p></p> <p><p><strong>Result:</strong> Energy is <strong>absorbed</strong>; the process is <strong>endothermic</strong>.</p></p> <p><p><strong>Conclusion:</strong> Electron affinity of N is <strong>negative</strong>.</p></p> <hr /> <h3>Option C: $\mathrm{O} \rightarrow \mathrm{O}^{2-}$</h3> <p><p><strong>Electronic Configuration of O:</strong> $1s^2\,2s^2\,2p^4$</p></p> <p><p><strong>Analysis:</strong></p></p> <p><p><strong>First Electron Affinity (O to O⁻):</strong> Exothermic (energy released).</p></p> <p><p><strong>Second Electron Affinity (O⁻ to O²⁻):</strong> Endothermic because adding an electron to a negative ion requires energy due to electron-electron repulsion.</p></p> <p><p><strong>Overall Process (O to O²⁻):</strong> The second step dominates, making the overall process <strong>endothermic</strong>.</p></p> <p><p><strong>Conclusion:</strong> Electron affinity for forming $\mathrm{O}^{2-}$ is <strong>negative</strong>.</p></p> <hr /> <h3>Option D: $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$</h3> <p><p><strong>Electronic Configuration of Na:</strong> $1s^2\,2s^2\,2p^6\,3s^1$</p></p> <p><p><strong>Analysis:</strong></p></p> <p><p>Adding an electron fills the $3s$ orbital.</p></p> <p><p>Sodium tends to lose an electron to form $\mathrm{Na}^+$, not gain one.</p></p> <p><p>However, adding an electron is still an exothermic process due to the low energy of the $3s$ orbital.</p></p> <p><p><strong>Result:</strong> Energy is <strong>released</strong>; the process is <strong>exothermic</strong>.</p></p> <p><p><strong>Conclusion:</strong> Electron affinity of Na is <strong>positive</strong>.</p></p> <hr /> <h3>Option E: $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$</h3> <p><p><strong>Electronic Configuration of Al:</strong> $1s^2\,2s^2\,2p^6\,3s^2\,3p^1$</p></p> <p><p><strong>Analysis:</strong></p></p> <p><p>Adding an electron to the $3p$ orbital.</p></p> <p><p>The process releases energy due to the addition of an electron to a partially filled orbital.</p></p> <p><p><strong>Result:</strong> Energy is <strong>released</strong>; the process is <strong>exothermic</strong>.</p></p> <p><p><strong>Conclusion:</strong> Electron affinity of Al is <strong>positive</strong>.</p></p> <hr /> <h3>Final Conclusion</h3> <p><p><strong>Negative Electron Affinity Values (Endothermic Processes):</strong></p></p> <p><p><strong>Option A:</strong> Beryllium ($\mathrm{Be}$)</p></p> <p><p><strong>Option B:</strong> Nitrogen ($\mathrm{N}$)</p></p> <p><p><strong>Option C:</strong> Oxygen forming $\mathrm{O}^{2-}$</p></p> <p><strong>Therefore, the correct answer is:</strong></p> <p><strong>Option D: A, B, and C only</strong></p> <hr /> <p><strong>Answer:</strong> <strong>Option D</strong></p>