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Least difference between IE2 and IE1 wil...

Least difference between IE2 and IE1 will be found in:

A

Li

B

B

C

Be

D

C

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The correct Answer is:
To determine which element has the least difference between its first ionization energy (IE1) and second ionization energy (IE2), we need to analyze the elements given in the options: lithium (Li), beryllium (Be), boron (B), and carbon (C). ### Step-by-Step Solution: 1. **Understand Ionization Energies**: - **First Ionization Energy (IE1)**: The energy required to remove the outermost electron from a neutral atom in the gaseous state. - **Second Ionization Energy (IE2)**: The energy required to remove an electron from a positively charged ion (after the first electron has been removed). 2. **Consider the Electron Configuration**: - **Lithium (Li)**: 1s² 2s¹ - **Beryllium (Be)**: 1s² 2s² - **Boron (B)**: 1s² 2s² 2p¹ - **Carbon (C)**: 1s² 2s² 2p² 3. **Analyze the Removal of Electrons**: - For **Li**, after removing one electron (IE1), it becomes Li⁺ (1s²). Removing another electron (IE2) from a stable noble gas configuration (1s²) requires significantly more energy. - For **Be**, after removing one electron (IE1), it becomes Be⁺ (1s² 2s¹). Removing another electron (IE2) from a filled subshell (1s²) is also quite high. - For **B**, after removing one electron (IE1), it becomes B⁺ (1s² 2s²). The second electron (IE2) is removed from the 2p subshell, which is less stable than a filled subshell. - For **C**, after removing one electron (IE1), it becomes C⁺ (1s² 2s² 2p¹). The second electron (IE2) is removed from the 2p subshell, similar to boron. 4. **Comparing the Differences**: - The difference between IE1 and IE2 is generally smaller when the second electron is removed from a less stable configuration (like p-orbitals) compared to a stable filled subshell (like s-orbitals). - **Li** has a large difference because it goes from a filled shell to a stable noble gas configuration. - **Be** also has a large difference due to the stability of the filled 1s subshell. - **B** and **C** have smaller differences since they are removing electrons from p-orbitals. 5. **Conclusion**: - The least difference between IE2 and IE1 will be found in **Boron (B)** because it has a more stable configuration after the removal of the first electron compared to lithium and beryllium, and the second electron is removed from a less stable p-orbital. ### Final Answer: The least difference between IE2 and IE1 will be found in **Boron (B)**.
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