The correct order of decreasing second ionization energy of Li, Be, Ne, C, B

• A. Ne > B > Li > C > Be
• B. Li > Ne > C > B > Be
• C. Ne > C > B > Be > Li
• D. Li > Ne > B > C > Be

The second ionization energy refers to the energy required to remove the second electron from an atom. The order of decreasing second ionization energy for the elements Li, Be, Ne, C, and B can be understood by considering their electronic configurations and the stability of the resulting ions.

Factors Influencing Ionization Energy

• Atomic Size: Smaller atoms generally have higher ionization energies.
• Nuclear Charge: A higher positive charge in the nucleus increases ionization energy.
• Electron Configuration: Atoms with a stable electron configuration (like noble gases) have higher ionization energies.

Analyzing the Elements

Let's look at the elements in question:

• Li (Lithium): After losing one electron, it has a stable configuration, but the second electron is easier to remove.
• Be (Beryllium): More stable after losing two electrons, leading to a higher second ionization energy.
• Ne (Neon): As a noble gas, it has a very high second ionization energy due to its stable configuration.
• C (Carbon): Intermediate ionization energy as it has a half-filled p subshell.
• B (Boron): Lower than carbon but higher than lithium due to its electron configuration.

Correct Order of Decreasing Second Ionization Energy

Based on the above analysis, the correct order is:

Ne > Be > C > B > Li

Thus, the answer is not listed in the options provided. However, if we consider the closest match, it would be option C: Ne > C > B > Be > Li, but Be should actually be higher than C in the correct order.