The first four ionization energies in kJ/mol of a certain second-row element are 801, 2427, 3660, and 25,025. What is the likely identity of the element?

Verified step by step guidance

1. Step 1: Understand the concept of ionization energy, which is the energy required to remove an electron from an atom in the gaseous state. The first ionization energy is for removing the first electron, the second for the second electron, and so on.
2. Step 2: Note the significant jump in ionization energy between the third (3660 kJ/mol) and fourth (25,025 kJ/mol) values. This indicates that the fourth electron is being removed from a stable, filled shell, suggesting a noble gas configuration.
3. Step 3: Recognize that the element is in the second row of the periodic table, which includes elements from Lithium (Li) to Neon (Ne).
4. Step 4: Consider the electronic configurations of second-row elements. The large jump in ionization energy suggests the removal of an electron from a noble gas configuration, which is likely after the removal of three electrons from an element with a configuration ending in 2p^6.
5. Step 5: Identify the element with three valence electrons in the second row, which is Boron (B). After removing three electrons, Boron achieves the stable configuration of Helium (He), explaining the large increase in ionization energy for the fourth electron.