Answer:
1. Neon
2. Oxygen
3. Sulfur
4. Aluminum
Step-by-step explanation:
The elements ranked by increasing ionization energy are as follows:
1. Neon
2. Oxygen
3. Sulfur
4. Aluminum
Step-by-step explanation:
Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. It is influenced by factors such as atomic size, nuclear charge, and electron shielding.
1. Neon (Print)
Neon, with the atomic number 10, has the lowest ionization energy among the given elements. This is because neon has a full valence shell of electrons (2s^22p^6), making it highly stable. The removal of an electron from a stable configuration requires a significant amount of energy.
2. Oxygen (Print)
Oxygen, with the atomic number 8, has a higher ionization energy than neon. Oxygen has six valence electrons (2s^22p^4) and tends to gain two electrons to achieve a stable octet configuration. The removal of an electron from oxygen requires more energy compared to neon due to the increased nuclear charge and decreased atomic size.
3. Sulfur (Print)
Sulfur, with the atomic number 16, has a higher ionization energy than oxygen. Sulfur has six valence electrons (3s^23p^4) and shares similarities with oxygen in terms of electron configuration and chemical behavior. However, sulfur has a larger atomic size and lower nuclear charge than oxygen, resulting in a slightly lower ionization energy.
4. Aluminum (Print)
Aluminum, with the atomic number 13, has the highest ionization energy among the given elements. Aluminum has three valence electrons (3s^23p^1) and tends to lose these electrons to achieve a stable configuration. The removal of an electron from aluminum requires more energy compared to neon, oxygen, and sulfur due to its smaller atomic size and higher nuclear charge.