Three atoms have the following electron configurations: (a) 1s2 2s2 2p6 3s2 3p1 (b) 1s2 2s2 2p6 3s2 3p5 (c) 1s2 2s2 2p6 3s2 3p6 4s1. Which of the three has the largest Ei1? Which has the smallest Ei4?

Verified step by step guidance

1. Step 1: Identify the elements based on their electron configurations. For (a) 1s^2 2s^2 2p^6 3s^2 3p^1, the element is Aluminum (Al). For (b) 1s^2 2s^2 2p^6 3s^2 3p^5, the element is Chlorine (Cl). For (c) 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1, the element is Potassium (K).
2. Step 2: Understand the concept of first ionization energy (Ei1), which is the energy required to remove the outermost electron from a neutral atom in the gaseous state. Generally, ionization energy increases across a period and decreases down a group in the periodic table.
3. Step 3: Compare the first ionization energies (Ei1) of the identified elements. Chlorine (Cl) is located further to the right in the periodic table compared to Aluminum (Al) and Potassium (K), indicating it has the highest Ei1 due to increased nuclear charge and smaller atomic radius.
4. Step 4: Understand the concept of fourth ionization energy (Ei4), which is the energy required to remove the fourth electron after the first three have been removed. This is generally higher for elements with a stable electron configuration after the third ionization.
5. Step 5: Compare the fourth ionization energies (Ei4) of the identified elements. Potassium (K) has a single electron in the 4s orbital, and after the first electron is removed, it achieves a noble gas configuration, making subsequent ionizations more difficult. Therefore, Potassium (K) is likely to have the smallest Ei4.