To determine which ion was formed by providing the second ionization energy to remove an electron, we need to consider the typical ionization process for each element listed.
Step-by-step solution:
1. Calcium (Ca):
- Calcium is an alkaline earth metal located in Group 2 of the periodic table.
- It typically forms a +2 ion (Ca²⁺) by losing two electrons.
- The first ionization energy removes one electron, resulting in Ca⁺.
- The second ionization energy removes a second electron from Ca⁺, resulting in Ca²⁺.
- Therefore, the ion formed by providing the second ionization energy to remove an electron from calcium is Ca²⁺.
2. Nitrogen (N):
- Nitrogen is a nonmetal located in Group 15 of the periodic table.
- It typically gains electrons to achieve a stable electron configuration, forming a -3 ion (N³⁻).
- This process doesn't involve removing electrons but rather adding them, so second ionization energy is irrelevant here.
3. Iron (Fe):
- Iron is a transition metal that can lose multiple electrons to form different ions.
- Common oxidation states for iron are +2 (Fe²⁺) and +3 (Fe³⁺).
- It requires significant energy to remove electrons sequentially, but the second ionization energy specifically refers to going from Fe⁺ to Fe²⁺.
- However, Fe³⁺ refers to the third ionization, not the second.
4. Sulfur (S):
- Sulfur is a nonmetal located in Group 16 of the periodic table.
- It typically gains two electrons to form a -2 ion (S²⁻).
- Similar to nitrogen, this process involves gaining electrons, not removing them.
Given the context and the requirements for second ionization energy, the only suitable candidate is Calcium, forming the ion Ca²⁺ after the second electron is removed.
So, the ion formed by providing the second ionization energy to remove an electron is:
[tex]\[ \boxed{Ca^{2+}} \][/tex]
