Which of the following has the smallest radius Cl- S2- Ca2+ K+ Ar
The Correct Answer and Explanation is :
To determine which ion has the smallest radius, we need to understand how ion size is influenced by several factors: nuclear charge, electron-electron repulsion, and electron configuration.
The elements and ions in the given list are:
- Cl⁻ (Chloride ion): Chlorine has 17 protons in the nucleus and gains one electron to form Cl⁻, giving it a total of 18 electrons.
- S²⁻ (Sulfide ion): Sulfur has 16 protons and gains two electrons to form S²⁻, giving it a total of 18 electrons.
- Ca²⁺ (Calcium ion): Calcium has 20 protons and loses two electrons to form Ca²⁺, giving it a total of 18 electrons.
- K⁺ (Potassium ion): Potassium has 19 protons and loses one electron to form K⁺, giving it a total of 18 electrons.
- Ar (Argon atom): Argon has 18 protons and 18 electrons, making it a neutral atom.
Since all of these species (except Argon) have the same number of electrons (18), we can compare their ionic radii based on the effective nuclear charge (Z_eff), which is the net positive charge experienced by an electron in an atom or ion. The higher the effective nuclear charge, the more strongly electrons are attracted to the nucleus, and the smaller the ion becomes.
Analyzing the ions:
- Cl⁻: It has 17 protons and 18 electrons. The addition of one electron causes an increase in electron-electron repulsion, but the effective nuclear charge is relatively lower than the other ions.
- S²⁻: It has 16 protons and 18 electrons. The greater electron-electron repulsion (due to the extra electron compared to Cl⁻) causes a larger radius than Cl⁻.
- Ca²⁺: It has 20 protons and 18 electrons. With more protons, the effective nuclear charge is stronger, pulling the electrons closer to the nucleus, resulting in a smaller radius than both Cl⁻ and S²⁻.
- K⁺: It has 19 protons and 18 electrons. Similar to Ca²⁺, but with slightly fewer protons, the effective nuclear charge is still strong, but not as much as in Ca²⁺, leading to a slightly larger radius than Ca²⁺.
- Ar: Being neutral with 18 protons and 18 electrons, the electron-electron repulsion is balanced by the effective nuclear charge, resulting in a smaller radius than Cl⁻ and S²⁻ but larger than Ca²⁺.
Conclusion:
Ca²⁺ has the smallest radius because it has the highest effective nuclear charge, which pulls the electrons closest to the nucleus.
Now, I’ll generate an image to visually compare their radii.
Here’s the diagram comparing the ionic radii of Cl⁻, S²⁻, Ca²⁺, K⁺, and Ar. You can see how their sizes relate to their nuclear charge and electron-electron repulsion. As discussed, Ca²⁺ has the smallest radius among these species.
