Which of the following isoelectronic ions has the lowest ionization energy?

Options

A. `K^(+)`
B. `Cl^(-)`
C. `Ca^(2+)`
D. `S^(2-)`

Answer

To determine which isoelectronic ion has the lowest ionization energy, we can follow these steps:

### Step 1: Understand Isoelectronic Species
Isoelectronic ions are ions that have the same number of electrons. In this case, we have the ions K⁺, Cl⁻, Ca²⁺, and S²⁻. All these ions have 18 electrons.

### Step 2: Determine the Number of Protons
Next, we need to identify the number of protons in each of these ions:
- K⁺ (Potassium) has 19 protons.
- Cl⁻ (Chlorine) has 17 protons.
- Ca²⁺ (Calcium) has 20 protons.
- S²⁻ (Sulfur) has 16 protons.

### Step 3: Analyze Effective Nuclear Charge
The effective nuclear charge (Z_eff) experienced by the electrons is influenced by the number of protons. The more protons there are, the greater the effective nuclear charge, which means the electrons are held more tightly to the nucleus.

### Step 4: Compare Sizes of Ions
Since all the ions have the same number of electrons (18), the ion with the least number of protons will have the largest size due to the weaker attraction between the nucleus and the electrons.

- K⁺: 19 protons → smaller size
- Cl⁻: 17 protons → larger size than K⁺
- Ca²⁺: 20 protons → smallest size
- S²⁻: 16 protons → largest size

### Step 5: Identify Ion with Lowest Ionization Energy
The ion with the largest size (S²⁻) will have the lowest ionization energy because it has the weakest attraction between the nucleus and the outermost electrons. Therefore, it will require the least amount of energy to remove an electron.

### Conclusion
The isoelectronic ion with the lowest ionization energy is **S²⁻**.

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