To determine which of the following isoelectronic ions has the lowest ionization energy, we will analyze the given ions: K⁺, Cl⁻, Ca²⁺, and S²⁻.
### Step 1: Identify the number of electrons in each ion
- **K⁺ (Potassium ion)**: Potassium (K) has an atomic number of 19, meaning it has 19 electrons. K⁺ has lost one electron, so it has 18 electrons.
- **Cl⁻ (Chloride ion)**: Chlorine (Cl) has an atomic number of 17, meaning it has 17 electrons. Cl⁻ has gained one electron, resulting in 18 electrons.
- **Ca²⁺ (Calcium ion)**: Calcium (Ca) has an atomic number of 20, meaning it has 20 electrons. Ca²⁺ has lost two electrons, leaving it with 18 electrons.
- **S²⁻ (Sulfide ion)**: Sulfur (S) has an atomic number of 16, meaning it has 16 electrons. S²⁻ has gained two electrons, resulting in 18 electrons.
**Conclusion**: All four ions (K⁺, Cl⁻, Ca²⁺, S²⁻) are isoelectronic, as they all contain 18 electrons.
### Step 2: Understand the concept of ionization energy
Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. It is influenced by two main factors:
- **Nuclear charge**: The more protons in the nucleus, the higher the nuclear charge, which holds the electrons more tightly.
- **Size of the ion**: A larger ion has its outermost electrons further from the nucleus, meaning they are held less tightly and can be removed more easily.
### Step 3: Compare the sizes of the isoelectronic ions
The size of isoelectronic ions can be compared based on their nuclear charge:
- **S²⁻**: Has the least nuclear charge (16 protons) and the most electrons (18), resulting in the largest size.
- **Cl⁻**: Has 17 protons, so it is smaller than S²⁻.
- **K⁺**: Has 19 protons, making it smaller than Cl⁻.
- **Ca²⁺**: Has the highest nuclear charge (20 protons), making it the smallest among these ions.
**Order of size (from largest to smallest)**: S²⁻ > Cl⁻ > K⁺ > Ca²⁺
### Step 4: Determine the ionization energy trend
Since ionization energy is inversely proportional to the size of the ion, the larger the ion, the lower the ionization energy. Therefore, the ion with the largest size (and thus the lowest ionization energy) is S²⁻.
### Final Answer
Among the given isoelectronic ions, **S²⁻ has the lowest ionization energy**.
