Which one of the following sets of ions represents a collection of isoelectronic species?

Options

A. `K^(+) ,Cl^(-) ,Ca^(2 +) , Sc^(3+)`
B. `Ba^(2+) , Sr^(2 +) ,K^(+) ,S^(2-)`
C. `N^(3-) ,O^(2-) , F^(-) , S^(2-)`
D. `Li^(+) ,Na^(+) ,Mg ^(2+) ,Ca^(2+)`

Explanation

To determine which set of ions represents a collection of isoelectronic species, we need to identify ions that have the same number of electrons. Let's analyze each option step by step.

### Step 1: Understand Isoelectronic Species
Isoelectronic species are atoms or ions that have the same number of electrons. This can occur with different elements or ions.

### Step 2: Analyze Each Option

#### Option 1: K⁺, Cl⁻, Ca²⁺, Sc³⁺
1. **K⁺ (Potassium ion)**:
- Atomic number of K = 19
- K loses one electron:
- Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ (18 electrons)

2. **Cl⁻ (Chloride ion)**:
- Atomic number of Cl = 17
- Cl gains one electron:
- Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ (18 electrons)

3. **Ca²⁺ (Calcium ion)**:
- Atomic number of Ca = 20
- Ca loses two electrons:
- Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ (18 electrons)

4. **Sc³⁺ (Scandium ion)**:
- Atomic number of Sc = 21
- Sc loses three electrons:
- Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ (18 electrons)

**Conclusion for Option 1**: All four ions have the same electron configuration (18 electrons), thus they are isoelectronic.

#### Option 2: Ba²⁺, Sr²⁺, K⁺, S²⁻
1. **Ba²⁺ (Barium ion)**:
- Atomic number of Ba = 56
- Ba loses two electrons:
- Electron configuration: [Xe] 6s² (54 electrons)

2. **Sr²⁺ (Strontium ion)**:
- Atomic number of Sr = 38
- Sr loses two electrons:
- Electron configuration: [Kr] 5s² (36 electrons)

3. **K⁺ (Potassium ion)**:
- Already calculated: 18 electrons

4. **S²⁻ (Sulfide ion)**:
- Atomic number of S = 16
- S gains two electrons:
- Electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ (18 electrons)

**Conclusion for Option 2**: The ions do not have the same number of electrons, thus they are not isoelectronic.

#### Option 3: N³⁻, O²⁻, F⁻, S²⁻
1. **N³⁻ (Nitride ion)**:
- Atomic number of N = 7
- N gains three electrons:
- Electron configuration: 1s² 2s² 2p⁶ (10 electrons)

2. **O²⁻ (Oxide ion)**:
- Atomic number of O = 8
- O gains two electrons:
- Electron configuration: 1s² 2s² 2p⁶ (10 electrons)

3. **F⁻ (Fluoride ion)**:
- Atomic number of F = 9
- F gains one electron:
- Electron configuration: 1s² 2s² 2p⁶ (10 electrons)

4. **S²⁻ (Sulfide ion)**:
- Already calculated: 18 electrons

**Conclusion for Option 3**: N³⁻, O²⁻, and F⁻ are isoelectronic (10 electrons), but S²⁻ is not. Thus, this option is incorrect.

#### Option 4: Li⁺, Na⁺, Mg²⁺, Ca²⁺
1. **Li⁺ (Lithium ion)**:
- Atomic number of Li = 3
- Li loses one electron:
- Electron configuration: 1s² (2 electrons)

2. **Na⁺ (Sodium ion)**:
- Atomic number of Na = 11
- Na loses one electron:
- Electron configuration: 1s² 2s² 2p⁶ (10 electrons)

3. **Mg²⁺ (Magnesium ion)**:
- Atomic number of Mg = 12
- Mg loses two electrons:
- Electron configuration: 1s² 2s² 2p⁶ (10 electrons)

4. **Ca²⁺ (Calcium ion)**:
- Already calculated: 18 electrons

**Conclusion for Option 4**: The ions do not have the same number of electrons, thus they are not isoelectronic.

### Final Answer
The correct option that represents a collection of isoelectronic species is **Option 1: K⁺, Cl⁻, Ca²⁺, Sc³⁺**.