In which of the following species both cation and Anion have same number of electrons

Options

A. `K^+, F^-`
B. `S^(2-), Na^+`
C. `Cu^(2+), I^-`
D. `AI^(3+), F^-`

Answer

To determine which of the given species has both the cation and anion with the same number of electrons, we will analyze each species step by step.

### Step 1: Identify the atomic numbers of the elements
- K (Potassium): Atomic number = 19
- S (Sulfur): Atomic number = 16
- Na (Sodium): Atomic number = 11
- Cu (Copper): Atomic number = 29
- I (Iodine): Atomic number = 53
- Al (Aluminum): Atomic number = 13
- F (Fluorine): Atomic number = 9

### Step 2: Calculate the number of electrons in each ion
1. **K⁺ (Potassium cation)**
- K has 19 electrons, loses 1 electron → 19 - 1 = 18 electrons.

2. **S⁻ (Sulfur anion)**
- S has 16 electrons, gains 1 electron → 16 + 1 = 17 electrons.

3. **S²⁻ (Sulfide anion)**
- S has 16 electrons, gains 2 electrons → 16 + 2 = 18 electrons.

4. **Na⁺ (Sodium cation)**
- Na has 11 electrons, loses 1 electron → 11 - 1 = 10 electrons.

5. **Cu²⁺ (Copper cation)**
- Cu has 29 electrons, loses 2 electrons → 29 - 2 = 27 electrons.

6. **I⁻ (Iodine anion)**
- I has 53 electrons, gains 1 electron → 53 + 1 = 54 electrons.

7. **Al³⁺ (Aluminum cation)**
- Al has 13 electrons, loses 3 electrons → 13 - 3 = 10 electrons.

8. **F⁻ (Fluorine anion)**
- F has 9 electrons, gains 1 electron → 9 + 1 = 10 electrons.

### Step 3: Compare the number of electrons in cations and anions
- K⁺: 18 electrons, S⁻: 17 electrons (not the same)
- S²⁻: 18 electrons, Na⁺: 10 electrons (not the same)
- Cu²⁺: 27 electrons, I⁻: 54 electrons (not the same)
- Al³⁺: 10 electrons, F⁻: 10 electrons (same)

### Conclusion
The species where both the cation and anion have the same number of electrons is **Al³⁺ and F⁻**, both having 10 electrons.

### Final Answer
The correct answer is **Al³⁺ and F⁻**.

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