Pick up the correct statement about the following pairs
I : Ar and `S^(2-)`, II: Fe and `Ni^(2+)`
(At. No. Ar = 18, S = 16, Fe = 26, Ni = 28)

Options

A. Both of these pairs contain isoelectronic species
B. Only pair (I) is isoelectronic
C. Only pair (II) is isoelectronic
D. None of the pair is isoelectronic.

Explanation

To solve the question, we need to determine whether the given pairs of elements are isoelectronic. Isoelectronic species are atoms or ions that have the same number of electrons.

### Step-by-Step Solution:

**Step 1: Analyze the first pair (Ar and S²⁻)**

- **Argon (Ar)**: The atomic number of Argon is 18. This means that in its neutral state, Argon has 18 electrons.

- **Sulfur (S)**: The atomic number of Sulfur is 16. The notation S²⁻ indicates that sulfur has gained 2 electrons. Therefore, the total number of electrons in S²⁻ is:
\[ 16 \text{ (atomic number)} + 2 \text{ (gained electrons)} = 18 \text{ electrons} \]

**Conclusion for the first pair**: Both Argon and S²⁻ have 18 electrons, making them isoelectronic.

---

**Step 2: Analyze the second pair (Fe and Ni²⁺)**

- **Iron (Fe)**: The atomic number of Iron is 26. In its neutral state, Iron has 26 electrons.

- **Nickel (Ni)**: The atomic number of Nickel is 28. The notation Ni²⁺ indicates that Nickel has lost 2 electrons. Therefore, the total number of electrons in Ni²⁺ is:
\[ 28 \text{ (atomic number)} - 2 \text{ (lost electrons)} = 26 \text{ electrons} \]

**Conclusion for the second pair**: Both Iron and Ni²⁺ have 26 electrons, making them isoelectronic.

---

### Final Conclusion:
Both pairs (Ar and S²⁻, Fe and Ni²⁺) are isoelectronic because they contain the same number of electrons.