The correct order of decreasing ionic radii among the following isoelectronic species is

Options

A. ` Ca^(2+) gt K^(+) gt S2^(–) gt Cl^(– )`
B. `Cl^(–) gt S^(2–) gt Ca^(2+) gt K^(+ )`
C. ` S^(2–) gt Cl^(–) gt K^(+) gt Ca^(2+ )`
D. `S^(2–) gt Cl^(–) gt K^(+) gt Ca^(2+ )`

Answer

To determine the correct order of decreasing ionic radii among the given isoelectronic species, we need to follow these steps:

### Step 1: Identify the Isoelectronic Species
Isoelectronic species are atoms or ions that have the same number of electrons. In this case, we are given four species: sulfur (S), chlorine (Cl), potassium (K), and calcium (Ca).

### Step 2: Determine the Number of Electrons in Each Species
- **Sulfur (S)**: Atomic number = 16, with a charge of -2.
- Electrons = 16 + 2 = 18
- **Chlorine (Cl)**: Atomic number = 17, with a charge of -1.
- Electrons = 17 + 1 = 18
- **Potassium (K)**: Atomic number = 19, with a charge of +1.
- Electrons = 19 - 1 = 18
- **Calcium (Ca)**: Atomic number = 20, with a charge of +2.
- Electrons = 20 - 2 = 18

All four species have 18 electrons, confirming they are isoelectronic.

### Step 3: Analyze the Ionic Charges
The ionic radius is influenced by the charge of the ion:
- Anions (negatively charged ions) have larger radii than cations (positively charged ions).
- The more negative the charge, the larger the ionic radius.
- Conversely, the more positive the charge, the smaller the ionic radius.

### Step 4: Order the Species Based on Charge
- **Sulfur (S²⁻)**: -2 charge (largest radius)
- **Chlorine (Cl⁻)**: -1 charge
- **Potassium (K⁺)**: +1 charge
- **Calcium (Ca²⁺)**: +2 charge (smallest radius)

### Step 5: Write the Final Order
Based on the analysis, the order of decreasing ionic radii is:
**S²⁻ > Cl⁻ > K⁺ > Ca²⁺**

### Final Answer
The correct order of decreasing ionic radii among the given isoelectronic species is:
**S²⁻ > Cl⁻ > K⁺ > Ca²⁺**

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