Among the following which species is/are paramagnetic
(i) `Sr^(2+)` (ii) `Fe^(3+)` (iii) `Co^(2+)` (iv) `S^(2-)` (v) `Pb^(2+)`

Options

A. i, iv, v
B. i, ii, iii
C. ii, iii
D. iv, v

Answer

To determine which of the given species are paramagnetic, we need to analyze their electronic configurations and identify the presence of unpaired electrons. Paramagnetic species have one or more unpaired electrons, while diamagnetic species have all electrons paired.

Let's evaluate each species step by step:

### Step 1: Analyze `Sr^(2+)`
- **Atomic Number of Strontium (Sr)**: 38
- **Electronic Configuration of Sr**: `[Kr] 5s^2`
- **For `Sr^(2+)`**: It loses 2 electrons from the outermost shell.
- **Electronic Configuration of `Sr^(2+)`**: `[Kr]` (which is a noble gas configuration)
- **Unpaired Electrons**: 0
- **Conclusion**: `Sr^(2+)` is **diamagnetic**.

### Step 2: Analyze `Fe^(3+)`
- **Atomic Number of Iron (Fe)**: 26
- **Electronic Configuration of Fe**: `[Ar] 4s^2 3d^6`
- **For `Fe^(3+)`**: It loses 3 electrons (2 from 4s and 1 from 3d).
- **Electronic Configuration of `Fe^(3+)`**: `[Ar] 3d^5`
- **Unpaired Electrons**: 5 (in 3d subshell)
- **Conclusion**: `Fe^(3+)` is **paramagnetic**.

### Step 3: Analyze `Co^(2+)`
- **Atomic Number of Cobalt (Co)**: 27
- **Electronic Configuration of Co**: `[Ar] 4s^2 3d^7`
- **For `Co^(2+)`**: It loses 2 electrons from 4s.
- **Electronic Configuration of `Co^(2+)`**: `[Ar] 3d^7`
- **Unpaired Electrons**: 3 (in 3d subshell)
- **Conclusion**: `Co^(2+)` is **paramagnetic**.

### Step 4: Analyze `S^(2-)`
- **Atomic Number of Sulfur (S)**: 16
- **Electronic Configuration of S**: `[Ne] 3s^2 3p^4`
- **For `S^(2-)`**: It gains 2 electrons.
- **Electronic Configuration of `S^(2-)`**: `[Ne] 3s^2 3p^6` (same as Argon)
- **Unpaired Electrons**: 0
- **Conclusion**: `S^(2-)` is **diamagnetic**.

### Step 5: Analyze `Pb^(2+)`
- **Atomic Number of Lead (Pb)**: 82
- **Electronic Configuration of Pb**: `[Xe] 6s^2 4f^{14} 5d^{10} 6p^2`
- **For `Pb^(2+)`**: It loses 2 electrons from the outermost shell.
- **Electronic Configuration of `Pb^(2+)`**: `[Xe] 6s^2 4f^{14} 5d^{10} 6p^0`
- **Unpaired Electrons**: 0
- **Conclusion**: `Pb^(2+)` is **diamagnetic**.

### Final Conclusion:
The species that are **paramagnetic** among the given options are:
- `Fe^(3+)`
- `Co^(2+)`

### Summary of Results:
- `Sr^(2+)`: Diamagnetic
- `Fe^(3+)`: Paramagnetic
- `Co^(2+)`: Paramagnetic
- `S^(2-)`: Diamagnetic
- `Pb^(2+)`: Diamagnetic

### Answer:
The correct answer is `Fe^(3+)` and `Co^(2+)`.

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