Select the correct option, among Sc(III) , Ti(IV), Pd(II) and Cu(II) ions

Options

A. all are paramagnetic
B. all are diamagnetic
C. Sc(III), Ti(IV) are paramagnetic and Pd(III), Cu(II) are diamagnetic
D. Sc(III), Ti(IV) are diamagnetic and Pd(II), Cu(II) are paramagnetic.

Answer

To determine the magnetic nature of the given ions (Sc(III), Ti(IV), Pd(II), and Cu(II)), we need to analyze their electronic configurations and the presence of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers:**
- Scandium (Sc) has an atomic number of 21.
- Titanium (Ti) has an atomic number of 22.
- Palladium (Pd) has an atomic number of 46.
- Copper (Cu) has an atomic number of 29.

2. **Determine the Electronic Configurations:**
- Sc(III): The electronic configuration of Sc is [Ar] 3d¹ 4s². In the +3 oxidation state, it loses 3 electrons (2 from 4s and 1 from 3d), resulting in [Ar] 3d⁰.
- Ti(IV): The electronic configuration of Ti is [Ar] 3d² 4s². In the +4 oxidation state, it loses 4 electrons (2 from 4s and 2 from 3d), resulting in [Ar] 3d⁰.
- Pd(II): The electronic configuration of Pd is [Kr] 4d¹⁰ 5s⁰. In the +2 oxidation state, it loses 2 electrons (from 4d), resulting in [Kr] 4d⁸ 5s⁰.
- Cu(II): The electronic configuration of Cu is [Ar] 3d¹⁰ 4s¹. In the +2 oxidation state, it loses 2 electrons (1 from 4s and 1 from 3d), resulting in [Ar] 3d⁹.

3. **Count the Number of Unpaired Electrons:**
- Sc(III): [Ar] 3d⁰ → 0 unpaired electrons (diamagnetic).
- Ti(IV): [Ar] 3d⁰ → 0 unpaired electrons (diamagnetic).
- Pd(II): [Kr] 4d⁸ 5s⁰ → 2 unpaired electrons (paramagnetic).
- Cu(II): [Ar] 3d⁹ → 1 unpaired electron (paramagnetic).

4. **Determine the Magnetic Nature:**
- Sc(III): Diamagnetic (no unpaired electrons).
- Ti(IV): Diamagnetic (no unpaired electrons).
- Pd(II): Paramagnetic (2 unpaired electrons).
- Cu(II): Paramagnetic (1 unpaired electron).

### Conclusion:
The ions that are paramagnetic (having unpaired electrons) among the given options are **Pd(II)** and **Cu(II)**.

### Final Answer:
The correct options are **Pd(II)** and **Cu(II)**.

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