The pair of ions which do not have diamag netic nature

Options

A. `Cu^(1+) and Zn^(2+)`
B. `Sc^(3+) and Ti^(4+)`
C. `Ca^(2+) and Zn^(2+)`
D. `V^(2+) and Fe^(2+)`

Answer

To determine the pair of ions that do not have diamagnetic nature, we need to analyze the electronic configurations of the given ions and check for the presence of unpaired electrons. Here’s the step-by-step solution:

### Step 1: Understand Diamagnetic and Paramagnetic Nature
- **Diamagnetic**: Substances that do not have unpaired electrons and are repelled by a magnetic field.
- **Paramagnetic**: Substances that have unpaired electrons and are attracted by a magnetic field.

### Step 2: Analyze Each Ion
We will analyze the given ions for their electronic configurations and check for unpaired electrons.

#### Pair 1: Cuprous (Cu⁺) and Zinc (Zn²⁺)
- **Cuprous (Cu⁺)**:
- Atomic number of Cu = 29
- Electronic configuration: [Ar] 3d¹⁰ 4s¹
- For Cu⁺: [Ar] 3d¹⁰ 4s⁰ (no unpaired electrons, **diamagnetic**)

- **Zinc (Zn²⁺)**:
- Atomic number of Zn = 30
- Electronic configuration: [Ar] 3d¹⁰ 4s²
- For Zn²⁺: [Ar] 3d¹⁰ 4s⁰ (no unpaired electrons, **diamagnetic**)

**Conclusion**: Both Cu⁺ and Zn²⁺ are diamagnetic.

#### Pair 2: Scandium (Sc³⁺) and Titanium (Ti⁴⁺)
- **Scandium (Sc³⁺)**:
- Atomic number of Sc = 21
- Electronic configuration: [Ar] 3d¹ 4s²
- For Sc³⁺: [Ar] 3d⁰ 4s⁰ (no unpaired electrons, **diamagnetic**)

- **Titanium (Ti⁴⁺)**:
- Atomic number of Ti = 22
- Electronic configuration: [Ar] 3d² 4s²
- For Ti⁴⁺: [Ar] 3d⁰ 4s⁰ (no unpaired electrons, **diamagnetic**)

**Conclusion**: Both Sc³⁺ and Ti⁴⁺ are diamagnetic.

#### Pair 3: Calcium (Ca²⁺) and Zinc (Zn²⁺)
- **Calcium (Ca²⁺)**:
- Atomic number of Ca = 20
- Electronic configuration: [Ar] 4s²
- For Ca²⁺: [Ar] 4s⁰ (no unpaired electrons, **diamagnetic**)

- **Zinc (Zn²⁺)**: (already analyzed above)
- Zn²⁺ is also diamagnetic.

**Conclusion**: Both Ca²⁺ and Zn²⁺ are diamagnetic.

#### Pair 4: Vanadium (V²⁺) and Iron (Fe²⁺)
- **Vanadium (V²⁺)**:
- Atomic number of V = 23
- Electronic configuration: [Ar] 3d³ 4s²
- For V²⁺: [Ar] 3d³ 4s⁰ (3 unpaired electrons, **paramagnetic**)

- **Iron (Fe²⁺)**:
- Atomic number of Fe = 26
- Electronic configuration: [Ar] 3d⁶ 4s²
- For Fe²⁺: [Ar] 3d⁶ 4s⁰ (4 unpaired electrons, **paramagnetic**)

**Conclusion**: Both V²⁺ and Fe²⁺ are paramagnetic.

### Final Answer
The pair of ions that do not have diamagnetic nature is **Vanadium (V²⁺) and Iron (Fe²⁺)**.