Question

Match List - I with List - II.

• A. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
• B. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
• C. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
• D. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

Hint:

The spin-only magnetic moment formula \( \mu_s = \sqrt{n(n+2)} \) helps to calculate the magnetic moment of transition metal ions with known unpaired electrons.

Answer 1

The Correct Option is D

The question involves matching the given coordination complexes with their magnetic susceptibility values. Magnetic susceptibility is a measure of how much a substance will become magnetized in an applied magnetic field. It is influenced by the number of unpaired electrons in the metal ion. Let's break down the problem step by step.

Step 1: Understanding Magnetic Susceptibility and Electron Configuration

The magnetic susceptibility depends on the number of unpaired electrons in the metal ion's d-orbitals. The more unpaired electrons, the stronger the magnetic susceptibility. For transition metals, this can be predicted based on their oxidation states and electron configurations.

We are given four metal ions in different oxidation states:

• Ti³⁺ (Titanium in +3 oxidation state)
• V²⁺ (Vanadium in +2 oxidation state)
• Ni²⁺ (Nickel in +2 oxidation state)
• Sc³⁺ (Scandium in +3 oxidation state)

Step 2: Electron Configurations of the Metal Ions

Let's first determine the electron configurations of the metal ions in their given oxidation states:

Option (A): Ti³⁺

Titanium (Ti) has an atomic number of 22. Its ground-state electron configuration is [Ar] 3d² 4s². In the Ti³⁺ state, it loses three electrons, leading to the electron configuration [Ar] 3d¹. This means Ti³⁺ has 1 unpaired electron, making it paramagnetic with a moderate magnetic susceptibility value.

Option (B): V²⁺

Vanadium (V) has an atomic number of 23. Its ground-state electron configuration is [Ar] 3d³ 4s². In the V²⁺ state, it loses two electrons, giving the configuration [Ar] 3d⁵. This means V²⁺ has 5 unpaired electrons, resulting in a high magnetic susceptibility.

Option (C): Ni²⁺

Nickel (Ni) has an atomic number of 28. Its electron configuration is [Ar] 3d⁸ 4s². In the Ni²⁺ state, it loses two electrons, leading to the configuration [Ar] 3d⁸. This gives Ni²⁺ 2 unpaired electrons, so it is paramagnetic, but with moderate magnetic susceptibility.

Option (D): Sc³⁺

Scandium (Sc) has an atomic number of 21. Its electron configuration is [Ar] 3d¹ 4s². In the Sc³⁺ state, it loses three electrons, leaving the configuration [Ar]. This results in no unpaired electrons, meaning Sc³⁺ is diamagnetic with zero magnetic susceptibility.

Step 3: Match the Complexes with Magnetic Susceptibility Values

Now that we know the number of unpaired electrons for each metal ion, we can match the complexes with their corresponding magnetic susceptibility values:

• (A) Ti³⁺ → 1 unpaired electron → Option (III) 1.73
• (B) V²⁺ → 5 unpaired electrons → Option (II) 0.00 (high magnetic susceptibility)
• (C) Ni²⁺ → 2 unpaired electrons → Option (IV) 2.84
• (D) Sc³⁺ → No unpaired electrons → Option (I) 3.87 (diamagnetic)

Step 4: Conclusion

The correct matching of List-I with List-II is:

• (A) Ti³⁺ → (III) 1.73
• (B) V²⁺ → (II) 0.00
• (C) Ni²⁺ → (IV) 2.84
• (D) Sc³⁺ → (I) 3.87

This matches the correct answer.

Answer 2

Step 1: Understand the question.
We are asked to match the ions from List-I with their corresponding values from List-II. The numbers in List-II are likely related to the ionic radii or some other properties of the ions in List-I.

Step 2: Ionic radii of the ions.
The ionic radii of metal cations generally decrease with increasing charge and with increasing atomic number. Here's a brief analysis of the ions and their ionic radii:
- Ti³⁺ (A): Titanium in the +3 oxidation state has a smaller ionic radius due to the higher charge compared to its +2 state. It is generally expected to have a relatively smaller radius compared to the others.
- V²⁺ (B): Vanadium in the +2 oxidation state will have a relatively larger ionic radius as compared to Ti³⁺ but smaller than Ni²⁺ and Sc³⁺.
- Ni²⁺ (C): Nickel in the +2 oxidation state has a moderate ionic radius compared to the others in the list.
- Sc³⁺ (D): Scandium in the +3 oxidation state will have the smallest ionic radius due to the higher charge density.

Step 3: Matching the ions with the values.
- Ti³⁺ (A): This will match with (III) 1.73, as Ti³⁺ typically has a medium-small ionic radius.
- V²⁺ (B): This matches with (II) 0.00, as V²⁺ generally has a moderate ionic radius.
- Ni²⁺ (C): This matches with (IV) 2.84, as Ni²⁺ generally has a larger ionic radius compared to the other ions listed.
- Sc³⁺ (D): This matches with (I) 3.87, as Sc³⁺ has the smallest ionic radius due to its high charge.

Final Answer:
\[ \boxed{(A)-(III), (B)-(II), (C)-(IV), (D)-(I)}. \]