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.