Question

Consider the hydrated ions of Ti²⁺, V²⁺, Ti³⁺, and Sc³⁺. The correct order of their spin-only magnetic moment is:
1. V²⁺ < Ti² + < Ti³⁺ < Sc³⁺
2. Sc³⁺ < Ti³⁺ < Ti²⁺ < V²⁺
3. Ti³⁺ < Ti²⁺ < Sc³⁺ < V²⁺
4. Sc³⁺ < Ti³⁺ < V²⁺ < Ti²⁺

Answer 1

Correct Answer - Option 2 : Sc³⁺ < Ti³⁺ < Ti²⁺ < V²⁺

Concept:

We know that, the formula used to calculate the spin-only magnetic moment can be written based on the number of unpaired electrons, n. Since for each unpaired electron, n=1 then the formula is clearly related and the answer obtained must be identical.

\(\mu = \sqrt {n\left( {n + 2} \right)} Bm\)

Where n = number of unpaired electrons.

and Bm = Bohr magneton

The electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals. Electronic configuration describes each electron is moving independently in an orbital, in an average field created by all other orbitals.

Electronic configurations of the given transition metal ions are:

Sc³⁺ (Z = 21) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s²

Where Z = atomic number

Ti²⁺ (Z = 22) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²

Ti³⁺ (Z = 22) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²

V²⁺ (Z = 23) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

Calculation:

Since, magnetic moment is directly proportional to the number of unpaired electrons.

Sc³⁺ → 3d⁰

Here, n = 0 \(\Rightarrow \mu = \sqrt {0\left( {0 + 2} \right)} = \;0\;Bm\)

Ti²⁺ → 3d²

Here, n = 2 \(\Rightarrow \mu = \sqrt {2\left( {2 + 2} \right)} = \sqrt {2 \times 4} = \sqrt 8 \;Bm\)

Ti³⁺ → 3d¹

Here, n = 1 \(\Rightarrow \mu = \sqrt {1\left( {1 + 2} \right)} = \sqrt {1 \times 3} = \sqrt 3 \;Bm\)

V²⁺ → 3d³

Here, n = 3 \(\Rightarrow \mu = \sqrt {3\left( {3 + 2} \right)} = \sqrt {3 \times 5} = \sqrt {15} \;Bm\)

The correct increasing order of magnetic moment is Sc³⁺ < Ti³⁺ < Ti²⁺ < V²⁺ because they have 0, 1, 2 and 3 unpaired electrons respectively.