Arrange the following in the increasing order of their spin-only magnetic moment values in free state

(A) $Mn^{3+}$
(B) $Fe^{3+}$
(C) $Sc^{3+}$
(D) $Cr^{3+}$

Choose the correct answer from the options given below:

(C) (D) (A) (B)

The correct answer is Option (3) → (C) (D) (A) (B)

Core Concept:

Spin-only magnetic moment depends on number of unpaired electrons:

μ = √[n(n+2)]

More unpaired electrons → higher magnetic moment

Step 1: Find Electronic Configuration

Sc³⁺ → Sc = 3d¹4s² → Sc³⁺ = 3d⁰ → 0 unpaired

Cr³⁺ → Cr = 3d⁵4s¹ → Cr³⁺ = 3d³ → 3 unpaired

Mn³⁺ → Mn = 3d⁵4s² → Mn³⁺ = 3d⁴ → 4 unpaired

Fe³⁺ → Fe = 3d⁶4s² → Fe³⁺ = 3d⁵ → 5 unpaired

Step 2: Compare Magnetic Moment

More unpaired electrons → higher μ

Sc³⁺ (0) < Cr³⁺ (3) < Mn³⁺ (4) < Fe³⁺ (5)

Option-wise Explanation

Option (C) Sc³⁺

Has no unpaired electrons, so it has the lowest magnetic moment.

Option (D) Cr³⁺

Has three unpaired electrons, giving moderate magnetic moment.

Option (A) Mn³⁺

Has four unpaired electrons, so magnetic moment is higher than Cr³⁺.

Option (B) Fe³⁺

Has five unpaired electrons, giving the highest magnetic moment.

Final Increasing Order:

(C) < (D) < (A) < (B)