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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C) (D) (A) (B)
The spin-only magnetic moment depends only on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
In the free ion, the 3d electrons occupy separate orbitals before pairing (Hund’s rule), so up to five d electrons are all unpaired. Find the 3d count from the atomic number, removing the 4s electrons first:
(A) \( \ce{Mn^3+} \): Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing three electrons gives \( \mathrm{3d^{4}} \), n = 4, \( \mu = \sqrt{24} = 4.90 \) BM.
(B) \( \ce{Fe^3+} \): Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing three electrons gives \( \mathrm{3d^{5}} \), n = 5, \( \mu = \sqrt{35} = 5.92 \) BM.
(C) \( \ce{Sc^3+} \): Sc (Z = 21) is \( \mathrm{[Ar]\,3d^{1}\,4s^{2}} \); losing three electrons gives \( \mathrm{3d^{0}} \), n = 0, \( \mu = 0 \).
(D) \( \ce{Cr^3+} \): Cr (Z = 24) is \( \mathrm{[Ar]\,3d^{5}\,4s^{1}} \); losing three electrons gives \( \mathrm{3d^{3}} \), n = 3, \( \mu = \sqrt{15} = 3.87 \) BM.
In increasing order of magnetic moment:
\( \displaystyle \ce{Sc^3+}\,(0) \lt \ce{Cr^3+}\,(3.87) \lt \ce{Mn^3+}\,(4.90) \lt \ce{Fe^3+}\,(5.92) \)
Shortcut: The magnetic moment rises with n, so there is no need to work out the square roots: compare n = 0, 3, 4, 5.