‘Spin only’ magnetic moment is same for which of the following ions?

A. Ti³⁺                B. Cr²⁺

C. Mn²⁺             D. Fe²⁺

E. Sc³⁺

Choose the most appropriate answer from the options given below.

To find which of the given ions have the same 'spin-only' magnetic moment, we need to determine the number of unpaired electrons ($n$) in each ion. The 'spin-only' magnetic moment (${\mu }_s$) is given by the formula:

${\mu }_s=\sqrt{n(n+2)}B.M.$

where $n$ is the number of unpaired electrons. Therefore, ions with the same number of unpaired electrons will have the same 'spin-only' magnetic moment.

Let's determine the electronic configuration and the number of unpaired electrons for each of the given transition metal ions (using their atomic numbers: Sc = 21, Ti = 22, Cr = 24, Mn = 25, Fe = 26):

A. Ti³⁺:
Atomic number of Ti = 22. Neutral Ti electronic configuration: [Ar] 3d² 4s².
For Ti³⁺ (removal of 3 electrons: two from 4s and one from 3d): [Ar] 3d¹.
Number of unpaired electrons ($n$) = 1.

B. Cr²⁺:
Atomic number of Cr = 24. Neutral Cr electronic configuration: [Ar] 3d⁵ 4s¹.
For Cr²⁺ (removal of 2 electrons: one from 4s and one from 3d): [Ar] 3d⁴.
Number of unpaired electrons ($n$) = 4.

C. Mn²⁺:
Atomic number of Mn = 25. Neutral Mn electronic configuration: [Ar] 3d⁵ 4s².
For Mn²⁺ (removal of 2 electrons from 4s): [Ar] 3d⁵.
Number of unpaired electrons ($n$) = 5.

D. Fe²⁺:
Atomic number of Fe = 26. Neutral Fe electronic configuration: [Ar] 3d⁶ 4s².
For Fe²⁺ (removal of 2 electrons from 4s): [Ar] 3d⁶.
According to Hund's rule, the 6 electrons in the 3d subshell are distributed as: 5 spin-up (singly occupied orbitals) and 1 spin-down (paired orbital).
Number of unpaired electrons ($n$) = 4.

E. Sc³⁺:
Atomic number of Sc = 21. Neutral Sc electronic configuration: [Ar] 3d¹ 4s².
For Sc³⁺ (removal of 3 electrons: two from 4s and one from 3d): [Ar] 3d⁰.
Number of unpaired electrons ($n$) = 0.

Comparing the results:
Both Cr²⁺ (B) and Fe²⁺ (D) have exactly 4 unpaired electrons ($n=4$).
Thus, their spin-only magnetic moments are identical:

${\mu }_s=\sqrt{4(4+2)}=\sqrt{24}\approx 4.90B.M.$

Therefore, the 'spin-only' magnetic moment is the same for B and D only.