Step 1: Understanding the Concept:
The 'spin-only' magnetic moment ($\mu$) depends solely on the number of unpaired electrons ($n$) in an ion.
The formula is: $\mu = \sqrt{n(n+2)}$ B.M. (Bohr Magnetons).
If two ions have the same number of unpaired electrons, they will have the same magnetic moment.
Step 2: Key Formula or Approach:
Identify the electronic configuration of each transition metal ion and count the number of unpaired electrons in the $d$-orbitals.
Step 3: Detailed Explanation:
Atomic Numbers: Sc(21), Ti(22), Cr(24), Mn(25), Fe(26).
A. $Ti^{3+$:}
Ground state Ti: $[Ar] 3d^2 4s^2$.
$Ti^{3+}$ (remove 3 electrons): $[Ar] 3d^1$.
Unpaired electrons ($n$) = 1.
B. $Cr^{2+$:}
Ground state Cr: $[Ar] 3d^5 4s^1$ (Exception).
$Cr^{2+}$ (remove 2 electrons): $[Ar] 3d^4$.
Unpaired electrons ($n$) = 4.
C. $Mn^{2+$:}
Ground state Mn: $[Ar] 3d^5 4s^2$.
$Mn^{2+}$ (remove 2 electrons): $[Ar] 3d^5$.
Unpaired electrons ($n$) = 5.
D. $Fe^{2+$:}
Ground state Fe: $[Ar] 3d^6 4s^2$.
$Fe^{2+}$ (remove 2 electrons): $[Ar] 3d^6$.
Configuration in $d$-orbitals: $\uparrow \downarrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$.
Unpaired electrons ($n$) = 4.
E. $Sc^{3+$:}
Ground state Sc: $[Ar] 3d^1 4s^2$.
$Sc^{3+}$ (remove 3 electrons): $[Ar] 3d^0$.
Unpaired electrons ($n$) = 0.
Comparing results: Both $Cr^{2+}$ (B) and $Fe^{2+}$ (D) have 4 unpaired electrons.
Therefore, their magnetic moment will be $\sqrt{4(4+2)} = \sqrt{24} \approx 4.89$ B.M.
Step 4: Final Answer:
Ions B and D have the same magnetic moment.