Question

'Spin only' magnetic moment is same for which of the following ions? A. \(Ti^{3+}\)
B. \(Cr^{2+}\)
C. \(Mn^{2+}\)
D. \(Fe^{2+}\)
E. \(Sc^{3+}\)
Choose the most appropriate answer from the options given below:

• A. B and D only
• B. A and E only
• C. B and C only
• D. A and D only

Hint:

For \(d^n\) configuration: \[ n \leq 5 \Rightarrow \text{Unpaired electrons} = n \] \[ n > 5 \Rightarrow \text{Unpaired electrons} = 10-n \] This shortcut helps in rapid calculation of magnetic moments.

Answer 1

The Correct Option is A

Concept: The spin-only magnetic moment depends on the number of unpaired electrons present in the atom or ion. Formula: \[ \mu = \sqrt{n(n+2)} \, BM \] where:
• \(n\) = number of unpaired electrons
• BM = Bohr Magneton If two ions possess the same number of unpaired electrons, they will have the same magnetic moment.

Step 1: Finding the electronic configuration of each ion.
\(Ti^{3+}\) Atomic number of Ti = 22 Electronic configuration: \[ Ti = [Ar]3d^24s^2 \] For \(Ti^{3+}\), remove 3 electrons: \[ Ti^{3+} = [Ar]3d^1 \] Number of unpaired electrons: \[ n = 1 \] \(Cr^{2+}\) Atomic number of Cr = 24 Electronic configuration: \[ Cr = [Ar]3d^54s^1 \] Removing two electrons: \[ Cr^{2+} = [Ar]3d^4 \] According to Hund's rule: \[ n = 4 \] \(Mn^{2+}\) Atomic number of Mn = 25 \[ Mn = [Ar]3d^54s^2 \] Removing two electrons: \[ Mn^{2+} = [Ar]3d^5 \] All electrons remain unpaired. \[ n = 5 \] \(Fe^{2+}\) Atomic number of Fe = 26 \[ Fe = [Ar]3d^64s^2 \] Removing two electrons: \[ Fe^{2+} = [Ar]3d^6 \] In \(3d^6\), four electrons remain unpaired. \[ n = 4 \] \(Sc^{3+}\) Atomic number of Sc = 21 \[ Sc = [Ar]3d^14s^2 \] Removing three electrons: \[ Sc^{3+} = [Ar]3d^0 \] \[ n = 0 \]

Step 2: Comparing the number of unpaired electrons.
\[ Ti^{3+} \Rightarrow n=1 \] \[ Cr^{2+} \Rightarrow n=4 \] \[ Mn^{2+} \Rightarrow n=5 \] \[ Fe^{2+} \Rightarrow n=4 \] \[ Sc^{3+} \Rightarrow n=0 \] Thus: \[ Cr^{2+} \text{ and } Fe^{2+} \] have equal unpaired electrons. Hence both possess same magnetic moment. Final Answer: \[ \boxed{\text{B and D only}} \]