Question

Which from following elements in respective oxidation state develops highest spin only magnetic moment?

• A. $Mn^{2+}$
• B. $Ti^{3+}$
• C. $Cu^{2+}$
• D. $Ni^{2+}$

Hint:

For 3d transition metals, the absolute maximum number of unpaired electrons any ion can have is 5. Ions with a $d^5$ configuration ($Mn^{2+}$ and $Fe^{3+}$) will always have the highest possible spin-only magnetic moments ($\sim 5.9$ BM).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We must determine which of the provided transition metal ions possesses the highest spin-only magnetic moment ($\mu$). This directly translates to finding the ion with the highest number of unpaired electrons.

Step 2: Key Formula or Approach:
The spin-only magnetic moment ($\mu$) is calculated using the formula:
$$\mu = \sqrt{n(n + 2)} \text{ Bohr Magnetons (BM)}$$
Where $n$ represents the number of unpaired electrons. The moment strictly increases as the number of unpaired electrons increases.

Step 3: Detailed Explanation:
Let's deduce the valence d-orbital electron configuration for each ion to count the unpaired electrons:
- (b) $Ti^{3+}$: Titanium is $[Ar] 4s^2 3d^2$. Losing 3 $e^-$ gives $3d^1$. It has 1 unpaired electron. ($\mu \approx 1.73$ BM)
- (c) $Cu^{2+}$: Copper is $[Ar] 4s^1 3d^{10}$. Losing 2 $e^-$ gives $3d^9$. It has 1 unpaired electron. ($\mu \approx 1.73$ BM)
- (d) $Ni^{2+}$: Nickel is $[Ar] 4s^2 3d^8$. Losing 2 $e^-$ gives $3d^8$. It has 2 unpaired electrons. ($\mu \approx 2.83$ BM)
- (a) $Mn^{2+}$: Manganese is $[Ar] 4s^2 3d^5$. Losing 2 $e^-$ gives $3d^5$. It has 5 unpaired electrons. ($\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM)
Because the $3d$ subshell has exactly 5 orbitals, having 5 electrons maximizes the number of unpaired spins according to Hund's Rule.

Step 4: Final Answer:
$Mn^{2+}$ has the highest number of unpaired electrons, giving it the highest magnetic moment. This matches option (a).