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 Concept:
The spin-only magnetic moment ($\mu$) depends on the number of unpaired electrons ($n$) and is calculated as $\mu = \sqrt{n(n+2)}$ BM. The more unpaired electrons, the higher the magnetic moment.

Step 2: Formula Application:
Determine $d$-electron configurations: - $Mn^{2+}$ ($Z=25$): $[Ar] 3d^5$ (5 unpaired electrons). - $Ti^{3+}$ ($Z=22$): $[Ar] 3d^1$ (1 unpaired electron). - $Cu^{2+}$ ($Z=29$): $[Ar] 3d^9$ (1 unpaired electron). - $Ni^{2+}$ ($Z=28$): $[Ar] 3d^8$ (2 unpaired electrons).

Step 3: Explanation:
$Mn^{2+}$ has the maximum possible number of unpaired electrons (5) in the $3d$ subshell. This gives it the highest magnetic moment ($\approx 5.92$ BM) compared to the others.

Step 4: Final Answer:
$Mn^{2+}$ has the highest spin-only magnetic moment.