Question

The transition metal ion with the highest magnetic moment is ________.

• A. $Fe^{2+}$
• B. $Mn^{2+}$
• C. $Ni^{2+}$
• D. $Co^{2+}$
• E. $Cr^{2+}$

Hint:

$d^5$ configuration in transition metals usually results in the highest spin-only magnetic moment.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The magnetic moment of a transition metal ion arises primarily from the spin of its unpaired electrons (this is the "spin-only" approximation). The magnitude of this magnetic moment is directly related to the number of unpaired electrons. To find the ion with the highest magnetic moment, we need to find the one with the most unpaired electrons.
Step 2: Key Formula or Approach
The spin-only magnetic moment, \(\mu\), is calculated by the formula: \[ \mu = \sqrt{n(n+2)} \text{ Bohr Magnetons (BM)} \] where \(n\) is the number of unpaired electrons. Since \(\mu\) increases as \(n\) increases, we only need to find the value of \(n\) for each ion. This requires writing their electronic configurations.
Step 3: Detailed Explanation
We will write the electronic configuration of each ion and count its unpaired d-electrons. Remember that for transition metal ions, electrons are first removed from the outermost \(s\)-orbital (e.g., 4s) before being removed from the \(d\)-orbital.
(A) Fe\(^{2+}\): Neutral Fe (Z=26) is \([Ar] 3d^6 4s^2\). Removing two electrons gives Fe\(^{2+}\) with the configuration \([Ar] 3d^6\). In the five d-orbitals, the six electrons are arranged as one pair and four unpaired electrons. (\(\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow\)). So, \(n=4\).
(B) Mn\(^{2+}\): Neutral Mn (Z=25) is \([Ar] 3d^5 4s^2\). Removing two electrons gives Mn\(^{2+}\) with the configuration \([Ar] 3d^5\). The five d-electrons occupy the five d-orbitals singly, with parallel spins (Hund's rule). (\(\uparrow, \uparrow, \uparrow, \uparrow, \uparrow\)). So, \(n=5\).
(C) Ni\(^{2+}\): Neutral Ni (Z=28) is \([Ar] 3d^8 4s^2\). Removing two electrons gives Ni\(^{2+}\) with the configuration \([Ar] 3d^8\). The eight d-electrons are arranged as three pairs and two unpaired electrons. (\(\uparrow\downarrow, \uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow\)). So, \(n=2\).
(D) Co\(^{2+}\): Neutral Co (Z=27) is \([Ar] 3d^7 4s^2\). Removing two electrons gives Co\(^{2+}\) with the configuration \([Ar] 3d^7\). The seven d-electrons are arranged as two pairs and three unpaired electrons. (\(\uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow, \uparrow\)). So, \(n=3\).
(E) Cr\(^{2+}\): Neutral Cr (Z=24) has an exceptional configuration, \([Ar] 3d^5 4s^1\). Removing two electrons (one from 4s, one from 3d) gives Cr\(^{2+}\) with the configuration \([Ar] 3d^4\). The four d-electrons are all unpaired. (\(\uparrow, \uparrow, \uparrow, \uparrow, \_\)). So, \(n=4\).
Comparison: The number of unpaired electrons is highest for Mn\(^{2+}\) (\(n=5\)). Therefore, it will have the highest magnetic moment.
Step 4: Final Answer
The transition metal ion with the highest magnetic moment is Mn\(^{2+}\).