Question

Out of Mn\(^{2+}\) and Zn\(^{2+}\), which ion will be more paramagnetic and why? (Atomic numbers: Mn = 25, Zn = 30)

Hint:

To compare paramagnetism, first write the electron configuration of each ion, remembering that transition metals lose electrons from the outer s subshell before the d subshell when forming a cation. Then check how many d electrons remain unpaired; a subshell that is exactly half filled keeps every electron unpaired, while a completely filled subshell has none.

Answer 1

Electronic configurations: Mn (Z = 25): \[ [Ar]\,3d^5 4s^2 \Rightarrow Mn^{2+} = 3d^5 \] Zn (Z = 30): \[ [Ar]\,3d^{10}4s^2 \Rightarrow Zn^{2+} = 3d^{10} \]
Magnetic nature:

• Mn\(^{2+}\): 5 unpaired electrons → highly paramagnetic
• Zn\(^{2+}\): No unpaired electrons → diamagnetic

\[ \therefore Mn^{2+} \text{ is more paramagnetic. \]

Answer 2

Concept:

• Paramagnetism can be measured quantitatively using the spin-only magnetic moment formula, $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons; a larger magnetic moment means stronger paramagnetism, and $\mu = 0$ means the species is diamagnetic.
• For transition metal atoms, electrons are removed from the outermost s subshell before the d subshell when a cation forms, since the ns electrons sit at a higher energy than the (n-1)d electrons once the d subshell starts filling.
• Within a set of degenerate d orbitals, each orbital receives one electron before any orbital receives a second, so a half-filled d subshell has the maximum possible number of unpaired electrons.

Step 1: Write the ground-state configuration of each neutral atom.
Mn (Z = 25): $[Ar]\,3d^5\,4s^2$
Zn (Z = 30): $[Ar]\,3d^{10}\,4s^2$

Step 2: Remove two electrons from the 4s subshell first to form each 2+ ion.
$Mn^{2+} = [Ar]\,3d^5$, with each of the five 3d orbitals holding one electron.
$Zn^{2+} = [Ar]\,3d^{10}$, with all five 3d orbitals completely filled in pairs.

Step 3: Count unpaired electrons and apply the spin-only magnetic moment formula.
For $Mn^{2+}$: $n = 5$, so $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.
For $Zn^{2+}$: $n = 0$, so $\mu = \sqrt{0(0+2)} = 0$ BM.

Step 4: Compare the two magnetic moment values.
A magnetic moment of about $5.92$ BM for $Mn^{2+}$ is far greater than the $0$ BM of $Zn^{2+}$, confirming that $Mn^{2+}$ is paramagnetic while $Zn^{2+}$ is diamagnetic.

Final Answer: $Mn^{2+}$ is more paramagnetic than $Zn^{2+}$, since it has 5 unpaired 3d electrons giving a magnetic moment of about 5.92 BM, while $Zn^{2+}$ has a completely filled $3d^{10}$ configuration with zero unpaired electrons.