Question

Which of the following elements in their respective oxidation states does not develop spin only magnetic moment? [$\text{Ti}\ (Z = 22)$, $\text{Zn}\ (Z = 30)$, $\text{V}\ (Z = 23)$, $\text{Cu}\ (Z = 29)$]

• A. $\text{Cu}^{2+}$
• B. $\text{Zn}^{2+}$
• C. $\text{Ti}^{3+}$
• D. $\text{V}^{3+}$

Hint:

Save time by remembering that Group 12 ions ($\text{Zn}^{2+}$, $\text{Cd}^{2+}$, $\text{Hg}^{2+}$) always form stable, closed-shell $\text{d}^{10}$ electron configurations. Because they have no single, unpaired electrons, they are consistently diamagnetic with zero magnetic moment!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks us to identify which transition metal ion lacks a spin-only magnetic moment in its common oxidation state.

Step 2: Key Formula or Approach:
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}\ \text{BM}$, where $n$ is the number of unpaired electrons. If a chemical ion possesses a completely empty ($d^0$) or completely filled ($d^{10}$) d-orbital subshell, the number of unpaired electrons is zero ($n = 0$), meaning it cannot develop a spin-only magnetic moment.

Step 3: Detailed Explanation:
Let us systematically derive the valence d-orbital configurations for each listed ion:
(A) $\text{Cu}^{2+}$: Neutral Copper is $[\text{Ar}]\, 3\text{d}^{10}\, 4\text{s}^1$. Losing two electrons gives a $3\text{d}^9$ configuration. This leaves $n = 1$ unpaired electron, so it has a magnetic moment.
(B) $\text{Zn}^{2+}$: Neutral Zinc is $[\text{Ar}]\, 3\text{d}^{10}\, 4\text{s}^2$. Losing two valence electrons from the outer $4\text{s}$ orbital gives a $3\text{d}^{10}$ configuration. All 10 electrons are completely paired within the d-orbitals ($n = 0$). Thus, $\mu = 0\ \text{BM}$.
(C) $\text{Ti}^{3+}$: Neutral Titanium is $[\text{Ar}]\, 3\text{d}^2\, 4\text{s}^2$. Losing three electrons yields a $3\text{d}^1$ configuration ($n = 1$ unpaired electron).
(D) $\text{V}^{3+}$: Neutral Vanadium is $[\text{Ar}]\, 3\text{d}^3\, 4\text{s}^2$. Losing three electrons yields a $3\text{d}^2$ configuration ($n = 2$ unpaired electrons).
Therefore, $\text{Zn}^{2+}$ has no unpaired electrons and does not develop a spin-only magnetic moment, matching option (B).

Step 4: Final Answer:
$\text{Zn}^{2+}$ does not develop a spin-only magnetic moment, which corresponds to option (B).