Question

Which of the following pairs of ions will have same spin only magnetic moment values within the pair?
• [A.] \( \text{Zn}^{2+}, \text{Ti}^{2+} \)
• [B.] \( \text{Cr}^{2+}, \text{Fe}^{2+} \)
• [C.] \( \text{Ti}^{3+}, \text{Cu}^{2+} \)
• [D.] \( \text{V}^{2+}, \text{Cu}^{+} \) Choose the correct answer from the options given below:

• A. C and D only
• B. A and D only
• C. A and B only
• D. B and C only

Hint:

To quickly find the number of unpaired electrons for a \( 3\text{d}^m \) configuration where \( m > 5 \), simply subtract the number of electrons from 10 (\( n = 10 - m \)). For example, for \( \text{Fe}^{2+} \) (\( 3\text{d}^6 \)), \( n = 10 - 6 = 4 \). This allows an instant matching shortcut with \( \text{Cr}^{2+} \) (\( 3\text{d}^4 \)) without drawing orbital diagrams!

Answer 1

The Correct Option is D

Concept: The spin-only magnetic moment (\( \mu \)) of a transition metal ion depends exclusively on the number of its unpaired electrons (\( n \)). It is calculated using the formula: \[ \mu = \sqrt{n(n + 2)} \text{ B.M.} \] where B.M. stands for Bohr Magneton. For two ions to exhibit the exact same spin-only magnetic moment value, they must possess the identical number of unpaired electrons (\( n \)) in their valence \( \text{d} \)-orbitals.

Step 1: Analyzing pair A: \( \text{Zn}^{2+} \) and \( \text{Ti}^{2+} \).

• \( \text{Zn} \) (Atomic Number = 30): Electronic configuration is \( [\text{Ar}] 3\text{d}^{10} 4\text{s}^2 \). For \( \text{Zn}^{2+} \), configuration is \( 3\text{d}^{10} \). All electrons are completely paired, so \( n = 0 \).
• \( \text{Ti} \) (Atomic Number = 22): Electronic configuration is \( [\text{Ar}] 3\text{d}^2 4\text{s}^2 \). For \( \text{Ti}^{2+} \), configuration is \( 3\text{d}^2 \). According to Hund's rule, these electrons are unpaired, so \( n = 2 \). Since \( 0 \neq 2 \), pair A does not have the same magnetic moment.

Step 2: Analyzing pair B: \( \text{Cr}^{2+} \) and \( \text{Fe}^{2+} \).

• \( \text{Cr} \) (Atomic Number = 24): Electronic configuration is \( [\text{Ar}] 3\text{d}^5 4\text{s}^1 \). For \( \text{Cr}^{2+} \), configuration is \( 3\text{d}^4 \). Arranging these in five degenerate d-orbitals yields \( n = 4 \) unpaired electrons.
• \( \text{Fe} \) (Atomic Number = 26): Electronic configuration is \( [\text{Ar}] 3\text{d}^6 4\text{s}^2 \). For \( \text{Fe}^{2+} \), configuration is \( 3\text{d}^6 \). Arranging 6 electrons results in one paired orbital and four singly occupied orbitals, giving \( n = 4 \) unpaired electrons. Since both have \( n = 4 \), pair B has the same magnetic moment.

Step 3: Analyzing pair C: \( \text{Ti}^{3+} \) and \( \text{Cu}^{2+} \).

• For \( \text{Ti}^{3+} \), configuration is \( 3\text{d}^1 \), which gives \( n = 1 \) unpaired electron.
• \( \text{Cu} \) (Atomic Number = 29): Electronic configuration is \( [\text{Ar}] 3\text{d}^{10} 4\text{s}^1 \). For \( \text{Cu}^{2+} \), configuration is \( 3\text{d}^9 \). Arranging 9 electrons leaves exactly one orbital singly occupied, giving \( n = 1 \) unpaired electron. Since both have \( n = 1 \), pair C has the same magnetic moment.

Step 4: Analyzing pair D: \( \text{V}^{2+} \) and \( \text{Cu}^{+} \).

• \( \text{V} \) (Atomic Number = 23): Electronic configuration is \( [\text{Ar}] 3\text{d}^3 4\text{s}^2 \). For \( \text{V}^{2+} \), configuration is \( 3\text{d}^3 \), giving \( n = 3 \) unpaired electrons.
• For \( \text{Cu}^{+} \), configuration is \( 3\text{d}^{10} \), which means all electrons are paired up, giving \( n = 0 \). Since \( 3 \neq 0 \), pair D does not have the same magnetic moment. Therefore, only statements B and C are correct, which corresponds to option (D).