Question

Identify the metal ions among $Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+}$ and $Ti^{2+}$ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is ____________.

Hint:

For 3d series metal ions, the number of unpaired electrons in high spin octahedral complexes is: $d^1(1), d^2(2), d^3(3), d^4(4), d^5(5), d^6(4), d^7(3), d^8(2), d^9(1), d^{10}(0)$.

Answer 1

Correct Answer: 7

Step 1: Understanding the Concept:
Magnetic moment ($\mu$) is calculated as $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.
If $\mu>3.0$ BM:
For $n=2$, $\mu = \sqrt{8} \approx 2.83$ BM.
For $n=3$, $\mu = \sqrt{15} \approx 3.87$ BM.
Therefore, "more than 3.0 BM" implies the ion must have $n \ge 3$ unpaired electrons.

Step 2: Key Formula or Approach:
Determine the electronic configuration and number of unpaired electrons ($n$) in high spin octahedral field:
1. $Ti^{2+}$ ($Z=22$): $[Ar] 3d^{2} \implies n = 2$.
2. $V^{3+}$ ($Z=23$): $[Ar] 3d^{2} \implies n = 2$.
3. $Fe^{2+}$ ($Z=26$): $[Ar] 3d^{6} \implies$ High spin: $t_{2g}^{4} e_{g}^{2} \implies n = 4$.
4. $Co^{2+}$ ($Z=27$): $[Ar] 3d^{7} \implies$ High spin: $t_{2g}^{5} e_{g}^{2} \implies n = 3$.
5. $Ni^{2+}$ ($Z=28$): $[Ar] 3d^{8} \implies n = 2$.

Step 3: Detailed Explanation:
The metal ions with $n \ge 3$ are $Fe^{2+}$ and $Co^{2+}$.
Number of unpaired electrons in $Fe^{2+}$ ($d^{6}$ high spin) = 4.
Number of unpaired electrons in $Co^{2+}$ ($d^{7}$ high spin) = 3.
Sum of unpaired electrons = $4 + 3 = 7$.

Step 4: Final Answer:
The required sum is 7.