Determining Metal Ions with Magnetic Moment > 3.0 BM

The objective is to identify metal ions from the list ($\text{Co}^{2+}, \text{Ni}^{2+}, \text{Fe}^{2+}, \text{V}^{3+}, \text{Ti}^{2+}$) that possess a spin-only magnetic moment ($\mu_s$) greater than $3.0\text{ BM}$ when forming high spin octahedral complexes. We then calculate the sum of unpaired electrons ($n$) for these specific ions.

Step 1: Identify $d^n$ Configurations

First, determine the $d^n$ electronic configuration for each metal ion involved:

• $\text{Co}^{2+}$: $d^7$
• $\text{Ni}^{2+}$: $d^8$
• $\text{Fe}^{2+}$: $d^6$
• $\text{V}^{3+}$: $d^2$
• $\text{Ti}^{2+}$: $d^2$

Step 2: Calculate Unpaired Electrons ($n$) and Magnetic Moment ($\mu_s$)

For high spin octahedral complexes, we determine the number of unpaired electrons ($n$) using crystal field theory. The spin-only magnetic moment is calculated using:

$ \mu_s = \sqrt{n(n+2)}\text{ BM} $

We need ions where $\mu_s > 3.0\text{ BM}$. This inequality holds true when the number of unpaired electrons $n \ge 3$ (since $n=2$ yields $\mu_s \approx 2.83\text{ BM}$ and $n=3$ yields $\mu_s \approx 3.87\text{ BM}$).

Note: Standard assignments for high spin octahedral $d^7$ configurations typically yield $n=3$.

Step 3: Identify Ions Meeting the Criteria

The metal ions from the list that satisfy the condition $\mu_s > 3.0\text{ BM}$ are:

• $\text{Co}^{2+}$ (with $n=3$)
• $\text{Fe}^{2+}$ (with $n=4$)

Step 4: Sum the Unpaired Electrons

Finally, calculate the sum of the unpaired electrons ($n$) for the identified ions ($\text{Co}^{2+}$ and $\text{Fe}^{2+}$):

$ \text{Sum of unpaired electrons} = n_{\text{Co}^{2+}} + n_{\text{Fe}^{2+}} = 3 + 4 = 7 $

The sum is 7.