JEE Chemistry 2026 Question with Solution

Given: The metal ions are $Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+}$ and $Ti^{2+}$. We need the ions having spin-only magnetic moment greater than $3.0 \; \text{BM}$ and then find the sum of unpaired electrons in their high spin octahedral complexes.

Find: The required numerical sum.

For spin-only magnetic moment,

where $n$ is the number of unpaired electrons.

If $n=2$,

If $n=3$,

So, magnetic moment more than $3.0 \; \text{BM}$ means the ion must have at least $3$ unpaired electrons.

Now determine the electronic configuration and unpaired electrons for each ion in high spin octahedral field:

• $Ti^{2+} : [Ar]3d^2$, so unpaired electrons $=2$
• $V^{3+} : [Ar]3d^2$, so unpaired electrons $=2$
• $Fe^{2+} : [Ar]3d^6$, high spin $t_{2g}^4 e_g^2$, so unpaired electrons $=4$
• $Co^{2+} : [Ar]3d^7$, high spin $t_{2g}^5 e_g^2$, so unpaired electrons $=3$
• $Ni^{2+} : [Ar]3d^8$, so unpaired electrons $=2$

Therefore, the ions with more than $3.0 \; \text{BM}$ magnetic moment are $Fe^{2+}$ and $Co^{2+}$.

Hence, the required sum is

Therefore, the required sum is $7$.