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

### Core Logic Let's check the d$d$-electron configurations and number of unpaired electrons (n$n$) in high spin octahedral complexes: - textV^3+ rightarrow 3d^2$\text{V}^{3+} \rightarrow 3d^2$ (n = 2, mu = sqrt8 approx 2.83 text BM$n = 2, \mu = \sqrt{8} \approx 2.83 \text{ BM}$) - textTi^2+ rightarrow 3d^2$\text{Ti}^{2+} \rightarrow 3d^2$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$) - textNi^2+ rightarrow 3d^8$\text{Ni}^{2+} \rightarrow 3d^8$ (n = 2, mu = 2.83 text BM$n = 2, \mu = 2.83 \text{ BM}$) - textFe^2+ rightarrow 3d^6$\text{Fe}^{2+} \rightarrow 3d^6$ (n = 4, mu = sqrt24 approx 4.9 text BM > 3 text BM$n = 4, \mu = \sqrt{24} \approx 4.9 \text{ BM} > 3 \text{ BM}$) - textCo^2+ rightarrow 3d^7$\text{Co}^{2+} \rightarrow 3d^7$ (n = 3, mu = sqrt15 approx 3.87 text BM > 3 text BM$n = 3, \mu = \sqrt{15} \approx 3.87 \text{ BM} > 3 \text{ BM}$) ### Step 1: Summing Unpaired Electrons Only textFe^2+$\text{Fe}^{2+}$ (n = 4$n = 4$) and textCo^2+$\text{Co}^{2+}$ (n = 3$n = 3$) have magnetic moments > 3.0 text BM$> 3.0 \text{ BM}$. Sum of unpaired electrons = 4 + 3 = 7$= 4 + 3 = 7$. ### Pattern Recognition Sees: high spin octahedral complexes and spin-only magnetic moment threshold checks. Trap: Misidentifying high-spin versus low-spin electron pairing in d^6$d^6$ or d^7$d^7$ configurations. ### Chapter Mix Class 12 Chemistry: Coordination Compounds