Calculate the number of unpaired electrons in the following gaseous ions: $\mathrm{Mn}^{3+}, \mathrm{Cr}^{3+}, \mathrm{V}^{3+}$ and $\mathrm{Ti}^{3+}$. Which one of these is the most stable in aqueous solution.

Question

Calculate the number of unpaired electrons in the following gaseous ions: $\mathrm{Mn}^{3+}, \mathrm{Cr}^{3+}, \mathrm{V}^{3+}$ and $\mathrm{Ti}^{3+}$. Which one of these is the most stable in aqueous solution.,

MARKS App · Accepted Answer

$\mathrm{Mn}^{3+}=3 d^4=4$ unpaired electrons, $\mathrm{Cr}^{3+}=3 d^3=3$ electrons, $\mathrm{V}^{3+}=3 d^2=2$ electrons, $\mathrm{Ti}^{3+}=3 d^1=1$ electron.Out of these, $\mathrm{Cr}^{3+}$ is most stable in aqueous solution because of half-filled $\mathrm{t}_{2 \mathrm{~g}}$ level.

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