Practicing Success
Among the following series of transition metal ions, the one where all metal ions have 3d2 electronic configuration is: |
$Ti^{+}, V^{4+}, Cr^{6+}, Mn^{7+}$ $Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}$ $Ti^{3+}, V^{2+}, Cr^{3+}, Mn^{4+}$ $Ti^{4+}, V^{3+}, Cr^{2+}, Mn^{3+}$ |
$Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}$ |
The correct answer is (2), $Ti^{2+}, V{3+}, Cr^{4+}, \text{ and } Mn^{5+}$ because these ions all have a \(3d^2\) electronic configuration. The 3d orbital can hold up to 10 electrons. The ions in the series $Ti^{2+}, V{3+}, Cr^{4+},\text{ and } Mn^{5+}$ have 2 electrons in the 3d orbital, so this is the only series where all the ions have a 3d2 electronic configuration. The other series in question have different numbers of electrons in the 3d orbital. For example, \(Ti^+\) has 1 electrons in the 3d orbital, \(V^{4+}\) has 1 electron in the 3d orbital, \(Cr^{6+}\) has 0 electrons in the 3d orbital, and Mn7+ has 0 electron in the 3d orbital. Here is a table of the electronic configurations of the ions in question:
As you can see, the only ions with a \(3d^2\) electronic configuration are $Ti^{2+}, V{3+}, Cr^{4+},\text{ and } Mn^{5+}$. This is why the answer is (2) $Ti^{2+}, V{3+}, Cr^{4+},\text{ and } Mn^{5+}$. |