Among the following series of transition metal ions, the one where al…

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:

Ion	Electronic Configuration
Ti+	[Ar] 4s1 3d1
V4+	[Ar] 4s0 3d1
Cr6+	[Ar] 4s0 3d0
Mn7+	[Ar] 4s0 3d0
Ti2+	[Ar] 4s0 3d2
V3+	[Ar] 4s0 3d2
Cr4+	[Ar] 4s0 3d2
Mn5+	[Ar] 4s0 3d2

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) $Ti^{2+}, V{3+}, Cr^{4+},\text{ and } Mn^{5+}$.