Which of the following ions is paramagnetic in nature?

\(V^{2+}\)

The correct answer is option 4. \(V^{2+}\).

To determine which of the given ions is paramagnetic, we need to examine the electronic configurations and the presence of unpaired electrons in each ion. A paramagnetic ion has one or more unpaired electrons, which make it attractable to a magnetic field.

1. \(Sc^{3+}\) (Scandium ion)

Scandium (Sc) has an atomic number of 21, so its ground-state electronic configuration is: \([Ar] 4s^2 3d^1\).

For \(Sc^{3+}\), it loses 3 electrons: two from the 4s orbital and one from the 3d orbital. Thus, the electronic configuration for \(Sc^{3+}\) is: \([Ar]\), which means it has a completely filled \(3d^0\) and \(4s^0\) orbitals.

Unpaired Electrons:None.

Magnetic Nature: Diamagnetic (not paramagnetic).

2. \(Ti^{4+}\) (Titanium ion)

Titanium (Ti) has an atomic number of 22, so its ground-state electronic configuration is: \([Ar] 4s^2 3d^2\).

For \(Ti^{4+}\), it loses 4 electrons: two from the 4s orbital and two from the 3d orbital. Thus, the electronic configuration for \(Ti^{4+}\) is: \([Ar]\), which means it has a completely filled \(3d^0\) and \(4s^0\) orbitals.

Unpaired Electrons: None.

Magnetic Nature: Diamagnetic (not paramagnetic).

3. \(Zn^{2+}\) (Zinc ion)

Zinc (Zn) has an atomic number of 30, so its ground-state electronic configuration is: \([Ar] 4s^2 3d^{10}\).

For \(Zn^{2+}\), it loses 2 electrons from the 4s orbital. Thus, the electronic configuration for \(Zn^{2+}\) is: \([Ar] 3d^{10}\).

Unpaired Electrons: None.

Magnetic Nature: Diamagnetic (not paramagnetic).

4. \(V^{2+}\) (Vanadium ion)

Vanadium (V) has an atomic number of 23, so its ground-state electronic configuration is: \([Ar] 4s^2 3d^3\).

For \(V^{2+}\), it loses 2 electrons: both from the 4s orbital. Thus, the electronic configuration for \(V^{2+}\) is: \([Ar] 3d^3\).

Unpaired Electrons: There are 3 unpaired electrons in the \(3d^3\) configuration.

Magnetic Nature: Paramagnetic (due to the presence of unpaired electrons).

Summary:

\(Sc^{3+}\), \(Ti^{4+}\), and \(Zn^{2+}\) are diamagnetic because they have no unpaired electrons.

\(V^{2+}\) is paramagnetic because it has unpaired electrons.

Therefore, the ion that is paramagnetic in nature is \(V^{2+}\).