Transition metal ions zero unpaired electrons are:

A. \(Sc^{3+}\)

B. \(Ti^{3+}\)

C. \(Cu^{2+}\)

D. \(Zn^{2+}\)

E. \(Mn^{2+}\)

Choose the correct answer from the options given below:

A and D only

The correct answer is option 3. A and D only.

Let us go through each transition metal ion in detail to understand the number of unpaired electrons they have:

Electronic Configurations and Unpaired Electrons

A. \(Sc^{3+}\) (Scandium Ion):

The atomic number of scandium is 21, so its ground state electronic configuration is:

\(\text{Sc: [Ar] 3d^1 4s^2}\)

For \(Sc^{3+}\): Scandium loses three electrons to form the \(Sc^{3+}\) ion. The electrons are removed first from the \(4s\) orbital and then from the \(3d\) orbital.

\(\text{Sc}^{3+}: [Ar]\)

Unpaired Electrons: The \(Sc^{3+}\) ion has the electronic configuration of argon, which has no unpaired electrons.

B. \(Ti^{3+}\) (Titanium Ion):

The atomic number of titanium is 22, so its ground state electronic configuration is:

\(\text{Ti: [Ar] 3d^2 4s^2}\)

For \(Ti^{3+}\): Titanium loses three electrons to form the \(Ti^{3+}\) ion. The electrons are removed from the \(4s\) orbital first and then from the \(3d\) orbital.

\(\text{Ti}^{3+}: [Ar] 3d^1\)

Unpaired Electrons: The \(Ti^{3+}\) ion has one unpaired electron in the \(3d\) orbital.

C. \(Cu^{2+}\) (Copper Ion):

The atomic number of copper is 29, so its ground state electronic configuration is:

\(\text{Cu: [Ar] 3d^{10} 4s^1}\)

For \(Cu^{2+}\): Copper loses two electrons to form the \(Cu^{2+}\) ion. The electrons are removed first from the \(4s\) orbital and then from the \(3d\) orbital.

\(\text{Cu}^{2+}: [Ar] 3d^9\)

Unpaired Electrons: The \(Cu^{2+}\) ion has one unpaired electron in the \(3d\) orbital.

D. \(Zn^{2+}\) (Zinc Ion):

The atomic number of zinc is 30, so its ground state electronic configuration is:

\(\text{Zn: [Ar] 3d^{10} 4s^2}\)

For \(Zn^{2+}\): Zinc loses two electrons to form the \(Zn^{2+}\) ion. These electrons are removed from the \(4s\) orbital.

\(\text{Zn}^{2+}: [Ar] 3d^{10}\)

Unpaired Electrons: The \(Zn^{2+}\) ion has a fully filled \(3d\) orbital, so it has no unpaired electrons.

E. \(Mn^{2+}\) (Manganese Ion):

The atomic number of manganese is 25, so its ground state electronic configuration is:

\(\text{Mn: [Ar] 3d^5 4s^2}\)

For \(Mn^{2+}\): Manganese loses two electrons to form the \(Mn^{2+}\) ion. The electrons are removed from the \(4s\) orbital.

\(\text{Mn}^{2+}: [Ar] 3d^5\)

Unpaired Electrons: The \(Mn^{2+}\) ion has five unpaired electrons in the \(3d\) orbitals (since each of the five \(3d\) orbitals has one electron)

Summary:

The ions with zero unpaired electrons are:

\(Sc^{3+}\): Configuration is \( [Ar] \), with no electrons in the \(3d\) orbitals.

\(Zn^{2+}\): Configuration is \( [Ar] 3d^{10} \), with all \(3d\) orbitals fully occupied.

Thus, the correct options with zero unpaired electrons are 3. A and D only