Which of the following transition metal ion is colourless?

\(Sc^{3+}\)

We are given four transition metal ions \(Sc^{3+}, V^{2+}, Mn^{2+}\) and \(Co^{3+}\). The elements or ions exhibit colours due to the presence of unpaired electrons in the d-orbital.

We know that the atomic number of scandium (Sc) is 21. Thus, the electronic configuration of scandium is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\)

When we say \(Sc^{3+}\) or scandium ion, two electrons are removed from the 4s orbital of scandium, and one electron is removed from the 3d orbital. Thus, the electronic configuration of \(Sc^{3+}\) is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^0\)

From the electronic configuration, we can see that there are no electrons in the d- orbital so, there will be no d-d transition, and hence it is colourless.

We know that the atomic number of vanadium (V) is 23. Thus, the electronic configuration of vanadium is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\)

When we say \(V^{2+}\) or vanadium ion, two electrons are removed from the valence 4s orbital. Thus, the electronic configuration of \(V^{2+}\) is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^3\)

From the electronic configuration, we can see that the electrons in the valence d-orbital of \(V^{2+}\) are unpaired. Thus, \(V^{2+}\) ion is not colourless.

We know that the atomic number of manganese (Mn) is 25. Thus, the electronic configuration of manganese is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\)

When we say \(Mn^{2+}\) or manganese ion, two electrons are removed from the valence 4s orbital. Thus, the electronic configuration of \(Mn^{2+}\) is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^5\)

From the electronic configuration, we can see that the electrons in the valence d-orbital of \(Mn^{2+}\) are unpaired. Thus, \(Mn^{2+}\) ion is not colourless.

We know that the atomic number of cobalt (Co) is 27. Thus, the electronic configuration of cobalt is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7\)

When we say \(Co^{3+}\) or cobalt ion, two electrons are removed from the 4s orbital and one electron is removed from the 3d orbital. Thus, the electronic configuration of \(Co^{3+}\) is as follows:

\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^6\)

From the electronic configuration, we can see that the electrons in the valence d-orbital of \(Co^{2+}\) are unpaired. Thus, \(Co^{2+}\) ion is not colourless.