The correct answer is option 3. (B), (C), and (E) only.
The color exhibited by transition metal ions is often associated with the presence of unpaired electrons in their d orbitals. Let's analyze each ion mentioned in the options:
(A) \(Cu^+\) (Copper(I)): Copper(I) has a completely filled \(3d^{10}\) electronic configuration, with no unpaired electrons. Since there are no unpaired electrons to absorb visible light, \(Cu^+\) is not expected to exhibit color.
(B) \(Ni^{2+}\) (Nickel(II)): Nickel(II) has a \(3d^8\) electronic configuration with two unpaired electrons. The presence of unpaired electrons allows for absorption of certain wavelengths of light, and \(Ni^{2+}\) is likely to exhibit color.
(C) \(Mn^{2+}\) (Manganese(II)): Manganese(II) has a \(3d^5\) electronic configuration with five unpaired electrons. The presence of multiple unpaired electrons suggests that \(Mn^{2+}\) may absorb visible light and exhibit color.
(D) \(Sc^{3+}\) (Scandium(III)): Scandium(III) has a \(3d^0\) electronic configuration with no unpaired electrons. Without unpaired electrons, \(Sc^{3+}\) is not expected to absorb visible light and, therefore, is not likely to be colored.
(E) \(Fe^{3+}\) (Iron(III)): Iron(III) has a \(3d^5\) electronic configuration with five unpaired electrons. The presence of unpaired electrons makes \(Fe^{3+}\) likely to absorb visible light and display color.
In summary, the ions that are likely to be colored due to the presence of unpaired electrons are \(Ni^{2+}\), \(Mn^{2+}\), and \(Fe^{3+}\). Therefore, the correct answer is (3) (B), (C), and (E) only. |
