Which of the following pairs of ion is colorless a)Ti+3,cu+3 b)Sc+3,Zn+2 c)Co+3,Fe+3 d)Ni+2,V+3

To determine which pair of ions is colorless, we need to consider the electronic configurations and oxidation states of the ions.

a) Ti+3, Cu+3: Titanium (Ti) in its +3 oxidation state has the electronic configuration [Ar]3d^1. Copper (Cu) in its +3 oxidation state has the electronic configuration [Ar]3d^7. Both ions have partially filled d orbitals, which can lead to the absorption of certain wavelengths of light and the appearance of color. Therefore, this pair is not colorless.

b) Sc+3, Zn+2: Scandium (Sc) in its +3 oxidation state has the electronic configuration [Ar]3d^0. Zinc (Zn) in its +2 oxidation state has the electronic configuration [Ar]3d^10. Both ions have completely filled d orbitals, which means they do not absorb any visible light and appear colorless. Therefore, this pair is colorless.

c) Co+3, Fe+3: Cobalt (Co) in its +3 oxidation state has the electronic configuration [Ar]3d^6. Iron (Fe) in its +3 oxidation state has the electronic configuration [Ar]3d^5. Both ions have partially filled d orbitals, which can lead to the absorption of certain wavelengths of light and the appearance of color. Therefore, this pair is not colorless.

d) Ni+2, V+3: Nickel (Ni) in its +2 oxidation state has the electronic configuration [Ar]3d^8. Vanadium (V) in its +3 oxidation state has the electronic configuration [Ar]3d^2. Both ions have partially filled d orbitals, which can lead to the absorption of certain wavelengths of light and the appearance of color. Therefore, this pair is not colorless.

Based on the analysis above, the pair of ions that is colorless is b) Sc+3, Zn+2.