In following pairs, the one in which both transition metal ions are colourless is

We need to find the pair where both transition metal ions are colourless. The colour of a transition metal ion arises from d-d electronic transitions. An ion will be colourless if it has either a $$d^0$$ (empty d-orbitals) or $$d^{10}$$ (completely filled d-orbitals) configuration, because in either case no d-d transition is possible.

Let us examine each option:

Option A: $$Sc^{3+}$$ and $$Zn^{2+}$$

Scandium (Sc) has atomic number 21 with configuration $$[Ar]\,3d^1\,4s^2$$. So $$Sc^{3+}$$ has configuration $$[Ar]$$ — that is $$3d^0$$. With no d-electrons, no d-d transition is possible, so $$Sc^{3+}$$ is colourless.

Zinc (Zn) has atomic number 30 with configuration $$[Ar]\,3d^{10}\,4s^2$$. So $$Zn^{2+}$$ has configuration $$[Ar]\,3d^{10}$$. With completely filled d-orbitals, no d-d transition is possible, so $$Zn^{2+}$$ is colourless.

Both ions are colourless.

Option B: $$Ti^{4+}$$ and $$Cu^{2+}$$

$$Ti^{4+}$$ is $$3d^0$$ (colourless), but $$Cu^{2+}$$ is $$3d^9$$ which allows d-d transitions, making it blue/coloured. Not both colourless.

Option C: $$V^{2+}$$ and $$Ti^{3+}$$

$$V^{2+}$$ is $$3d^3$$ and $$Ti^{3+}$$ is $$3d^1$$. Both have partially filled d-orbitals, so both are coloured.

Option D: $$Zn^{2+}$$ and $$Mn^{2+}$$

$$Zn^{2+}$$ is $$3d^{10}$$ (colourless), but $$Mn^{2+}$$ is $$3d^5$$. Although $$Mn^{2+}$$ is only faintly coloured (all transitions are spin-forbidden), it is still not truly colourless. Moreover, strictly speaking it has partially filled d-orbitals.

Hence, the correct answer is Option A: $$Sc^{3+}, Zn^{2+}$$.