Principle for colour in transition metal ions

Transition metal ions are generally coloured due to $d$-$d$ transitions.

• If the ion has partially filled $d$-orbitals ($d^1$ to $d^9$), it is usually coloured.
• If the ion has empty ($d^0$) or completely filled ($d^{10}$) $d$-orbitals, it is generally colourless.

Check each ion in the options

Option A: $\mathrm{Sc}^{3+},\ \mathrm{Zn}^{2+}$

• Scandium: $\mathrm{Sc} = [\mathrm{Ar}]\,3d^1 4s^2$ $\mathrm{Sc}^{3+} = [\mathrm{Ar}]\,3d^0$ So, $\mathrm{Sc}^{3+}$ is colourless.

• Zinc: $\mathrm{Zn} = [\mathrm{Ar}]\,3d^{10}4s^2$ $\mathrm{Zn}^{2+} = [\mathrm{Ar}]\,3d^{10}$ So, $\mathrm{Zn}^{2+}$ is colourless.

Hence, both are colourless.

Option B: $\mathrm{Ti}^{4+},\ \mathrm{Cu}^{2+}$

• Titanium: $\mathrm{Ti} = [\mathrm{Ar}]\,3d^2 4s^2$ $\mathrm{Ti}^{4+} = [\mathrm{Ar}]\,3d^0$ So, $\mathrm{Ti}^{4+}$ is colourless.

• Copper: $\mathrm{Cu} = [\mathrm{Ar}]\,3d^{10}4s^1$ $\mathrm{Cu}^{2+} = [\mathrm{Ar}]\,3d^9$ $d^9$ means partially filled $d$-orbitals, so $\mathrm{Cu}^{2+}$ is coloured.

Hence, option B is not correct.

Option C: $\mathrm{V}^{2+},\ \mathrm{Ti}^{3+}$

• Vanadium: $\mathrm{V} = [\mathrm{Ar}]\,3d^3 4s^2$ $\mathrm{V}^{2+} = [\mathrm{Ar}]\,3d^3$ So, $\mathrm{V}^{2+}$ is coloured.

• Titanium: $\mathrm{Ti}^{3+} = [\mathrm{Ar}]\,3d^1$ So, $\mathrm{Ti}^{3+}$ is coloured.

Hence, option C is not correct.

Option D: $\mathrm{Zn}^{2+},\ \mathrm{Mn}^{2+}$

• $\mathrm{Zn}^{2+}$ is $3d^{10}$, hence colourless.
• Manganese: $\mathrm{Mn} = [\mathrm{Ar}]\,3d^5 4s^2$ $\mathrm{Mn}^{2+} = [\mathrm{Ar}]\,3d^5$ Since it has partially filled $d$-orbitals, $\mathrm{Mn}^{2+}$ is coloured.

Hence, option D is not correct.

Conclusion

The only pair in which both ions are colourless is: $\boxed{\text{A: } \mathrm{Sc}^{3+},\ \mathrm{Zn}^{2+}}$

Comparison with stored answer

Stored correct answer is A, which matches the derived answer.