Question

Which of the following ions can form coloured compounds?

• A. $S{c}^{3+}$
• B. $Z{n}^{2+}$
• C. $T{i}^{4+}$
• D. $F{e}^{2+}$

Answer 1

The correct option is D $F{e}^{2+}$

Transition metal ions form coloured compounds due to d-d transition of electons.
But if there is no unpaired electrons in the d orbitals then they are colourless.
Here $S{c}^{3+},Z{n}^{2+}$and $T{i}^{4+}$ have no unpaired electron in there d orbitals.So they form only colourless compounds.
$Z{n}^{2+}=\left[Ar\right]4s^{0}3d^{10}$
$S{c}^{3+}=\left[Ar\right]4s^{0}3d^0$
$T{i}^{4+}=\left[Ar\right]4s^{0}3d^0$
But $F{e}^{2+}$ has unpaired electrons in d orbitals and hence it can form coloured compounds.
$F{e}^{2+}=\left[Ar\right]4s^{0}3d^6$