Predict which of the following will be coloured in aqueous solution: $Ti^{3+}, V^{3+}, Cu^{+}, Sc^{3+}, Mn^{2+}, Fe^{3+},$ and $Co^{2+}$. Give reasons for each.

(N/A) Only the ions that have unpaired electrons in the $d$-orbital will be coloured due to $d-d$ transitions. The ions in which the $d$-orbital is empty $(d^0)$ or completely filled $(d^{10})$ will be colourless.

$Element. \text{Atomic Number}$	$Ionic \text{State}. \text{Electronic Configuration}$
$Ti. 22$	$Ti^{3+}. [Ar] 3d^{1}$
$V. 23$	$V^{3+}. [Ar] 3d^{2}$
$Cu. 29$	$Cu^{+}. [Ar] 3d^{10}$
$Sc. 21$	$Sc^{3+}. [Ar]$
$Mn. 25$	$Mn^{2+}. [Ar] 3d^{5}$
$Fe. 26$	$Fe^{3+}. [Ar] 3d^{5}$
$Co. 27$	$Co^{2+}. [Ar] 3d^{7}$

From the table,$Sc^{3+}$ $(d^0)$ and $Cu^{+}$ $(d^{10})$ are colourless. All other ions,namely $Ti^{3+}, V^{3+}, Mn^{2+}, Fe^{3+},$ and $Co^{2+}$,contain unpaired electrons in their $d$-orbitals and will be coloured in aqueous solution due to $d-d$ transitions.