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.
Only the ions that have electrons in $d$ -orbital will be coloured. The ions in which $d$ -orbital is empty will be colourless.
Element | Atomic Number | Ionic State | Electronic configuration in ionic state |
$Ti$ | $22$ | $T i^{3+}$ | $[ Ar ]\, 3 d^{1}$ |
$V$ | $23$ | $V^{3+}$ | $[ Ar ] \,3 d^{2}$ |
$Cu$ | $29$ | $Cu ^{+}$ | $[ Ar ] \,3 d^{10}$ |
$Sc$ | $21$ | $Sc ^{3+}$ | $[ Ar ]$ |
$Mn$ | $25$ | $Mn ^{2+}$ | $[ Ar ] \,3 d^{5}$ |
$Fe$ | $26$ | $Fe ^{3+}$ | $[ Ar ] \,3 d^{5}$ |
$Co$ | $27$ | $Co ^{2+}$ | $[ Ar ] \,3 d^{7}$ |
From the above table, it can be easily observed that only $Sc ^{3+}$ has an empty $d$ -orbital. All other ions, except $Sc ^{3+}$, will be coloured in aqueous solution because of $d-d$ transitions.
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