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$Ti ^{3+}, V ^{3+}, Cu ^{+},$ $Sc ^{3+}, Mn ^{2+}, Fe ^{3+},Co ^{2+}$ प्रत्येक के लिए कारण बताइए।

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.