Question

Which of the following ions will not coloured in aqueous solution?
$A.$ $N{i^{2 + }}$
$B.$ $F{e^{3 + }}$
$C.$ $T{i^{4 + }}$
$D.$ $C{u^{2 + }}$

Answer 1

All the ions are belonging to the $d -$ block element. The coloured of the metal and ion is characterized by its electronic configuration of the metal or its ion. The unpaired electron in the $d -$ orbital shows colour in aqueous solution whereas the paired electron does not show any colour in aqueous solution.

Complete step by step solution
The colour of the metal or its ion due to an unpaired electron and it can be determined by its electronic configuration of these ions. The electronic configuration of the ions is given below:
Electronic configuration of the $N{i^{2 + }}$ is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^8}$ . it has two unpaired electron in its $d -$ orbital. So, it shows colour in the aqueous solution.
Similarly, electronic configuration of the $F{e^{3 + }}$ is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^5}$ . It has five unpaired electrons in its $d -$ orbital. So, it shows colour in the aqueous solution.
Electronic configuration of the $T{i^{4 + }}$ is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^0}$ . It has no unpaired electron in its $d -$ orbital. So, it does not show any colour in the aqueous solution.
Electronic configuration of the $C{u^{2 + }}$ is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^9}$ . It has one unpaired electron in its $d -$ orbital. So, it shows colour in the aqueous solution.
Thus, only $T{i^{4 + }}$ ion will not colour in aqueous solution.
So, the only option $\left( C \right)$ is correct.

Note
Electronic configuration of the metals and its ions must be done according to the Aufbau’s principle. It states that the electron should be first filled in the lowest energy orbital then followed by higher energy orbital. When an electron absorbs energy it goes to higher level from lower level then it comes back higher to lower level and emits radiation in the form of $UV$ visible region. This is the phenomena of showing colour.