Question

Identify the oxidation states of titanium ($Z = 22$) and copper ($Z = 29$) in their colourless compounds:
A.$T{i^{3 + }},C{u^{2 + }}$
B.$T{i^{2 + }},C{u^{2 + }}$
C.$T{i^{4 + }},C{u^ + }$
D.$T{i^{4 + }},C{u^{2 + }}$

Answer 1

Hint: The transition metal ions having completely filled or completely empty d - orbitals are colourless. All transition metals possess one or more electrons in their $(n - 1)d$ sub shell.

Complete step by step answer:
The colour of the transition metal ion is due to the d-d transitions taking place between the splitted d-orbitals.When visible light fall on these ions, the unpaired electrons present in the lower energy orbitals get transferred or promoted into the higher energy d-orbitals due to the absorption of visible light. But the energy difference between these d- sub shells are small, only a certain wavelength gets absorbed. The remaining white light gets transmitted and the compound appears coloured.
In transition metal ions ,if the (n-1)d sub shell is either empty or completely filled ,that means there are no unpaired electrons in the outer shell, the d-d transition will not take place. So these ions appear colourless.
Here, we have Ti(Atomic.no.22) and Cu(Atomic.no.29).The outer electronic configuration of Ti is $3{d^2}4{s^2}$ and that of Cu is $3{d^{10}}4{s^1}$. The oxidation states of Ti are $ + 2$,$ + 3$ and $ + 4$ and that of Cu is $ + 1$,$ + 2$ and $ + 3$.
 Out of all the possible ions of Ti, the $T{i^{4 + }}$ ion has a completely empty ($3{d^0}$) d-orbital. So the d-d transition is not possible in it, and thus $T{i^{4 + }}$ appears colourless.
Similarly out of $C{u^ + }$, $C{u^{2 + }}$ and $C{u^{3 + }}$ ions, the outermost d-subshell of $C{u^ + }$ ion is completely filled ($3{d^{10}}$). Here no electron is available for the d-d transition and hence the $C{u^ + }$ ion appears as colourless.

Hence the option (C) is the correct answer.

Additional information:
In case of s and p block elements the energy required to promote the s or p electron to a higher energy level is very high and it does not correspond to the wavelength of the visible light and also their d-orbital is either completely filled or missing. Because of these reasons the s and p block elements are usually colourless.

Note:
Thus an ion is coloured if it contains unpaired electrons in its d- sub shell and if the ion does not contain any unpaired electron in its d- subshell it appears colourless. In order to determine if the metal ion is colourless, one should remember the electronic configuration. Here, the electronic configuration of copper ($3{d^{10}}4{s^1}$) has an anomaly, its 4s orbital is not completely filled. This occurs in order to attain a completely filled subshell of 3d which is more stable than $3{d^9}4{s^2}$ configuration.