Question

\[{\text{T}}{{\text{i}}^{2 + }}\] is purple while \[{\text{T}}{{\text{i}}^{4 + }}\] is colorless, because:
A.There is no crystal field effect in \[{\text{T}}{{\text{i}}^{4 + }}\]
B.\[{\text{T}}{{\text{i}}^{2 + }}\] has \[{\text{ 3}}{{\text{d}}^2}{\text{ }}\]configuration
C.\[{\text{T}}{{\text{i}}^{4 + }}\] has \[{\text{ 3}}{{\text{d}}^0}{\text{ }}\]configuration
D.\[{\text{T}}{{\text{i}}^{4 + }}\] is very small cation when compared to \[{\text{T}}{{\text{i}}^{2 + }}\] and hence does not absorb any radiation.

Answer 1

Hint: First of all we need to write the electronic configuration of both the ions. The one which has all the electrons paired will not show d-d transition and hence will be colourless. In the compound which has unpaired electrons will show d-d transition and the complex formed will have colour.

Complete step by step answer:
Titanium is a d block element that belongs to group number four. Its atomic number is 22 and it is 3d series elements. Electronic configuration of titanium can be written as:
\[{\text{ }}\left[ {{\text{Ar}}} \right]4{{\text{s}}^2}3{{\text{d}}^2}{\text{ }}\]
We have used argon because it is the preceding inert gas.
After removing 2 electrons the electronic configuration of \[{\text{T}}{{\text{i}}^{2 + }}\] will be:
\[{\text{ }}\left[ {{\text{Ar}}} \right]4{{\text{s}}^0}3{{\text{d}}^2}{\text{ }}\]
As we can see that there are two unpaired electrons that are present in the d orbital and hence it can undergo the transition to show colour.
After removing 2 more electrons from \[{\text{T}}{{\text{i}}^{2 + }}\] the electronic configuration of \[{\text{T}}{{\text{i}}^{4 + }}\] will be:
\[{\text{ }}\left[ {{\text{Ar}}} \right]4{{\text{s}}^0}3{{\text{d}}^0}{\text{ }}\]
Here \[{\text{T}}{{\text{i}}^{4 + }}\] has no electron in d orbital and no d-d transition will be possible. Hence the compound formed will be colourless.

So, the correct option is C.

Note:
We know that all the d orbital in an element is degenerate, that is they have the same energy. In case of complexes due to crystal field splitting the d orbital splits. Three sets of d orbital have lower energy and the two sets of d orbital have high energy. This creates a difference in energy and hence the electron excited from one d orbital to another. When the electron returns back and energy falls in the visible region then we see the colour of complex.