Question

Which of the following compounds is expected to be coloured?
A.\[{\mathbf{A}}{{\mathbf{g}}_2}{\mathbf{S}}{{\mathbf{O}}_4}\]
B.\[{\mathbf{Cu}}{{\mathbf{F}}_2}\]
C.\[{\mathbf{Mg}}{{\mathbf{F}}_2}\]
D.\[{\mathbf{CuCl}}\]

Answer 1

Hint: Main phenomenon which makes a compound colourful is the transition of electrons. If a compound contains a lone pair of electron or an unbounded electron, when we give it energy, the electron jumps to its higher states and acquires some energy while coming to the ground level the electrons emit that energy back and therefore emit colours.

Complete answer:
The simple phenomenon of colour is the same for every compound. That is the electron transition, electrons jump from its lower states to higher states when it is provided with energy and then comebacks to the ground state emitting the same energy that was absorbed by that electron. But in the case of a transition metal, there is more into it. All the compound which shows colour in the transition elements contains \[\left( {n - 1} \right)d\] orbital, \[\left( {n - 1} \right)d\] orbital is partially filled. That now the electrons which are unpaired in \[\left( {n - 1} \right)d\] orbital undergoes an electronic transition from one d orbital to the other d orbital, in d-d transition, the electrons absorb the energy is given to it and the remaining energy which is not absorbed by the electron is emitted as the colour. This emitted energy falls in the visible spectrum and we can see coloured ions of transition metal.
In the case \[Cu{F_2}\] copper is partially filled having configuration \[4{s^0}3{d^9}\], since the electrons in d orbital are unpaired they emit colours when provided with energy.
And hence option B is the correct answer.

Note:
Always remember that all transition elements do not show coloured ions. The phenomenon is only shown when the d orbital is partially filled and there are unpaired electrons, which on providing energy emits colour.