Question

The aqueous solution containing which one of the following ions will be colourless? [Atomic number of Sc=21, Fe=26, Ti=22, Mn=25]
A. $Sc^{3+}$
B. $Fe^{2+}$
C. $Ti^{3+}$
D. $Mn^{2+}$

Answer 1

We are going to write down the electronic configuration of each element along with their ions. The elements with unpaired electrons in the d subshell will be colored and the element without any unpaired electron in the d subshell will be colorless.

Complete step by step answer:
In order to solve this question we are going to write the electronic configuration of the ions in their current state. Now the aqueous solution appears to be colored because of the d subshell to d sub shell transition of unpaired electrons in the ions.

If we write the Electronic configuration of all the four ions :
(a) the electronic configuration of ${}_{21} Sc^{3+}$ is
${}_{21} Sc^{3+} =1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}$: : It has no unpaired electrons in d-subshell. Hence will be colorless.

(b) the electronic configuration of $Fe^{2+}$ is
${}_{26} Fe^{2+} =1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6}$: has 4 unpaired electrons, so will be colored

(c) the electronic configuration of $Ti^{3+}$ is
${}_{22} Ti^{3+} =1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{1}$: 1 unpaired electron in d-subshell hence colored.

(d) the electronic configuration of $Mn^{2+}$ is
${}_{25} Mn^{2+} =1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5}$ : 5 unpaired electrons, hence it will be colored.

**This proves that ${}_{21} Sc^{3+}$ is colorless in an aqueous solution and hence option A is correct.

Note : **
Student can make in writing the correct electronic configuration of the elements. Also an important concept here is that the color of aqueous solution is due to unpaired electrons in the d subshell and not in any other subshell.