Question

Calculate the number of unpaired electrons in the following gaseous ions: $M{n}^{3+}$, $C{r}^{3+}$, $V^{3+}$, and $T{i}^{3+}$. Which one of these is the most stable in aqueous solution?

Answer 1

Hint: For finding the number of unpaired electrons, then first we have to find the atomic number of the element then write the configuration in the ground state, then according to the oxidation state the subtract the number of electrons from the outer shell. The ion that will have a half-filled or fully configuration will be the most stable ion in water.

Complete answer:
We are given some ions and these all ions belong to the d-block or transition elements, and we have to find the number of unpaired electrons. For finding the number of unpaired electrons, then first we have to find the atomic number of the element then write the configuration in the ground state, then according to the oxidation state subtract the number of electrons from the outer shell.
Let us check all the ions one by one:
$M{n}^{3+}$, this is the ion of manganese and the atomic number of manganese is 25. The electronic configuration in group state will be:
$[Ar]3d^{5}4s^2$
In +3 oxidation state, the configuration will be:
$[Ar]3d^4$
So, there are 4 unpaired electrons.
$C{r}^{3+}$, this is the ion of chromium and the atomic number of chromium is 24. The electronic configuration in group state will be:
$[Ar]3d^{5}4s^1$
In +3 oxidation state, the configuration will be:
$[Ar]3d^3$
So, there are 3 unpaired electrons.
$V^{3+}$, this is the ion of vanadium and the atomic number of vanadium is 23. The electronic configuration in group state will be:
$[Ar]3d^{3}4s^2$
In +3 oxidation state, the configuration will be:
$[Ar]3d^2$
So, there are 2 unpaired electrons.
$T{i}^{3+}$, this is the ion of titanium and the atomic number of titanium is 22. The electronic configuration in group state will be:
$[Ar]3d^{2}4s^2$
In +3 oxidation state, the configuration will be:
$[Ar]3d^1$
So, there is 1 unpaired electron.
The most stable ion in the aqueous solution will be chromium ion ($C{r}^{3+}$) because it has three unpaired electrons and these unpaired electrons are filled in $t_{2}g$ orbital which makes it half-filled.

Note:
In the oxidation state, when the electrons are removed from the ground electronic configuration then the electrons will be first removed from the 4s orbital because it is the outermost orbital.