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Which of the following ions is the most stable in aqueous solution ?
(At. No are $Ti = 22$, $V = 23$, $Cr = 24$, $Mn = 25$)
(A) $M{n^{3 + }}$
(B) $C{r^{3 + }}$
(C) ${V^{3 + }}$
(D) $T{i^{3 + }}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: The ion having the most stable electronic configuration on the splitting of d orbitals in ${t_{2g}}$ and ${e_g}$ orbitals will decide the stability of ions in the aqueous solution. The d orbitals splitting in this have a stable configuration. In the presence of the field, the stable electronic configuration will be the deciding factor for stability.

Complete step by step answer:
We know that the most stable ions have a stable ionic configuration that is the full filled or half-filled electronic configuration. The d orbital when the splitting takes place in the presence of a field consists of ${t_{2g}}$ and ${e_g}$ levels. The lower energy level is called the ${t_{2g}}$ level and the high energy level is called the ${e_g}$ level. The completed filled ${t_{2g}}$level consists of six electrons and ${e_g}$ the level consists of four electrons. The half-filled configuration contains half of the electrons in both of them.
Now the atomic number of manganese is twenty-five. The electronic configuration of manganese is as follows
$Mn \to [Ar]4{s^2}3{d^5}$
Now it loses three electrons. So the electronic configuration of $M{n^{3 + }}$ is as follows
$M{n^{3 + }} \to [Ar]4{s^0}3{d^4}$
Out of the four electrons present in the d orbital three are in ${t_{2g}}$ and one electron is present in the ${e_g}$ orbital. Here the half-filled electronic configuration is in the lower energy level and in the higher energy level, not anything as half-filled or full-filled as one electron is present which reduces the stability. So it does not have the most stable configuration in an aqueous solution.
Similarly for chromium ion, the electronic configuration is $C{r^{3 + }} \to [Ar]4{s^0}3{d^3}$ . Here the configuration has a half-filled exceptionally stable configuration. For vanadium ion and titanium ion, the configurations are $[Ar]4{s^0}3{d^2}$ and $[Ar]4{s^0}3{d^1}$ respectively. These are not stable in aqueous solutions. So $C{r^{3 + }}$ is the most stable in an aqueous solution. $C{r^{3 + }}$

So, the correct answer is Option B.

Note: The distribution of electrons in the ${t_{2g}}$ orbital and ${e_g}$ orbital are there in such a way that the three electrons enter the former level first. It is due to the crystal field splitting energy(CFSE). In the presence of the field, the stable electronic configuration will be the deciding factor for stability.