JEE MAIN - Chemistry (2023 - 15th April Morning Shift - No. 2)
The complex with highest magnitude of crystal field splitting energy $\left(\Delta_{0}\right)$ is :
$\left[\mathrm{Fe}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$
$\left[\mathrm{Mn}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$
$\left[\mathrm{Cr}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$
$\left[\mathrm{Ti}\left(\mathrm{OH}_{2}\right)_{6}\right]^{3+}$
Explanation
The crystal field splitting energy ($\Delta_0$) is proportional to the electrostatic interaction between the metal ion and the ligands. This interaction is affected by the size of the metal ion, as well as the size of the ligands.
The size of the metal ion can be estimated by its ionic radius, which is the distance from the nucleus to the outermost electron shell. A smaller ionic radius corresponds to a greater electrostatic interaction with the ligands, resulting in a larger $\Delta_0$ value.
Out of the given options, the ionic radii of the metal ions are:
- $\mathrm{Ti}^{3+}$: 67 pm
- $\mathrm{Cr}^{3+}$: 62 pm
- $\mathrm{Mn}^{3+}$: 65 pm
- $\mathrm{Fe}^{3+}$: 65 pm
The smaller ionic radius of $\mathrm{Cr}^{3+}$ compared to the other metal ions indicates a stronger electrostatic interaction with the ligands, resulting in a higher crystal field splitting energy ($\Delta_0$).
Therefore, the correct answer is option (C) $\left[\mathrm{Cr}\left(\mathrm{OH}_2\right)_6\right]^{3+}$, as it has the highest tendency to attract ligands due to its smaller ionic radius.
The size of the metal ion can be estimated by its ionic radius, which is the distance from the nucleus to the outermost electron shell. A smaller ionic radius corresponds to a greater electrostatic interaction with the ligands, resulting in a larger $\Delta_0$ value.
Out of the given options, the ionic radii of the metal ions are:
- $\mathrm{Ti}^{3+}$: 67 pm
- $\mathrm{Cr}^{3+}$: 62 pm
- $\mathrm{Mn}^{3+}$: 65 pm
- $\mathrm{Fe}^{3+}$: 65 pm
The smaller ionic radius of $\mathrm{Cr}^{3+}$ compared to the other metal ions indicates a stronger electrostatic interaction with the ligands, resulting in a higher crystal field splitting energy ($\Delta_0$).
Therefore, the correct answer is option (C) $\left[\mathrm{Cr}\left(\mathrm{OH}_2\right)_6\right]^{3+}$, as it has the highest tendency to attract ligands due to its smaller ionic radius.
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