Statement I: \([Cr(H_2O)_6]^{3+}\) is more acidic compared to \([Fe(H_2O))6]^{3+}\)
Statement II: Both are inner orbital complexes
Answer & explanation
Correct answer: option 4
The answer is (4) Statement I is false but Statement II is correct.
Statement I is false. The acidity of a complex is determined by the stability of the resulting aquated metal ion. The more stable the aquated metal ion, the less acidic the complex.
In the case of \([Cr(H_2O)_6]^{3+}\) and \([Fe(H_2O)_6]^{3+}\), the aquated metal ions are \([Cr(H_2O)_6]^{2+}\) and \([Fe(H_2O)_6]^{2+}\). The \([Cr(H_2O)_6]^{2+}\) ion is less stable than the \([Fe(H_2O)_6]^{2+ ion because chromium has a smaller ionic radius than iron. This means that the chromium ion can be less easily hydrated, which makes the aquated ion less stable.
Therefore, \([Cr(H_2O)_6]^{3+}\) is less acidic than \([Fe(H_2O)_6]^{3+}\).
Statement II is correct. Both complexes are inner orbital complexes, but this is not the reason why \([Cr(H_2O)_6]^{3+}\) is less acidic than \([Fe(H_2O))6]^{3+}\). The reason for the difference in acidity is the stability of the aquated metal ions.