To determine the geometry of the ion formed when aluminium chloride is dissolved in an acidified aqueous solution, we need to understand what happens during this chemical process.
Concepts involved:
This leads to the formation of a complex ion in solution, namely, \(\text{[Al(H}_2\text{O})_6]^{3+}\).
Geometry deduction:
Conclusion:
Given the aforementioned chemical facts and reasoning, the ion formed in an acidified aqueous solution of aluminium chloride has an octahedral geometry.
Therefore, the correct answer is Octahedral.
Determine the Complex Ion Formed: In acidified aqueous solution, AlCl3 forms a complex ion, typically [Al(H2O)6]3+.
Analyze Geometry: - This complex ion has six ligands (water molecules) coordinated around the central aluminum ion. - Six ligands around a central atom generally form an octahedral geometry.
Conclusion: The geometry of [Al(H2O)6]3+ is octahedral.
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
