Step 1: Identifying the hybridization states of X and Y.
In the given reaction, Y represents the \({AlCl}_3\) complex in aqueous solution.
\({AlCl}_3\) in aqueous solution forms \({[Al(H}_2{O})_6]^{3+}\), where aluminium has an \( sp^3d^2 \) hybridization due to its octahedral geometry.
X represents the hydrated form of Al, where Al is surrounded by six water molecules, maintaining an octahedral coordination with \( sp^3d^2 \) hybridization. Step 2: Conclusion
Y (\({AlCl}_3\)) in aqueous solution undergoes hydration to form \({[Al(H}_2{O})_6]^{3+}\), which exhibits \( sp^3d^2 \) hybridization.
The correct answer is (4) \( sp^3, sp^3d^2 \).
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Approach Solution -2
To solve this problem, we need to determine the hybridization of aluminum in the complexes X and Y based on the given chemical reactions.
1. Analyzing the Reaction: The given reaction involves the conversion of complex Y (which contains AlCl3) into complex X by reacting with H2O and H+. The two different complexes indicate different hybridizations of the central aluminum atom, depending on the number of bonds and the type of ligands attached to aluminum.
2. Hybridization in Complex X: In the case of complex X (formed after reaction with H2O), aluminum is bonded to a water molecule. The hybridization of aluminum in this case is sp3d2. This is because aluminum typically uses the 3s, 3p, and 3d orbitals to form six bonds with ligands (in this case, possibly six equivalent bonds to water molecules), leading to an octahedral geometry.
3. Hybridization in Complex Y: In complex Y (with AlCl3), aluminum is bonded to three chloride ions, and it likely undergoes sp3 hybridization since it forms three bonds with the chloride ions in a planar geometry.
4. Final Answer: The hybridization of aluminum in complex Y is sp3, and in complex X is sp3d2.
Final Answer: The correct option is (D) sp3, sp3d2.
