
Consider the following Lewis acid-base reaction, \[AlC{l_3} + C{l^ - } \to AlCl_4^ - \]. Hybridization of \[Al\]in \[AlC{l_3}\]and \[AlCl_4^ - \]are respectively:
A. \[s{p^3},ds{p^2}\]
B. \[s{p^2},ds{p^2}\]
C. \[ds{p^2},s{p^3}d\]
D. \[s{p^2},s{p^3}\]
Answer
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Hint: The Lewis acid reacts with the Lewis base and then forms an acid-base compound that changes the p-orbital from lower to higher degree.
Complete Step by Step Solution:
Lewis acid-base reaction:
1. In Lewis acid-base reactions, a base donates electrons to acid and forms Lewis acid-base adduct.
2. In a Lewis acid-base adduct, a coordinate covalent bond is formed between the Lewis base and acid.
Hybridization:
1. Intermixing of orbitals with slightly different energies, resulting in a new set of orbitals with equally strong energies and shapes created through redistribution of energies.
2. For example, if there are one 2s and three 2p-orbitals of the element hybridize then there is a formation of \[s{p^3}\]hybrid orbitals.
Hybridization of \[Al\]in \[AlC{l_3}\]and \[AlCl_4^ - \]:
1. In the given reaction, \[AlC{l_3}\]acts as a Lewis acid and \[Al\]acts as a Lewis base. There is a formation acid-base adduct (\[AlCl_4^ - \]).
2. The aluminium trichloride at the excited state has one 2s and two 2p-orbitals so it has \[s{p^2}\]hybrid orbitals.
3. There are one 2s and three 2p-orbitals of aluminium tetrachloride at the excited state that forms \[s{p^3}\]hybrid orbitals.
4. The shape of the \[AlC{l_3}\]and \[AlCl_4^ - \]is a trigonal planar and tetrahedral structure.
Thus, the option (D) is correct.
Additional information:
1. There are basically three types of hybridization which are \[sp,s{p^2},s{p^3}\].
2. The types of hybridization also indicate the geometry of the compounds.
3. There are basically three types of hybridization that involve d-orbital that are \[ds{p^2},s{p^3}d,s{p^3}{d^2}\] but the last one has two different structures.
4. The structure of \[sp,s{p^2},s{p^3}\]are linear, triangular, and tetrahedral.
5. The structures of d-orbital hybridization are square planar, trigonal bipyramidal, square pyramidal, and octahedral.
Note: The number of hybrid orbitals is equal to the number of atomic orbitals which are to be hybridised and also hybridization orbitals are equivalent in energy and shape.
Complete Step by Step Solution:
Lewis acid-base reaction:
1. In Lewis acid-base reactions, a base donates electrons to acid and forms Lewis acid-base adduct.
2. In a Lewis acid-base adduct, a coordinate covalent bond is formed between the Lewis base and acid.
Hybridization:
1. Intermixing of orbitals with slightly different energies, resulting in a new set of orbitals with equally strong energies and shapes created through redistribution of energies.
2. For example, if there are one 2s and three 2p-orbitals of the element hybridize then there is a formation of \[s{p^3}\]hybrid orbitals.
Hybridization of \[Al\]in \[AlC{l_3}\]and \[AlCl_4^ - \]:
1. In the given reaction, \[AlC{l_3}\]acts as a Lewis acid and \[Al\]acts as a Lewis base. There is a formation acid-base adduct (\[AlCl_4^ - \]).
2. The aluminium trichloride at the excited state has one 2s and two 2p-orbitals so it has \[s{p^2}\]hybrid orbitals.
3. There are one 2s and three 2p-orbitals of aluminium tetrachloride at the excited state that forms \[s{p^3}\]hybrid orbitals.
4. The shape of the \[AlC{l_3}\]and \[AlCl_4^ - \]is a trigonal planar and tetrahedral structure.
Thus, the option (D) is correct.
Additional information:
1. There are basically three types of hybridization which are \[sp,s{p^2},s{p^3}\].
2. The types of hybridization also indicate the geometry of the compounds.
3. There are basically three types of hybridization that involve d-orbital that are \[ds{p^2},s{p^3}d,s{p^3}{d^2}\] but the last one has two different structures.
4. The structure of \[sp,s{p^2},s{p^3}\]are linear, triangular, and tetrahedral.
5. The structures of d-orbital hybridization are square planar, trigonal bipyramidal, square pyramidal, and octahedral.
Note: The number of hybrid orbitals is equal to the number of atomic orbitals which are to be hybridised and also hybridization orbitals are equivalent in energy and shape.
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