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Chapter 10: Problem 33

Describe the change in hybridization (if any) of the \(\mathrm{Al}\) atom in this reaction: $$ \mathrm{AlCl}_{3}+\mathrm{Cl}^{-} \longrightarrow \mathrm{AlCl}_{4}^{-} $$

Short Answer

Expert verified
The hybridization of the aluminum atom changes from \(sp^2\) to \(sp^3\) during the reaction.

Step by step solution

01

Identify the original hybridization state

In the beginning, the aluminum atom in the \(AlCl_3\) compound is making three bonds, which correspond to the three chlorine atoms. Since it has 3 sigma bonds and no lone pairs of electrons, this implies that the hybridization state of the aluminum atom is \(sp^2\).
02

Identify the final hybridization state

After receiving an extra \(Cl^-\) ion, the aluminum atom is now part of a \(AlCl_4^-\) ion and it is making four bonds, which correspond to the four chlorine atoms. In this case, considering 4 sigma bonds and no lone pairs of electrons, this implies that the hybridization state of the aluminum atom is now \(sp^3\).
03

Describe the change in hybridization

Comparing the initial state of hybridization (\(sp^2\)) with the final state of hybridization (\(sp^3\)), it can be concluded that the hybridization of the aluminum atom has changed from \(sp^2\) to \(sp^3\) as a result of this reaction.

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Most popular questions from this chapter

Predict the geometry of sulfur dichloride \(\left(\mathrm{SCl}_{2}\right)\) and the hybridization of the sulfur atom.

Predict the bond angles for these molecules: (a) \(\mathrm{BeCl}_{2},\) (b) \(\mathrm{BCl}_{3},\) (c) \(\mathrm{CCl}_{4},\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}\) (e) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (arrangement of atoms: \(\mathrm{ClHgHgCl}\) ), (f) \(\mathrm{SnCl}_{2}\) (g) \(\mathrm{H}_{2} \mathrm{O}_{2},\) (h) \(\mathrm{SnH}_{4}\).

Sketch the bond moments and resultant dipole moments for these molecules: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{PCl}_{3}, \mathrm{XeF}_{4}, \mathrm{PCl}_{5}\) \(\mathrm{SF}_{6}\)

List these molecules in order of increasing dipole moment: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CBr}_{4}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{HF}, \mathrm{NH}_{3}, \mathrm{CO}_{2}\)

Sketch the shapes of these molecular orbitals: \(\sigma_{1 s}\), \(\sigma_{1 s}^{\star}, \pi_{2 p},\) and \(\pi_{2 p}^{\star}\) How do their energies compare?
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