Question 19 of 32intermediate🔧 ApplyShort Answer2 marks

Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl₃ + Cl⁻ → AlCl₄⁻

Correct Answer

In AlCl3\text{AlCl}_3, the Al atom is bonded to three Cl atoms. To form three equivalent bonds, one ss and two pp-orbitals of the Al atom intermix, resulting in sp2sp^2 hybridisation with a trigonal planar geometry.

When Cl−\text{Cl}^- reacts with AlCl3\text{AlCl}_3, the Al atom accepts the lone pair from Cl−\text{Cl}^- into its vacant pp-orbital, forming a fourth bond. To accommodate four bonds, one ss and three pp-orbitals intermix, resulting in sp3sp^3 hybridisation with a tetrahedral geometry.

Thus, the hybridisation of the Al atom changes from sp2sp^2 in AlCl3\text{AlCl}_3 to sp3sp^3 in AlCl4−\text{AlCl}_4^-.

Exercise: EXERCISES | Q: 4.25 | (Page 34)
For More Understanding

Explanation

The Al atom in AlCl3 has 3 sigma bonds and an empty p-orbital, making it sp2 hybridised (similar to BCl3 mentioned in the text). When Cl- donates a lone pair to form AlCl4-, Al forms 4 sigma bonds, requiring one s and three p orbitals to mix, resulting in sp3 hybridisation (similar to CH4).

Solution Steps

  1. Step 1: Identify the hybridisation in AlCl3. Al forms 3 bonds using one s and two p orbitals, giving sp2sp^2 hybridisation.

  2. Step 2: Identify the hybridisation in AlCl4-. Al forms 4 bonds by accepting a lone pair from Cl-, using one s and three p orbitals, giving sp3sp^3 hybridisation.

  3. Step 3: Conclude that the hybridisation changes from sp2sp^2 to sp3sp^3.

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