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Write the formation of BCl3 by valence bond theory​

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User Haplo
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In valence bond theory, the formation of BCl3 (boron trichloride) can be explained by the overlap of atomic orbitals. Here's a simplified description:

1. **Atomic Orbitals:** Boron (B) has three valence electrons, and chlorine (Cl) has seven valence electrons. Boron's electron configuration is 1s²2s²2p¹, while chlorine's electron configuration is 1s²2s²2p⁶3s²3p⁵.

2. **Hybridization:** To form BCl3, boron undergoes hybridization. It promotes one of its 2p electrons to the empty 2pz orbital, resulting in sp² hybridization. This forms three equivalent sp² hybrid orbitals in a trigonal planar arrangement around boron.

3. **Overlap of Orbitals:** Each chlorine atom has one unpaired electron in its 3p orbital. These unpaired electrons overlap with the three sp² hybrid orbitals of boron, forming three sigma (σ) bonds. These sigma bonds result from head-on overlap between the orbitals.

4. **Molecular Geometry:** The overlapping of these orbitals and the formation of sigma bonds give BCl3 its trigonal planar molecular geometry.

So, in summary, BCl3 is formed by the overlap of boron's sp² hybrid orbitals with the unpaired 3p orbitals of three chlorine atoms, resulting in the formation of three sigma bonds and a trigonal planar molecular shape.

(. ❛ ᴗ ❛.)

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User Cudds
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