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For BBr₃, write about its electron geometry, molecule shape, bond angles around the central atom, and molecular polarity (polar or non-polar).
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Step 1**Electron Geometry of BBr₃:**
The electron geometry of BBr₃ can be determined using the Valence Shell Electron Pair Repulsion (VSEPR) theory. Boron (B) has 3 valence electrons, and each Bromine (Br) atom contributes 7 valence electrons, resulting in a total of 10 valence electrons. These electrons are arranged around the central boron atom in such a way that they repel each other as much as possible. This arrangement leads to a trigonal planar geometry with 120° bond angles. Explanation: \begin{array}{c} \text{Central atom: Boron (B)} \ \text{Number of electron pairs around B: 3 (2 from B-Br bonds and 1 lone pair)} \ \text{Electron geometry: Trigonal planar (120° bond angles)} \end{array}
Step 2**Molecular Shape of BBr₃:**
Since there are no lone pairs on the central boron atom in BBr₃, the molecular shape is identical to the electron geometry. Therefore, the molecular shape of BBr₃ is also trigonal planar. Explanation: \begin{array}{c} \text{Central atom: Boron (B)} \ \text{Number of electron pairs around B: 3 (2 from B-Br bonds and 1 lone pair)} \ \text{Molecular shape: Trigonal planar (120° bond angles)} \end{array}
Final Answer
- Electron geometry: Trigonal planar - Molecular shape: Trigonal planar - Bond angles around the central atom: 120° - Molecular polarity: Non-polar
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