Question

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{PF}_{3}\), (b) \(\mathrm{CH}_{3}{ }^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{4}^{-}(\mathrm{e}) \mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

Short answer

The Lewis structures, electron-domain geometries, and molecular geometries for the given molecules and ions are: (a) \(\mathrm{PF}_{3}\): - Lewis structure: P in the center with 3 F atoms and 1 lone pair. - Electron-domain geometry: Tetrahedral - Molecular geometry: Trigonal pyramidal (b) \(\mathrm{CH}_{3}{ }^{+}\): - Lewis structure: C in the center with 3 H atoms. - Electron-domain geometry: Trigonal planar - Molecular geometry: Trigonal planar (c) \(\mathrm{BrF}_{3}\): - Lewis structure: Br in the center with 3 F atoms and 2 lone pairs. - Electron-domain geometry: Octahedral - Molecular geometry: T-Shaped (d) \(\mathrm{ClO}_{4}^{-}\): - Lewis structure: Cl in the center with 4 O atoms and -1 charge on the extra Cl lone pair. - Electron-domain geometry: Tetrahedral - Molecular geometry: Tetrahedral (e) \(\mathrm{XeF}_{2}\): - Lewis structure: Xe in the center with 2 F atoms and 3 lone pairs. - Electron-domain geometry: Trigonal Bipyramidal - Molecular geometry: Linear (f) \(\mathrm{BrO}_{2}^{-}\): - Lewis structure: Br in the center with 2 O atoms, 1 extra electron, and -1 charge on Br. - Electron-domain geometry: Tetrahedral - Molecular geometry: Bent or V-shaped

Step-by-step solution

Find the number of valence electrons

P has 5 valence electrons and F has 7. There are 3 F atoms, so the total number of valence electrons = 5 + (3*7) = 26.

Draw the Lewis structure

Put P in the center and surround it with the 3 F atoms. Connect these atoms with single bonds. Each bond uses 2 electrons, so we have 2*3=6 electrons in bonds, leaving 20 electrons to be placed as lone pairs. Add 3 lone pairs (6 electrons) to each F atom, and one lone pair (2 electrons) on P, leaving 0 electrons. The formal charge on each atom is 0, and the Lewis structure is complete.

Determine electron-domain and molecular geometries

There are 4 electron domains (3 single bonds and 1 lone pair) around P. According to VSEPR, with 4 electron domains, the electron-domain geometry is tetrahedral. Considering only the bond positions, the molecular geometry is trigonal pyramidal. (b) \(\mathrm{CH}_{3}^{+}\)

Find the number of valence electrons

C has 4 valence electrons and H has 1. There are 3 H atoms, and we subtract 1 electron due to the +1 charge. The total number of valence electrons = 4 + (3*1) - 1 = 6.

Draw the Lewis structure

Put C in the center and surround it with the 3 H atoms. Connect these atoms with single bonds. Each bond uses 2 electrons, so we have 2*3=6 electrons in bonds, leaving 0 electrons. The formal charge on C is +1, and the Lewis structure is complete.

Determine electron-domain and molecular geometries

There are 3 electron domains (3 single bonds) around C. According to VSEPR, with 3 electron domains, the electron-domain and molecular geometries are both trigonal planar. (c) \(\mathrm{BrF}_{3}\) Follow steps 1-3 above to find the Lewis structure, electron-domain geometry, and molecular geometry. The final answer is: Lewis structure: Br in the center; 3 F atoms and 2 lone pairs around Br. Electron-domain geometry: Octahedral Molecular geometry: T-Shaped (d) \(\mathrm{ClO}_{4}^{-}\) Lewis structure: Cl in the center; 4 O atoms around Cl and -1 charge on the extra Cl lone pair. Electron-domain geometry: Tetrahedral Molecular geometry: Tetrahedral (e) \(\mathrm{XeF}_{2}\) Lewis structure: Xe in the center; 2 F atoms around Xe and 3 lone pairs. Electron-domain geometry: Trigonal Bipyramidal Molecular geometry: Linear (f) \(\mathrm{BrO}_{2}^{-}\) Lewis structure: Br in the center; 2 O atoms around Br, 1 extra electron, and -1 charge on Br. Electron-domain geometry: Tetrahedral Molecular geometry: Bent or V-shaped