\(\mathrm{BH}_{3}\) and \(\mathrm{PH}_{3}\) each contain four atoms, with the three hydrogens surrounding the central atom. Do the two molecules have the same electron-group geometry? Do they have the same molecular shape? Name the geometry and the shape for both molecules. You know from Problem \(6.59\) that \(\mathrm{PH}_{3}\) is nonpolar. What about \(\mathrm{BH}_{3} ?\) Is it polar or nonpolar? Explain.

To start, we need to draw the Lewis structures for both \(\mathrm{B}\) and \(\mathrm{P}\) atoms and find out the number of electron pairs surrounding them. In \(\mathrm{BH}_{3}\), the central \(\mathrm{B}\) atom has 3 valence electrons. Each \(\mathrm{H}\) atom contributes 1 electron, making a total of 6 electrons (3 from \(\mathrm{B}\) and 3 from the three \(\mathrm{H}\) atoms). In \(\mathrm{PH}_{3}\), the central \(\mathrm{P}\) atom has 5 valence electrons. Each \(\mathrm{H}\) atom contributes 1 electron, making a total of 8 electrons (5 from \(\mathrm{P}\) and 3 from the three \(\mathrm{H}\) atoms).

Now, let's determine the electron-group geometry for each molecule. In \(\mathrm{BH}_{3}\), there are 3 bonding electron pairs around the central atom and no lone pairs. These electron pairs will arrange themselves in a trigonal planar geometry to minimize the repulsion between them. In \(\mathrm{PH}_{3}\), there are 3 bonding electron pairs around the central atom and 1 lone pair. The lone pair takes the position of an electron group, so the electron-group geometry will be tetrahedral. Therefore, the electron-group geometries of \(\mathrm{BH}_{3}\) and \(\mathrm{PH}_{3}\) are different.

Now, let's determine the molecular shape, which is influenced by the electron-group geometry. In \(\mathrm{BH}_{3}\), the electron-group geometry is trigonal planar and all electron groups are bonding electron pairs. Therefore, the molecular shape of \(\mathrm{BH}_{3}\) is also trigonal planar. In \(\mathrm{PH}_{3}\), the electron-group geometry is tetrahedral, with one lone pair and three bonding electron pairs. The molecular shape is determined by the positions of the bonding electron pairs only, which take on a trigonal pyramidal shape in this case. Therefore, the molecular shapes of \(\mathrm{BH}_{3}\) and \(\mathrm{PH}_{3}\) are different.

To determine if \(\mathrm{BH}_{3}\) is polar or nonpolar, we need to consider the electronegativity difference between \(\mathrm{B}\) and \(\mathrm{H}\) and the molecular shape of the molecule. The electronegativity difference between \(\mathrm{B}\) and \(\mathrm{H}\) is small, so the \(\mathrm{B-H}\) bonds can be considered nonpolar. Furthermore, the molecular shape of \(\mathrm{BH}_{3}\) is trigonal planar, which has a symmetrical distribution of charge. Therefore, the polarity of the bonds cancels out, making \(\mathrm{BH}_{3}\) a nonpolar molecule. In conclusion: \(\mathrm{BH}_{3}\): Electron-group geometry - Trigonal planar. Molecular shape - Trigonal planar. Polarity - Nonpolar. \(\mathrm{PH}_{3}\): Electron-group geometry - Tetrahedral. Molecular shape - Trigonal pyramidal. Polarity - Nonpolar.