(a) Is the molecule BF \(_{3}\) polar or nonpolar? (b) If you react BF \(_{3}\) to make the ion \(\mathrm{BF}_{3}^{2-}\) , is this ion planar? (c) Does the molecule BF\(_{2}\)Cl have a dipole moment?

To determine if the BF\(_3\) molecule is polar or nonpolar, we must first examine its molecular geometry and the electronegativity of its atoms. 1. Draw the Lewis structure for the molecule BF\(_3\). Count the total number of valence electrons of the molecule (Boron has 3 valence electrons and each Fluorine atom has 7 valence electrons). The total number of valence electrons is 3 + 3 × 7 = 24. 2. Use the electron-pair geometry and VSEPR theory to determine the molecular geometry. For BF\(_3\), the central atom is Boron with three Fluorine atoms attached. In this case, the molecular geometry is trigonal planar with bond angles of 120°. 3. Calculate the electronegativity differences between the atoms. The electronegativity values for Boron and Fluorine are 2.0 and 3.98, respectively. The difference between the two is 1.98, which indicates the presence of polar bonds in the molecule. In conclusion, even though BF\(_3\) has polar bonds, its molecular geometry (trigonal planar) helps to cancel out the dipole moments, resulting in a nonpolar molecule.

To determine if the \(\mathrm{BF}_{3}^{2-}\) ion is planar, we need to examine its molecular geometry. 1. Calculate the total number of valence electrons for the ion, considering the charge. Boron has 3 valence electrons, each Fluorine atom has 7 valence electrons, and the ion has a charge of 2-, which gives two extra electrons. The total number of valence electrons is 3 + 3 × 7 + 2 = 26. 2. Draw the Lewis structure for the ion \(\mathrm{BF}_{3}^{2-}\). There will be a single bond between the Boron atom and each Fluorine atom, and an extra lone pair on each Fluorine atom, forming resonance structures. 3. Use the electron-pair geometry and VSEPR theory to determine its molecular geometry. The electron-pair geometry of the central Boron atom will be tetrahedral, with three Fluorine atoms and one lone pair. 4. By considering the molecular geometry (by ignoring the lone pair), we find that \(\mathrm{BF}_{3}^{2-}\) has a trigonal pyramidal geometry, which is not planar. In conclusion, the \(\mathrm{BF}_{3}^{2-}\) ion is not planar due to its trigonal pyramidal molecular geometry.

To determine if the BF\(_2\)Cl molecule has a dipole moment, we must consider its molecular geometry and the electronegativity of its atoms. 1. Calculate the total number of valence electrons for the BF\(_2\)Cl molecule (Boron has 3 valence electrons, each Fluorine atom has 7 valence electrons, and Chlorine has 7 valence electrons). The total number of valence electrons is 3 + 2 × 7 + 7 = 24. 2. Use the electron-pair geometry and VSEPR theory to determine its molecular geometry. The central atom is Boron, with two Fluorine atoms and one Chlorine atom attached to it. The molecular geometry is also trigonal planar, with bond angles of 120°. 3. Calculate the electronegativity differences between the atoms. The difference between Boron and Chlorine (2.0 and 3.16, respectively) is 1.16, and the difference between Boron and Fluorine was calculated in step (a) as 1.98. These differences indicate the presence of polar bonds in the molecule. In conclusion, since the BF\(_2\)Cl molecule has polar bonds and its molecular geometry is not symmetrical (due to the different electronegativity values of Fluorine and Chlorine), the molecule has a dipole moment.