Consider the following \(\mathrm{XF}_{4}\) ions: $\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+},\( and \)\mathrm{AlF}_{4}^{-}$ (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electron-domain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry? (d) Which of the ions will exhibit a see-saw molecular geometry?

To determine the number of valence electrons around the central atom for each ion, we will sum the valence electrons contributed by each atom and add or subtract the charge of the ion. - \(\mathrm{PF}_{4}^{-}\): P has 5 valence electrons, each F has 7, and the ion has a -1 charge. Total valence electrons: \(5 + 4(7) - 1 = 33\). - \(\mathrm{BrF}_{4}^{-}\): Br has 7 valence electrons, each F has 7, and the ion has a -1 charge. Total valence electrons: \(7 + 4(7) - 1 = 41\). - \(\mathrm{ClF}_{4}^{+}\): Cl has 7 valence electrons, each F has 7, and the ion has a +1 charge. Total valence electrons: \(7 + 4(7) + 1 = 35\). - \(\mathrm{AlF}_{4}^{-}\): Al has 3 valence electrons, each F has 7, and the ion has a -1 charge. Total valence electrons: \(3 + 4(7) - 1 = 31\).

Now we will determine how many electron pairs surround the central atom for each ion. - \(\mathrm{PF}_{4}^{-}\): With 33 valence electrons, P uses 8 electrons in bonding with 4 F atoms. It leaves 25 electrons for the central P atom, meaning 12 lone pairs (an octet). - \(\mathrm{BrF}_{4}^{-}\): With 41 valence electrons, Br uses 8 electrons in bonding with 4 F atoms. It leaves 33 electrons for the central Br atom, meaning 16 lone pairs (more than an octet). - \(\mathrm{ClF}_{4}^{+}\): With 35 valence electrons, Cl uses 8 electrons in bonding with 4 F atoms. It leaves 27 electrons for the central Cl atom, meaning 13 lone pairs (more than an octet). - \(\mathrm{AlF}_{4}^{-}\): With 31 valence electrons, Al uses 8 electrons in bonding with 4 F atoms. It leaves 23 electrons for the central Al atom, meaning 11 lone pairs (an octet). #a) Answer#: The ions \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) have more than an octet of electrons around the central atom. #b) Identification of ions with the same electron-domain and molecular geometries#

We will determine the electron-domain geometries of each ion by considering the number of electron domains around the central atom (lone pairs + bonding pairs). - \(\mathrm{PF}_{4}^{-}\): 4 bonding pairs + 1 lone pair = 5 electron domains. The electron-domain geometry is trigonal bipyramidal. - \(\mathrm{BrF}_{4}^{-}\): 4 bonding pairs + 2 lone pairs = 6 electron domains. The electron-domain geometry is octahedral. - \(\mathrm{ClF}_{4}^{+}\): 4 bonding pairs + 1 lone pair = 5 electron domains. The electron-domain geometry is trigonal bipyramidal. - \(\mathrm{AlF}_{4}^{-}\): 4 bonding pairs + 0 lone pairs = 4 electron domains. The electron-domain geometry is tetrahedral.

Now we will determine the molecular geometries of each ion by considering the arrangement of atoms in space. - \(\mathrm{PF}_{4}^{-}\): With 4 bonding pairs and 1 lone pair in a trigonal bipyramidal arrangement, the molecular geometry is see-saw. - \(\mathrm{BrF}_{4}^{-}\): With 4 bonding pairs and 2 lone pairs in an octahedral arrangement, the molecular geometry is square planar. - \(\mathrm{ClF}_{4}^{+}\): With 4 bonding pairs and 1 lone pair in a trigonal bipyramidal arrangement, the molecular geometry is see-saw. - \(\mathrm{AlF}_{4}^{-}\): With 4 bonding pairs and 0 lone pairs in a tetrahedral arrangement, the molecular geometry is tetrahedral. #b) Answer#: The ion \(\mathrm{AlF}_{4}^{-}\) has the same electron-domain and molecular geometries (both tetrahedral). #c) Identification of ions with an octahedral electron-domain geometry# #c) Answer#: From our previous analysis, we can see that the ion \(\mathrm{BrF}_{4}^{-}\) has an octahedral electron-domain geometry. #d) Identification of ions with a see-saw molecular geometry# #d) Answer#: From our previous analysis, we can see that the ions \(\mathrm{PF}_{4}^{-}\) and \(\mathrm{ClF}_{4}^{+}\) exhibit a see-saw molecular geometry.