The Group \(5 \mathrm{A}\) (15) elements can form molecules or ions that involve three, five, or six covalent bonds; \(\mathrm{NH}_{3}, \mathrm{AsCl}_{5},\) and \(\mathrm{PF}_{6}^{-}\) are examples. Draw the Lewis structure for each of these substances, and predict the molecular structure and hybridization for each. Why doesn't \(\mathrm{NF}_{5}\) or \(\mathrm{NCl}_{6}^{-}\) form?

First, we need to draw the Lewis structures for NH3, AsCl5, and PF6-. To do this, we count the total number of valence electrons for each compound, and distribute them to form bonds and satisfy the octet rule for each atom. For NH3 (ammonia): 1. Nitrogen has 5 valence electrons, and each hydrogen atom has 1 valence electron. The total valence electrons for ammonia are 5 (from N) + 3*1 (from H) = 8 electrons. 2. Nitrogen forms three single covalent bonds with three hydrogen atoms, using 6 valence electrons. 3. The remaining 2 valence electrons form a lone pair on nitrogen. The Lewis structure for NH3 is: H | H-N-H : (the lone pair is represented by two dots) For AsCl5 (arsenic pentachloride): 1. Arsenic has 5 valence electrons, and each chlorine atom has 7 valence electrons. The total valence electrons for AsCl5 are 5 (from As) + 5*7 (from Cl) = 40 electrons. 2. Arsenic forms single covalent bonds with five chlorine atoms, using 10 valence electrons. 3. Each chlorine atom has 3 lone pairs, using the remaining 30 valence electrons. The Lewis structure for AsCl5 is: Cl | Cl-As-Cl | Cl For PF6- (hexafluorophosphate ion): 1. Phosphorus has 5 valence electrons, each fluorine atom has 7 valence electrons, and the overall ion has an extra electron (due to the -1 charge). The total valence electrons for PF6- are 5 (from P) + 6*7 (from F) + 1(extra electron) = 48 electrons. 2. Phosphorus forms single covalent bonds with six fluorine atoms, using 12 valence electrons. 3. Each fluorine atom has 3 lone pairs, using the remaining 36 valence electrons. The Lewis structure for PF6- is: F \ F-P-F | \ F F

We will use the VSEPR theory to predict the molecular geometry for each compound, and then determine the corresponding hybridization. For NH3: 1. The nitrogen atom in NH3 has 3 bonding electron pairs and 1 lone pair. Therefore, the VSEPR electron pair arrangement is tetrahedral. The molecular geometry is trigonal pyramidal, due to the presence of 3 bonded atoms and 1 lone pair. 2. Hybridization for a tetrahedral electron pair arrangement is sp3. For AsCl5: 1. The arsenic atom in AsCl5 has 5 bonding electron pairs and no lone pair. The VSEPR electron pair arrangement and the molecular geometry are both trigonal bipyramidal. 2. Hybridization for a trigonal bipyramidal electron pair arrangement is sp3d. For PF6-: 1. The phosphorus atom in PF6- has 6 bonding electron pairs and no lone pair. The VSEPR electron pair arrangement and the molecular geometry are both octahedral. 2. Hybridization for an octahedral electron pair arrangement is sp3d2.

NF5 and NCl6- are not stable compounds because nitrogen, being in the 2nd period of the periodic table, has a smaller size and can only accommodate a maximum of 4 electron pairs (including bond pairs and lone pairs) in its valence shell, which corresponds to an sp3 hybridization. This is due to the unavailability of d orbitals in the 2nd period elements. In the hypothetical NF5 molecule, nitrogen would have to form 5 bond pairs, which is not possible due to its limited number of available orbitals. Similarly, in the hypothetical NCl6- ion, nitrogen would have to form 6 bond pairs, which is also not possible due to the same reasoning. Hence, NF5 and NCl6- do not form stable compounds.