The Group 5 \(\mathrm{A}\) 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 will draw the Lewis structures for NH₃, AsCl₅, and PF₆⁻. For NH₃: N has 5 valence electrons, and H has 1 valence electron. Since there are three H atoms, we have a total of 5 + (3×1) = 8 valence electrons. In NH₃, N forms three single bonds with three H atoms. N: •• | H:-(N)-H: | H: For AsCl₅: As has 5 valence electrons, and Cl has 7 valence electrons. Since there are five Cl atoms, we have a total of 5 + (5×7) = 40 valence electrons. In AsCl₅, As forms five single bonds with five Cl atoms. Cl: | Cl:-(As)-Cl: | Cl: For PF₆⁻: P has 5 valence electrons, F has 7 valence electrons, and since it has a negative charge, we add 1 more electron. Since there are six F atoms, we have a total of 5 + 1 + (6×7) = 48 valence electrons. In PF₆⁻, P forms six single bonds with six F atoms. F: | F:-(P)-F: | | F+-+-F:

Using VSEPR theory, we predict the molecular structures for each: NH₃: N is surrounded by three bonding pairs (with H) and one lone pair. So, its molecular structure is trigonal pyramidal. AsCl₅: As is surrounded by five bonding pairs (with Cl). So, its molecular structure is trigonal bipyramidal. PF₆⁻: P is surrounded by six bonding pairs (with F). So, its molecular structure is octahedral.

Now, we'll determine the hybridization for each central atom: NH₃: N has three sigma bonds and one lone pair, which gives it an electron group count of 4. Thus, N undergoes sp³ hybridization. AsCl₅: As has five sigma bonds, which gives it an electron group count of 5. Thus, As undergoes sp³d hybridization. PF₆⁻: P has six sigma bonds, which gives it an electron group count of 6. Thus, P undergoes sp³d² hybridization.

Nitrogen, which is in Group 5A, has only five valence electrons. Given its relatively small size and electronegativity, it can only expand its valence shell to accommodate four electron groups by forming three single bonds and having one lone pair (as seen in NH₃). Therefore, it cannot form NF₅ (which would require five single bonds) or NCl₆⁻ (which would require six single bonds). Nitrogen's inability to expand its valence shell beyond four electron groups is mainly due to the absence of available d-orbitals in its second energy level.