Write Lewis structures for \(\mathrm{NF}_{3}\) and \(\mathrm{PF}_{5}\). On the basis of hybrid orbitals, explain the fact that \(\mathrm{NF}_{3}, \mathrm{PF}_{3}\), and \(\mathrm{PF}_{5}\) are stable molecules, but \(\mathrm{NF}_{5}\) does not exist.

Understanding valence electrons is crucial for predicting how atoms will bond and form molecules. Valence electrons are the outermost electrons of an atom and are chiefly responsible for the chemical properties of the element. They are the electrons involved in bonding because their position in the atom allows them to easily interact with other atoms.

For example, nitrogen (\textbf{N}) has 5 valence electrons in its outer shell, while fluorine (\textbf{F}) has 7. It's these electrons that interact to form compounds, such as \textbf{NF}\(_3\). When writing the Lewis structure, a key task is accounting for all the valence electrons, making sure they are distributed in a way that satisfies the bonding needs of the constituent atoms.

In the case of \textbf{NF}\(_3\), nitrogen shares three of its valence electrons with three fluorine atoms, forming three single bonds, and keeps two as a lone pair. This sharing allows all the atoms involved to reach a more stable state.

The octet rule is a fundamental principle that guides the formation of stable molecular structures. It states that atoms tend to form bonds in such a way that they each end up with eight electrons in their valence shell, emulating the electron configuration of noble gases, which are inherently stable. However, there are exceptions to this rule, particularly for elements beyond the second period of the periodic table that can have more than eight electrons.

In the creation of \textbf{Lewis structures}, the octet rule is adhered to strictly for elements like nitrogen and fluorine. For example, in \textbf{NF}\(_3\), each fluorine achieves an octet by obtaining one electron from nitrogen. Conversely, nitrogen achieves an octet by sharing each of its valence electrons with different fluorine atoms and keeping a pair unshared. This sharing results in a stable structure. However, nitrogen’s inability to expand its octet explains why \textbf{NF}\(_5\) does not exist; it cannot form bonds with five fluorine atoms without having more than eight electrons in its valence shell, which is not favorable for nitrogen.

Hybrid orbitals provide insight into the actual shape and bonding capability of molecules, which can’t be fully explained by the octet rule alone. When an atom forms covalent bonds, its atomic orbitals can mix to create new orbitals—hybrid orbitals—that are degenerate in energy and oriented in specific geometries optimum for bonding.

In \textbf{PF}\(_5\), the phosphorus atom forms five bonds using sp^3d hybridization, which allows it to have an extended octet and accommodate 10 electrons in its valence shell. This hybridization gives \textbf{PF}\(_5\) a trigonal bipyramidal shape, with each fluorine atom bonded to the central phosphorus atom, thus explaining its stability.

In contrast, nitrogen in \textbf{NF}\(_3\) forms sp^3 hybrid orbitals, but since nitrogen cannot expand its octet like phosphorus, it can only form three sigma bonds with fluorine, resulting in a pyramid shape for \textbf{NF}\(_3\). The existence of sp^3d or other such hybrid orbitals allows atoms like phosphorus to form more bonds than would be predicted by the octet rule alone, while nitrogen is limited by its inability to form such extended orbital hybridizations.