Question

\(\mathrm{PCl}_{3}\) and \(\mathrm{PCl}_{5}\) both exists; \(\mathrm{NCl}_{3}\) exists but \(\mathrm{NCl}_{5}\) does not exist. It is due to (a) lower electronegativity of P than \(\mathrm{N}\) (b) lower tendency of \(\mathrm{N}\) to form covalent bond (c) availability of vacant d orbital in P but not in \(\mathrm{N}\) (d) statement is itself incorrect

Short answer

(c) availability of vacant d orbital in P but not in N

Step-by-step solution

Understanding the Compounds

First, let's understand the compounds given in the exercise. We have two phosphorus compounds, \(\mathrm{PCl}_3\) and \(\mathrm{PCl}_5\), and two nitrogen compounds, \(\mathrm{NCl}_3\) and \(\mathrm{NCl}_5\). We need to find out why \(\mathrm{NCl}_5\) does not exist while the other three compounds do.

Exploring the Valence Shell Electron Configuration

Phosphorus has the electronic configuration of \([Ne] 3s^2 3p^3\). It has 5 valence electrons and can expand its octet by using the available \(3d\) orbitals. This is why it can form \(\mathrm{PCl}_5\). Nitrogen, on the other hand, has an electronic configuration of \([He] 2s^2 2p^3\). It lacks available \(d\) orbitals in its valence shell at the second energy level, so it cannot expand its octet beyond 8 electrons.

Analyzing the Options

Option (a) suggests lower electronegativity of \(P\), which is not directly relevant to octet expansion. Option (b) suggests lower tendency of \(N\) to form covalent bonds, which is incorrect as \(N\) can form bonds effectively (e.g., \(\mathrm{NH}_3\)). Option (c) refers to the absence of \(d\) orbitals in \(N\), making it impossible for nitrogen to form \(\mathrm{NCl}_5\). Option (d) implies the statement is incorrect, which it isn't.

Conclusion

After analyzing each option, it becomes clear that the key factor is the availability of vacant \(d\) orbitals that phosphorus has and nitrogen lacks, which prevents \(\mathrm{NCl}_5\) from forming.

Key concepts

Valence Shell Electron Configuration

Understanding the valence shell electron configuration is crucial to explaining why certain compounds exist or don't exist. Phosphorus boasts an electron configuration of \([Ne] 3s^2 3p^3\), which means it has five electrons in its outer shell. This arrangement enables phosphorus to both achieve a stable structure like in \(\mathrm{PCl}_3\) or expand its bonding capacity through access to \(3d\) orbitals to form \(\mathrm{PCl}_5\). On the other hand, nitrogen's electron configuration is \([He] 2s^2 2p^3\), providing only five valence electrons and no access to additional orbitals for bonding expansion. This means nitrogen can't expand its valence shell to accommodate more than eight electrons.

Octet Expansion

Octet expansion is a concept that explains why certain elements can form more bonds than might initially seem possible. Elements from the third period and beyond, like phosphorus, can make use of \(d\) orbitals in chemical bonding. For phosphorus, having a capacity to expand the octet means it can form compounds like \(\mathrm{PCl}_5\). However, nitrogen, which resides in the second period, lacks vacant \(d\) orbitals. Consequently, it is restricted to an octet configuration, limiting it to compounds such as \(\mathrm{NCl}_3\) and not \(\mathrm{NCl}_5\).

Phosphorus Compounds

Phosphorus forms a range of compounds, with \(\mathrm{PCl}_3\) and \(\mathrm{PCl}_5\) being notable examples. In \(\mathrm{PCl}_3\), phosphorus forms three covalent bonds with chlorine, utilizing its five valence electrons. However, due to the availability of \(d\) orbitals, phosphorus can expand its valence shell to accommodate ten electrons, as seen in \(\mathrm{PCl}_5\). This makes phosphorus quite versatile in bonding, able to form more complex compounds thanks to its expanded octet potential.

Nitrogen Compounds

Nitrogen typically forms compounds where it can complete its octet, such as \(\mathrm{NCl}_3\) with three bonds, similar to \(\mathrm{NH}_3\). With its valence shell configuration, consisting of three \(p\) orbitals and no access to \(d\) orbitals, nitrogen cannot expand its octet. Hence, compounds like \(\mathrm{NCl}_5\) are not feasible. Nitrogen’s ability to form stable bonds is not diminished, only its ability to expand the octet, confining it to structures that respect the rule of eight.

D Orbitals

D orbitals, present from the third period onwards, expand the number of possible bonds an element can form. Phosphorus utilizes these \(d\) orbitals to form compounds like \(\mathrm{PCl}_5\), as these orbitals can be occupied by electrons when the element forms more bonds than its typical valence would suggest. However, nitrogen, being in the second period, doesn't have access to these \(d\) orbitals, limiting its bonding structure to an octet. This distinction underlies many of the differences in chemical behavior between phosphorus and nitrogen.

Covalent Bonding

Covalent bonding involves the sharing of electron pairs between atoms. For nitrogen and phosphorus, their electron configurations dictate how they form these bonds. Whereas nitrogen readily forms three covalent bonds in compounds like \(\mathrm{NH}_3\) and \(\mathrm{NCl}_3\), phosphorus can extend its bonding through valence shell expansion. This allows phosphorus to participate in more complex covalent bonding arrangements, as seen in \(\mathrm{PCl}_5\). Thus, the variability in their bonding is rooted in their electron shell makeup.

Electronegativity

Electronegativity measures an atom's tendency to attract a bonding pair of electrons. Phosphorus and nitrogen have different electronegativities, with nitrogen generally being more electronegative. However, this difference in electronegativity is not the main reason for nitrogen's inability to form \(\mathrm{NCl}_5\). Instead, it is the lack of \(d\) orbitals that limit nitrogen's bonding potential to compounds consistent with the octet rule. While electronegativity impacts how strong the bond will be, it does not alter the structural limitations set by the absence of \(d\) orbitals.