Problem 57 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{N}\) forms \(\mathrm{NCl}_{3}\) whereas \(\mathrm{P}\) can form both \(\mathrm{PCl}_{3}\) and \(\mathrm{PCl}_{5}\). Why? (a) \(\mathrm{N}\) atoms is larger than \(\mathrm{P}\) in size (b) P has low lying \(3 \mathrm{~d}\) orbitals, which can be used for bonding but \(\mathrm{N}\) does not have \(3 \mathrm{~d}\) orbitals in its valence shell (c) \(\mathrm{P}\) is more reactive towards \(\mathrm{Cl}\) than \(\mathrm{N}\) (d) None of these

Step-by-Step Solution

Answer

The correct answer is (b) because P has available \(3d\) orbitals, which N lacks.

Step 1: Understanding the Problem

The problem asks why phosphorus (P) can form both \(\mathrm{PCl}_3\) and \(\mathrm{PCl}_5\), whereas nitrogen (N) only forms \(\mathrm{NCl}_3\). To solve this, we need to consider the electronic structures and orbital availability of both nitrogen and phosphorus.

Step 2: Analyzing Orbital Availability

Nitrogen belongs to the second period and has an electronic configuration of \(1s^2 2s^2 2p^3\). Since it is in the second period, it lacks available \(d\) orbitals in its valence shell. Phosphorus belongs to the third period with an electronic configuration of \(1s^2 2s^2 2p^6 3s^2 3p^3\), and it has access to \(3d\) orbitals, allowing it to form bonds with additional chlorine atoms.

Step 3: Evaluating Given Answers

Option A claims nitrogen is larger in size than phosphorus; however, in the periodic table, atoms get larger as we move down a group, so \(\mathrm{P}>\mathrm{N}\) in size. Option C states phosphorus is more reactive than nitrogen, but this is not related to the ability to form \(PCl_5\). Option B states the availability of \(3d\) orbitals in phosphorus which nitrogen lacks, enabling \(\mathrm{PCl}_5\) formation. Option D suggests none of these is correct.

Step 4: Selecting the Correct Answer

Since the presence of low-lying \(3d\) orbitals in phosphorus explains the difference in their compounds, the correct option is B.

Key Concepts

Orbital Hybridization

When it comes to chemical bonding, orbital hybridization plays an essential role in determining the shapes and bond angles of molecules. Hybridization refers to the mixing of atomic orbitals to form new hybrid orbitals, which are used to form chemical bonds.
In the context of our problem, phosphorus (P) can use its valence orbitals to form both \( ext{PCl}_3\) and \( ext{PCl}_5\). In \( ext{PCl}_3\), phosphorus undergoes \(sp^3\) hybridization, mixing one \(s\) orbital and three \(p\) orbitals to form four \(sp^3\) hybrid orbitals. Each of these hybrid orbitals overlaps with a chlorine atom to form a bond.
For \(PCl_5\), phosphorus can utilize \(sp^3d\) hybridization due to the availability of \(3d\) orbitals. Here, one \(s\), three \(p\) and one \(d\) orbital mix to form five \(sp^3d\) hybrid orbitals, allowing the formation of five bonds with chlorine.
It's important to note that nitrogen (N), which doesn't have access to \(d\) orbitals due to its position in the second period of the periodic table, can only form \(sp^3\) hybridized molecules such as \( ext{NCl}_3\). This inability to expand its octet differentiates it from phosphorus' bonding capabilities.

Electronic Configuration

Understanding electronic configuration is crucial for deducing how elements bond and their capacity for hybridization. Electronic configuration outlines the distribution of electrons in an atom's orbitals. For nitrogen (N), the electronic configuration is \(1s^2 2s^2 2p^3\); this means nitrogen has five valence electrons and lacks available \(d\) orbitals due to its presence in the second period.
On the other hand, phosphorus (P) possesses an electronic configuration of \(1s^2 2s^2 2p^6 3s^2 3p^3\), indicating that phosphorus has five valence electrons and an empty set of \(3d\) orbitals in its valence shell. These \(3d\) orbitals provide additional room for electron accommodation, enabling phosphorus to form \(PCl_5\) in addition to \(PCl_3\).
By examining these configurations, we can see that phosphorus' ability to engage higher-energy \(d\) orbitals allows for expanded chemical bonding, whereas nitrogen is restricted to forming bonds that do not exceed an octet. Hence, electronic configuration directly influences the bonding properties and abilities of atoms in forming different chemical compounds.

Valence Electrons

Valence electrons are the electrons that reside in the outermost shell of an atom. They are primarily responsible for chemical bonding interactions. In the case of nitrogen and phosphorus, both elements have five valence electrons; however, their capabilities in forming compounds differ significantly due to the availability of orbitals.
Nitrogen's valence electrons are confined to the \(2s\) and \(2p\) orbitals, limiting its chemical bonding to configurations like \(sp^3\), thus forming \( ext{NCl}_3\) and not beyond the octet rule.
Phosphorus has more flexibility with its five valence electrons, utilizing not only \(3s\) and \(3p\) orbitals but also \(3d\) orbitals when forming compounds like \(PCl_5\). This is due to its position in the third period, where additional \(d\) orbitals become accessible.
The access to \(3d\) orbitals allows phosphorus to surpass the set eight-electron limit in its valence shell, effectively holding additional electrons and thus, forming more than three bonds as seen in \(PCl_5\). This capability reveals how valence electrons, along with orbital availability, define an element's potential to form diverse and complex compounds.