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Chapter 18: Problem 45

Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable \(P_{4}\) molecules instead of \(\mathrm{P}_{2}\) molecules?

Short Answer

Expert verified
Nitrogen forms a stable diatomic N2 gas due to its ability to effectively overlap p orbitals and form a strong triple bond, while phosphorus forms P4 molecules because its larger size and less effective orbital overlap are not conducive to strong multiple bonding as seen in N2, making P4 more stable than P2.

Step by step solution

01

Consider the Electron Configurations of Nitrogen and Phosphorus

Examine the electron configurations of nitrogen (N) and phosphorus (P) to understand their bonding tendencies. Nitrogen has the electron configuration `1s^2 2s^2 2p^3`, indicating it has three unpaired p-electrons that allow it to form three strong pi bonds with another nitrogen atom, resulting in a stable N2 molecule. Phosphorus has the electron configuration `1s^2 2s^2 2p^6 3s^2 3p^3`, and due to its larger size and energy considerations, it does not readily form multiple pi bonds with another phosphorus atom.
02

Understand Molecular Stability

Molecular stability relies on effective overlap of atomic orbitals to form strong bonds. The triple bond in nitrogen molecules is very strong due to effective sideways overlap of p orbitals, but in phosphorus atoms, the larger orbitals do not overlap as effectively for multiple bonding due to increased atomic radii and shielding effects.
03

Recognize the Differences in P2 and P4 Molecules

In theory, a molecule could exist, but it would require the effective overlap of p orbitals to form a strong triple bond, similar to nitrogen. However, phosphorus prefers to form P4 tetrahedral molecules, where each phosphorus atom forms three single bonds with other phosphorus atoms. This structure minimizes the repulsion between the lone pairs of electrons and maximizes orbital overlap under these constraints, leading to a more stable configuration than a P2 molecule could achieve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Bonding
Chemical bonding is the force that holds atoms together in molecules.
It stems from the attraction between the positively charged nuclei and negatively charged electrons of different atoms. Bonds can be covalent, where electrons are shared between atoms, or ionic, where electrons are transferred from one atom to another.
For nitrogen (N2), the atoms share six electrons, forming three covalent bonds, referred to as a triple bond, which is exceptionally strong and leads to high molecular stability.
Electron Configurations
The electron configuration of an atom describes the distribution of electrons in the atomic orbitals.
It is essential for predicting the chemical bonding behavior of an atom. Nitrogen, with an electron configuration allowing for three unpaired electrons, can form a triple bond in the N2 molecule, whereas phosphorus, due to larger atomic size and different electron distribution, does not naturally form strong pi bonds and instead forms P4 molecules.
Pi Bonds
Pi (π) bonds are a type of covalent bond formed by the sideways overlap of adjacent p-orbitals, enabling electrons to be shared across both nuclei.
Pi bonds are typically found in double and triple bonded atoms and contribute to the rigidity and planar structure of molecules. In the N2 molecule, the π bonds are significant for molecular stability, which is not commonly exhibited in phosphorus atoms.
Orbital Overlap
Orbital overlap is a condition where atomic orbitals occupy the same region of space, allowing electrons to be shared between atoms.
This overlap is maximized when the orbitals are properly oriented and of suitable energy. Nitrogen's small atomic size allows for effective overlap and strong triple bond formation, while phosphorus' larger orbitals lead to less effective overlap, precluding similar strong bond formation.
Atomic Radii
Atomic radius is the measure of the size of an atom.
Larger atomic radii in elements like phosphorus result in less effective orbital overlap compared to nitrogen, due to a greater distance between the valence electrons and the nucleus. This aspect of phosphorus contributes to its tendency to form P4 molecules instead of P2, as larger atoms require different bonding configurations to achieve stability.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule.
It helps determine the molecule's reactivity, polarity, color, phase of matter, and physical and chemical properties. The tetrahedral geometry of the P4 molecule is a result of phosphorus atoms seeking to minimize repulsion and achieve a stable geometric arrangement.
Lone Pair Repulsion
Lone pair repulsion is an aspect of the Valence Shell Electron Pair Repulsion (VSEPR) theory, which postulates that electron pairs around a central atom are arranged as far apart as possible to minimize repulsion.
Phosphorus has more electron lone pairs compared to nitrogen, which affects the molecular geometry. The P4 molecules arrange themselves in a manner that reduces this repulsion, influencing phosphorus' preference for this less reactive, more stable form over a P2 structure.

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Most popular questions from this chapter

Describe the hybridization of silicon and the molecular structure of the following molecules and ions: (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{SiH}\) (b) \(\operatorname{sio}_{4}^{4-}\) (c) \(\mathrm{Si}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{Si}(\mathrm{OH})_{4}\) (e) \(\operatorname{SiF}_{6}^{2-}\)

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Structure and General Properties of the Metalloids Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples. (a) \(\mathrm{GeH}_{4}\) (b) \(\mathrm{SbF}_{3}\) (c) \(\operatorname{Te}(\mathrm{OH})_{6}\) (d) \(\mathrm{H}_{2} \mathrm{Te}\) (e) \(\mathrm{GeF}_{2}\) (f) \(\mathrm{TeCl}_{4}\) (g) \(\operatorname{SiF}_{6}^{2-}\) (h) \(\mathrm{SbCl}_{5}\) (i) TeF \(_{6}\)

Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) \(\mathrm{PH}_{3}\) (b) \(\mathrm{PH}_{4}^{+}\) (c) \(\mathrm{P}_{2} \mathrm{H}_{4}\) (d) \(\mathrm{PO}_{4}^{3-}\) (e) \(\mathrm{PF}_{5}\)
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