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Nitrogen forms \(\mathrm{N}_{2}\), but phosphorous is converted into \(\mathrm{P}_{4}\) from \(\mathrm{P}\), the reason is (a) triple bond is present between phosphorous atom (b) \(\mathrm{p} \pi-\mathrm{p} \pi\) bonding is strong (c) p \(\pi-\mathrm{p} \pi\) bonding is weak (d) multiple bond is formed easily

Short Answer

Expert verified
The reason is that p π-p π bonding is weak (option c).

Step by step solution

01

Understanding Nitrogen's Structure

Nitrogen molecules ( N_{2}) possess a triple bond between the nitrogen atoms. This triple bond is very strong due to significant  p - p  overlapping, which results in a stable diatomic molecule. Nitrogen makes effective use of  p - p  multiple bonding.
02

Analyzing Phosphorus Structure

Phosphorus forms  P_{4}  tetrahedra instead of   P_2  because the   p - p  bonding in phosphorus is not as effective at forming stable multiple bonds as observed in nitrogen. The propensity for   p - p   bonding is considerably weaker due to larger atomic size and less effective orbital overlap.
03

Final Evaluation of Options

Based on the analysis, phosphorus forms   P_4   because the   p - p  multiple bonding is weak. As a result, simple covalent multiple bonding does not stabilize phosphorus, unlike in nitrogen. Therefore, the correct answer aligns with option (c).

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Key Concepts

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

Triple Bond in Nitrogen
When understanding why nitrogen forms \( \mathrm{N}_{2} \) molecules, it's crucial to explore the nature of the triple bond present between two nitrogen atoms. The triple bond in nitrogen is exceptionally robust due to the presence of one sigma (σ) bond and two pi (π) bonds. This combination makes the bond not only strong but also very stable.

This stability is a direct result of effective overlapping. Nitrogen atoms participate in \( p \text{-} p \) overlapping, which leads to the formation of both a sigma bond and two pi bonds. These pi bonds are strong because the orbitals overlap significantly, ensuring that electrons are shared efficiently.

The lower energy and high stability of this triple bond give nitrogen its diatomic nature. Nitrogen molecules become less reactive due to this bonding, which is why \( \mathrm{N}_{2} \) is a very inert gas under standard conditions.
p-p Overlapping
The concept of \( p \text{-} p \) overlapping is vital in understanding molecular bonding, particularly in non-metals like nitrogen and phosphorus. Overlapping occurs when atomic orbitals from different atoms come into close proximity, allowing for shared electrons.

In nitrogen, as mentioned, \( p \text{-} p \) overlapping is particularly effective. This efficiency is due to the smaller atomic size and less shielding effect, which allows the p orbitals to align properly and overlap significantly.

However, in phosphorus, the \( p \text{-} p \) overlapping is not as pronounced. Phosphorus atoms are larger and their orbitals are more diffused compared to nitrogen. This diffusion prevents effective overlapping, resulting in weaker bonding when it comes to forming multiple bonds. This is why phosphorus prefers other types of molecular structures.
Molecular Structure of Phosphorus
In contrast to nitrogen, phosphorus shows a distinct molecular structure by forming \( \mathrm{P}_{4} \) tetrahedra rather than a diatomic \( \mathrm{P}_{2} \). This difference lies in the nature and strength of phosphorus's molecular bonding. The inability to form strong \( p \text{-} p \) pi bonds in phosphorus means that alternative structures are more stable.

The formation of \( \mathrm{P}_{4} \) involves tetrahedral bonding where each phosphorus atom is bonded to three other phosphorus atoms. This creates a more stable molecular structure due to the ample use of sigma bonds, which are essential for stability in the absence of pi bonding.

The tetrahedral \( \mathrm{P}_{4} \) has lower energy compared to any potential diatomic structure, making it the preferred form of phosphorus at room temperature. This arrangement is a direct consequence of weaker \( p \text{-} p \) interactions and larger atomic sizes, dictating a preference for different bonding schemes.

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Most popular questions from this chapter

Among the following compounds the one that is polar and has the central atom with sp² hybridization is (a) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (b) \(\mathrm{SiF}_{4}\) (c) \(\mathrm{BF}_{3}\) (d) \(\mathrm{HClO}_{2}\)

Which of the pairs have identical values of bond order? (a) \(\mathrm{F}_{2}\) and \(\mathrm{Ne}_{2}\) (b) \(\mathrm{N}_{2}^{+}\)and \(\mathrm{O}_{2}^{+}\) (c) \(\mathrm{O}_{2}^{2-}\) and \(\mathrm{B}_{2}\) (d) \(\mathrm{C}_{2}\) and \(\mathrm{N}_{2}\)

When two oppositely charged ions approach each other, the ion smaller in size attracts outermost electrons of the other ion and repels its nuclear charge. The electron cloud of anion no longer remains symmetrical but is elongated towards the cation. Due to that, sharing of electrons occur between the two ions to some extent and the bond shows some covalent character. The value of dipole moment can be used for determining the amount of ionic character in a bond. Thus, percentage ionic character = \(\frac{\text { Experimental value of dipole moment }}{\text { Theoretical value of dipole moment }} \times 100\) The dipole moment of \(\mathrm{LiH}\) is \(1.964 \times 10^{-29} \mathrm{C} . \mathrm{m}\). and the interatomic distance between \(\mathrm{Li}\) and \(\mathrm{H}\) in this molecule is \(1.596 \AA\). What is the \% ionic character in \(\mathrm{LiH}\) ? (a) \(76.8 \%\) (b) \(60.25 \%\) (c) \(15.5 \%\) (d) \(26.2 \%\)

Match the following: List I (Hydridization) 1\. \(\mathrm{sp}^{2}\) 2\. \(\mathrm{sp}^{3}\) 3\. \(\mathrm{sp}\) \(4 .\) List II (Geometry of the molecule) (i) trigonal bipyramidal (ii) planar trigonal (iii) octahedral (iv) tetrahedral (v) linear

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}\) ? (a) \(\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}\) (b) \(\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}\) (c) \(\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}\) (d) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)

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