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White phosphorus \(\left(\mathrm{P}_{4}\right)\) has 1\. six P-P sigma bonds 2\. four P-P single bonds 3\. four lone pairs of electrons 4\. PPP angle of \(60^{\circ}\) (a) \(1,2,3\) (b) \(2,3,4\) (c) \(1,3,4\) (d) all are correct

Short Answer

Expert verified
Correct options are (1, 3, 4); answer is (c).

Step by step solution

01

Understand White Phosphorus Structure

White phosphorus consists of four phosphorus atoms arranged in a tetrahedral structure, commonly represented as \( \text{P}_4 \). The structure is compact and each phosphorus atom is bonded to the other three, forming a closed cage-like structure.
02

Analyze P-P Sigma Bonds

Each phosphorus atom in \( \text{P}_4 \) forms three sigma bonds with the adjacent phosphorus atoms. This means there are a total of 6 P-P sigma bonds (3 sigma bonds per atom, with 2 elements in each bond, leading to \( 3 \times 4 / 2 = 6 \)).
03

Analyze P-P Single Bonds

The structure of \( \text{P}_4 \) indicates that each P atom is connected to 3 others via single bonds. However, considering unique bonds without repetition means there are exactly 6 single bonds.
04

Explore Lone Pairs in White Phosphorus

Each phosphorus atom in \( \text{P}_4 \) has one lone pair. Since there are 4 phosphorus atoms, there are 4 lone pairs of electrons in the molecule.
05

Determine the PPP Bond Angle

The triangular face of the \( \text{P}_4 \) molecule forms an equilateral triangle, with an angle of \(60^{\circ}\) between any two P-P bonds at a vertex.
06

Verify Each Option

1 is true because \( \text{P}_4 \) has 6 sigma bonds. 2 is not quite accurate because of interpretation but can be understood based on counting methods as 6 not 4. 3 is correct with 4 lone pairs. 4 is accurate with the tetrahedral structure having \(60^{\circ}\) angles.

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Key Concepts

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

P-P Sigma Bonds
White phosphorus, also denoted as \( \text{P}_4 \), forms six P-P sigma bonds. A sigma bond is a type of covalent bond where two electrons are shared in an overlap of orbitals between two adjacent atoms. In the case of \( \text{P}_4 \), each phosphorus atom forms sigma bonds with three of its neighbors. This arrangement results in the formation of a closed cage-like tetrahedral structure.

Here's why the total sum of bonds is six: you have four phosphorus atoms, each making three bonds. Now, because each bond is shared between two atoms, we count each bond once. Hence the formula \( 3 \times 4 / 2 = 6 \) arrives at the six P-P sigma bonds. This bonding arrangement creates a stable and balanced molecular geometry, providing the characteristic properties of white phosphorus.
Lone Pairs of Electrons
Lone pairs of electrons play a substantial role in the molecular characteristics of white phosphorus. Each phosphorus atom in \( \text{P}_4 \) possesses one lone pair. Lone pairs are pairs of valence electrons that are not involved in covalent bonding and are instead localized on a single atom.

In white phosphorus, these lone pairs contribute to the molecule's reactivity and chemical behavior. With a total of four phosphorus atoms, the \( \text{P}_4 \) molecule contains four lone pairs of electrons altogether. These lone pairs are critically important as they affect the steric hindrance within the molecule, influencing how it interacts with other substances. Understanding these lone pairs helps to explain why white phosphorus is highly reactive, and why it must be handled with care in a controlled environment.
Tetrahedral Structure
The tetrahedral structure of white phosphorus \( (\text{P}_4) \) is fascinating and fundamental to its unique properties. The molecular geometry is arranged in a tetrahedron, with each phosphorus atom situated at a vertex and bonded to the other three. This gives the molecule its characteristic closed, cage-like shape.

The tetrahedral arrangement results in a bond angle of \(60^{\circ}\) between any two phosphorus atoms, significantly smaller than typical tetrahedral angles like the \(109.5^{\circ}\) found in carbon-based tetrahedral molecules. This angle contributes to the molecular strain within \( \text{P}_4 \), which is one of the reasons for its high reactivity. Understanding this structure sheds light on the behavior and applications of white phosphorus in industrial and chemical contexts. Its tetrahedral structure influences not only its chemical reactions but also its physical properties, such as its ability to glow upon exposure to oxygen.

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Most popular questions from this chapter

Which of the following statement(s) is correct here: (a) Hypo phosphorous acid shows reducing properties (b) Pyrophosphoric acid has although 4 - OH groups but it is a dibasic acid (c) Orthophosphorous acid, is obtained during disproportionation of hypo phosphorous acid (d) Meta phosphoric acid on heating gives \(\mathrm{P}_{4} \mathrm{O}_{10}\)

Compound \(\mathrm{X}\) on reduction with \(\mathrm{LiAlH}_{4}\) gives a hydride Y containing \(21.72 \%\) hydrogen along with other products. The compound \(Y\) reacts with air explosively resulting in boron trioxide. Identify \(\mathrm{Y}\). (a) \(\mathrm{B}_{4} \mathrm{H}_{10}\) (b) \(\mathrm{B}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}\) (d) both (a) and (b)

Regular use of which of the following fertilizers increases the acidity of soil? (a) urea (b) super phosphate of lime (c) ammonium sulphate (d) potassium nitrate

Identify the incorrect statement among the following: (a) silicon reacts with \(\mathrm{NaOH}\) (aq) in the presence of air to give \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Cl}_{2}\) reacts with excess of \(\mathrm{NH}_{3}\) to give \(\mathrm{N}_{2}\) and \(\mathrm{HCl}\) (c) \(\mathrm{Br}_{2}\) reacts with hot and strong \(\mathrm{NaOH}\) solution to give \(\mathrm{NaBr}, \mathrm{NaBrO}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) (d) ozone reacts with \(\mathrm{SO}_{2}\) to give \(\mathrm{SO}_{3}\)

\(\mathrm{SO}_{2}\) will be obtained in: (a) \(\mathrm{PbS}+\) air (b) \(\mathrm{FeS}_{2}+\) air (c) \(\mathrm{ZnS}+\mathrm{air}\) (d) \(\mathrm{H}_{2} \mathrm{~S}^{2}+\mathrm{O}_{3}\)

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