Phosphorus: Practice Questions (JEE/NEET & CBSE)
Phosphorus: Practice Question Set
This section provides a structured set of practice questions on the chemistry of Phosphorus (P), designed to reinforce key concepts for JEE/NEET and CBSE students.
Multiple Choice Questions (MCQs)
Q1. Which of the following statements is incorrect regarding white phosphorus? A. It is a soft, waxy solid. B. It is soluble in carbon disulphide (CS₂). C. It glows in the dark (chemiluminescence). D. It is less reactive than red phosphorus.
Solution: Correct Answer: D Explanation: White phosphorus is kinetically the most reactive allotrope of phosphorus due to its highly strained tetrahedral P₄ molecule (bond angle 60°), which leads to easy ring opening and high reactivity. Red phosphorus is polymeric and much less reactive.
Q2. The basicity of orthophosphorous acid (H₃PO₃) is: A. 1 B. 2 C. 3 D. 4
Solution: Correct Answer: B Explanation: The basicity of an oxyacid is determined by the number of ionizable P-OH bonds. In orthophosphorous acid (H₃PO₃), there are two P-OH bonds and one P-H bond. Only the hydrogen atoms attached to oxygen are ionizable. Therefore, H₃PO₃ is a dibasic acid. Its structure is O=P(OH)₂H.
Q3. When P₄O₁₀ is dissolved in cold water, the major product formed is: A. Phosphinic acid (H₃PO₂) B. Phosphonic acid (H₃PO₃) C. Pyrophosphoric acid (H₄P₂O₇) D. Orthophosphoric acid (H₃PO₄)
Solution: Correct Answer: D Explanation: Phosphorus pentoxide (P₄O₁₀) is the anhydride of orthophosphoric acid (H₃PO₄). When P₄O₁₀ reacts with cold water, it forms orthophosphoric acid. P₄O₁₀ + 6H₂O → 4H₃PO₄ P₄O₁₀ is also a strong dehydrating agent.
Assertion-Reason Questions
Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the choices given below. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, but R is not the correct explanation of A. C. A is true, but R is false. D. A is false, but R is true.
Q4. Assertion (A): Phosphine (PH₃) is a weaker base than ammonia (NH₃). Reason (R): The lone pair of electrons on phosphorus in PH₃ is less available for donation than that on nitrogen in NH₃ due to larger atomic size and less electronegativity of P compared to N.
Solution: Correct Answer: A Explanation:
- Assertion A is true: PH₃ is indeed a weaker base than NH₃.
- Reason R is true: Nitrogen is smaller and more electronegative than phosphorus. This means the lone pair on nitrogen is held in a smaller, more concentrated orbital, making it more readily available for donation and stronger bonding with a proton. In phosphorus, the lone pair is in a larger, more diffuse orbital, making it less available for donation and resulting in a weaker base.
- R is the correct explanation of A: The factors mentioned in the reason directly explain why PH₃ is a weaker base.
Q5. Assertion (A): Red phosphorus is polymeric in nature. Reason (R): Red phosphorus is formed by breaking the P-P bonds in white phosphorus and forming new P-P bonds to create a polymeric structure.
Solution: Correct Answer: A Explanation:
- Assertion A is true: Red phosphorus exists as a polymeric network structure, consisting of chains or networks of P₄ tetrahedral units linked together.
- Reason R is true: Red phosphorus is obtained by heating white phosphorus at 573 K in an inert atmosphere or by exposure to sunlight. This process involves the rupture of some P-P bonds in the individual P₄ tetrahedra of white phosphorus and the formation of new P-P bonds, linking these units into a polymeric structure.
- R is the correct explanation of A: The mechanism described in the reason directly leads to the polymeric nature of red phosphorus.
Short Answer Questions
Q6. Draw the structures of PCl₃ and PCl₅ in the gaseous state, and comment on their hybridization and geometry.
Model Answer:
Phosphorus Trichloride (PCl₃):
- Structure: Pyramidal
- Hybridization: sp³
- Geometry: Tetrahedral electron pair geometry around phosphorus, but trigonal pyramidal molecular geometry due to one lone pair of electrons.
- Bond Angle: Approximately 100° (less than 109.5° due to lone pair-bond pair repulsion).
graph TD
P -- Cl
P -- Cl
P -- Cl
style P fill:#f9f,stroke:#333,stroke-width:2px
style Cl fill:#c3f,stroke:#333,stroke-width:2px
subgraph PCl3
direction LR
P --- Cl1
P --- Cl2
P --- Cl3
P -- L(Lone Pair)
end(Representing a pyramidal structure with a central P, three Cl atoms, and a lone pair on P)
Phosphorus Pentachloride (PCl₅):
- Structure (Gaseous State): Trigonal bipyramidal
- Hybridization: sp³d
- Geometry: Trigonal bipyramidal. Three chlorine atoms occupy equatorial positions, and two chlorine atoms occupy axial positions.
- Bond Angles: 120° (equatorial-equatorial) and 90° (axial-equatorial). The axial bonds are slightly longer and weaker than the equatorial bonds due to greater repulsive interactions.
graph TD
P -- Cl(axial)
P -- Cl(axial)
P -- Cl(equatorial)
P -- Cl(equatorial)
P -- Cl(equatorial)
style P fill:#f9f,stroke:#333,stroke-width:2px
style Cl fill:#c3f,stroke:#333,stroke-width:2px
subgraph PCl5
direction LR
Cl_ax1 --- P --- Cl_ax2
P --- Cl_eq1
P --- Cl_eq2
P --- Cl_eq3
end(Representing a trigonal bipyramidal structure with central P, two axial Cl, and three equatorial Cl atoms)
Q7. Phosphinic acid (H₃PO₂) acts as a reducing agent, while phosphoric acid (H₃PO₄) does not. Explain this difference based on their structures.
Model Answer:
The reducing nature of phosphorus oxyacids is attributed to the presence of P-H bonds. These P-H bonds are not acidic but are responsible for the reducing properties because the hydrogen atom attached directly to phosphorus can be oxidized.
-
Phosphinic acid (H₃PO₂):
- Structure: O=P(OH)H₂
- It has two P-H bonds and one P-OH bond.
- The presence of two P-H bonds makes it a strong reducing agent. The oxidation state of phosphorus in H₃PO₂ is +1, which can be easily oxidized to higher oxidation states (e.g., +3 in H₃PO₃ or +5 in H₃PO₄).
- Example reaction: 4AgNO₃ + H₃PO₂ + 2H₂O → 4Ag + H₃PO₄ + 4HNO₃
-
Phosphoric acid (H₃PO₄):
- Structure: O=P(OH)₃
- It has three P-OH bonds but no P-H bonds.
- Since there are no P-H bonds, it does not exhibit reducing properties. The oxidation state of phosphorus in H₃PO₄ is +5, which is its maximum oxidation state, meaning it cannot be further oxidized.
High-Order Thinking Skills (HOTS) Question
Q8. When PCl₅ is heated, it sublimes at a low temperature, but upon stronger heating, it dissociates into PCl₃ and Cl₂. Explain the bonding in solid PCl₅ and gaseous PCl₅, and how this relates to its dissociation behavior.
Detailed Chemical Explanation:
1. Gaseous PCl₅:
- In the gaseous phase, PCl₅ exists as discrete trigonal bipyramidal molecules.
- Hybridization: The central phosphorus atom is sp³d hybridized.
- Geometry: It has a trigonal bipyramidal geometry with five P-Cl bonds.
- Bond Lengths: Not all P-Cl bonds are equivalent. There are three equatorial P-Cl bonds (120° apart, shorter, and stronger) and two axial P-Cl bonds (90° to equatorial, longer, and weaker). This difference arises from greater lone pair-bond pair repulsions (or bond pair-bond pair repulsions in the case of ideal sp³d hybridisation interpretation) and also due to the “bent rule” or higher s-character in equatorial bonds.
- Stability: The axial bonds are weaker and more susceptible to bond cleavage.
2. Solid PCl₅:
- In the solid state, PCl₅ does not exist as neutral PCl₅ molecules. Instead, it exists as an ionic solid composed of tetrahedral [PCl₄]⁺ cations and octahedral [PCl₆]⁻ anions.
- [PCl₄]⁺ cation: The central phosphorus atom is sp³ hybridized, and the ion has a tetrahedral geometry.
- [PCl₆]⁻ anion: The central phosphorus atom is sp³d² hybridized, and the ion has an octahedral geometry.
- Ionic Lattice: These ions are held together by strong electrostatic forces in a crystal lattice.
3. Dissociation Behavior:
- Sublimation (low temperature): At relatively low temperatures, the intermolecular forces (van der Waals forces) between discrete PCl₅ molecules in the gaseous phase are overcome, allowing PCl₅ to sublime. However, the ionic lattice in the solid state has a higher sublimation point due to strong electrostatic forces. When PCl₅ sublimes, it forms molecular PCl₅ in the gas phase.
- Dissociation (stronger heating): Upon stronger heating (higher temperatures), the thermal energy provided is sufficient to overcome the weaker axial P-Cl bonds in the trigonal bipyramidal PCl₅ molecules. This leads to the homolytic cleavage of these bonds, resulting in the dissociation of PCl₅ into PCl₃ and Cl₂. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) In PCl₃, the phosphorus is sp³ hybridized (pyramidal geometry), and in Cl₂, a covalent bond exists. This dissociation is favored at higher temperatures due to an increase in entropy.
Therefore, the different bonding arrangements in gaseous and solid PCl₅, particularly the presence of weaker axial bonds in the gaseous molecular form, explain its propensity to dissociate upon heating.