NEET Chemistry: The p-Block Elements (Nitrogen Family) – Detailed Notes and Practice Questions

Chapter 7A: The p-Block Elements (Nitrogen Family – Group 15)

1. Introduction to Group 15 Elements

• Group 15 elements (Nitrogen Family) include Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), and Bismuth (Bi).
• The general electronic configuration is ns2np3.
• The sum of the group number and the number of valence electrons is 15.
• Nitrogen and Phosphorus are non-metals, Arsenic and Antimony are metalloids, and Bismuth is a metal.

2. General Characteristics of Group 15 Elements

A. Electronic Configuration:

• ns2np3. (Half-filled p-orbitals, which contributes to their stability).
• N: [He]2s22p3
• P: [Ne]3s23p3
• As: [Ar]3d104s24p3
• Sb: [Kr]4d105s25p3
• Bi: [Xe]4f145d106s26p3

B. Atomic and Ionic Radii:

• Atomic and ionic radii generally increase down the group.
• There is a small increase from As to Sb, and a very small increase from Sb to Bi (due to poor shielding of 4f and 5d electrons in Bi, similar to lanthanoid contraction effects).

C. Ionization Enthalpy:

• Generally decreases down the group due to increasing atomic size and increasing shielding effect.
• Group 15 elements have higher ionization enthalpies compared to Group 14 elements in the corresponding periods due to their stable half-filled p-orbitals.
• Irregularities: IE1​ of Bi is slightly higher than Sb due to poor shielding of 4f electrons.

D. Electronegativity:

• Electronegativity generally decreases down the group.
• Nitrogen is the most electronegative element in this group.

E. Oxidation States:

• Common oxidation states are −3,+3, and +5.
• −3 Oxidation State:
  • Tendency to exhibit −3 oxidation state decreases down the group due to increasing atomic size and metallic character. Nitrogen and Phosphorus commonly show −3.
  • Examples: NH3​,PH3​.
• +3 and +5 Oxidation States:
  • The most common oxidation states are +3 and +5.
  • The stability of the +5 oxidation state decreases down the group, while the stability of the +3 oxidation state increases down the group. This is due to the inert pair effect.
  • Examples: PCl5​ (stable), BiCl5​ (unstable). BiCl3​ (stable), SbCl3​ (stable).
  • Bismuth typically forms compounds in the +3 oxidation state, e.g., BiCl3​. BiF5​ exists but is a strong oxidizing agent.

F. Allotropy:

• All elements of Group 15 exhibit allotropy except Bismuth.
• Nitrogen: Does not show allotropy under normal conditions (only exists as N2​ gas).
• Phosphorus: White phosphorus, Red phosphorus, Black phosphorus.
• Arsenic: Grey, Yellow, Black.
• Antimony: Metallic, Yellow, Explosive.

G. Metallic Character:

• Non-metallic character decreases down the group.
• Nitrogen, Phosphorus: Non-metals.
• Arsenic, Antimony: Metalloids.
• Bismuth: Metal.

3. Anomalous Behaviour of Nitrogen

Nitrogen differs from other members of its group due to:

1. Small size.
2. High electronegativity.
3. High ionization enthalpy.
4. Absence of d-orbitals in its valence shell. (This is the most significant reason for many differences).

• Ability to form pπ−pπ multiple bonds: Nitrogen forms triple bonds (N≡N) because of its small size. Other elements (P, As, Sb, Bi) due to larger size and diffused p-orbitals, do not form pπ−pπ multiple bonds effectively.
• Diatomic molecule (N2​): Nitrogen exists as a stable diatomic molecule with a triple bond. Others exist as polyatomic molecules (e.g., P4​).
• Maximum covalency: Nitrogen can exhibit a maximum covalency of 4 (e.g., in NH4+​). Other elements can extend their covalency beyond 4 due to the presence of vacant d-orbitals (e.g., PCl5​ where P shows covalency of 5, PF6−​ where P shows covalency of 6).
• Hydrogen Bonding: Nitrogen forms strong hydrogen bonds (e.g., in NH3​) due to its high electronegativity, which is not prominent in hydrides of other elements.
• Basicity of Hydrides: Basicity of hydrides decreases down the group (NH3​>PH3​>AsH3​>SbH3​>BiH3​).
• Reducing Nature of Hydrides: Reducing nature of hydrides increases down the group.

4. Important Compounds of Nitrogen

A. Dinitrogen (N2​):

• Preparation:
  • Laboratory: By heating a mixture of ammonium chloride and sodium nitrite. NH4​Cl(aq)+NaNO2​(aq)Heat​N2​(g)+2H2​O(l)+NaCl(aq)
  • From ammonium dichromate: (NH4​)2​Cr2​O7​Heat​N2​+Cr2​O3​+4H2​O
  • Very pure N2​: By thermal decomposition of sodium or barium azide. Ba(N3​)2​Heat​Ba+3N2​
• Properties: Colourless, odourless, non-toxic gas. Chemically unreactive at room temperature due to high bond dissociation enthalpy of N≡N triple bond. Reactive at high temperatures.
• Uses: In Haber’s process for NH3​ synthesis, inert atmosphere, cryosurgery.

B. Ammonia (NH3​):

• Preparation:
  • Laboratory: By heating ammonium salts with a base. 2NH4​Cl+Ca(OH)2​→CaCl2​+2NH3​+2H2​O
  • Haber’s Process (Industrial): N2​(g)+3H2​(g)⇌2NH3​(g) Conditions: High pressure (200atm), optimal temperature (700K), catalyst (Fe oxide with K2​O,Al2​O3​).
• Properties: Colourless gas with a pungent smell. Basic in nature (aqueous solution is alkaline). Forms hydrogen bonds.
• Structure: Pyramidal shape (sp$^3$ hybridized N with one lone pair).
• Uses: Fertilizers (urea, ammonium salts), nitric acid production, refrigerants.

C. Nitric Acid (HNO3​):

• Preparation:
  • Laboratory: By heating KNO3​ or NaNO3​ with concentrated H2​SO4​. NaNO3​+H2​SO4​→NaHSO4​+HNO3​
  • Ostwald’s Process (Industrial):
    1. Catalytic oxidation of ammonia by atmospheric oxygen. 4NH3​(g)+5O2​(g)Pt/Rh gauze catalyst,500K,9bar​4NO(g)+6H2​O(g)
    2. Nitric oxide reacts with oxygen to form nitrogen dioxide. 2NO(g)+O2​(g)⇌2NO2​(g)
    3. Nitrogen dioxide dissolves in water to form nitric acid. 3NO2​(g)+H2​O(l)→2HNO3​(aq)+NO(g)
• Properties: Strong oxidizing agent. Reacts with metals and non-metals (except noble metals like Au, Pt). Brown ring test for nitrates.
• Uses: Fertilizers, explosives, dyes, pickling of stainless steel.

D. Oxides of Nitrogen: (Different oxidation states of Nitrogen from +1 to +5)

• N2​O (Nitrous oxide) – +1, colourless, neutral, “laughing gas”
• NO (Nitric oxide) – +2, colourless, neutral
• N2​O3​ (Dinitrogen trioxide) – +3, blue solid, acidic
• NO2​ (Nitrogen dioxide) – +4, brown gas, acidic
• N2​O4​ (Dinitrogen tetroxide) – +4, colourless solid/liquid, acidic (dimer of NO2​)
• N2​O5​ (Dinitrogen pentoxide) – +5, colourless solid, acidic

5. Important Compounds of Phosphorus

A. Allotropic Forms of Phosphorus:

1. White Phosphorus (P4​):
   • Structure: Discrete tetrahedral P4​ molecules. Highly strained bonds (bond angle 60∘).
   • Properties: Translucent white waxy solid. Highly reactive, catches fire in air (chemiluminescence). Soluble in CS2​. Extremely poisonous.
   • Storage: Stored under water to prevent reaction with air.
2. Red Phosphorus:
   • Structure: Polymeric, network structure of P4​ units linked together. Less strained.
   • Properties: Less reactive than white P, stable in air. Insoluble in CS2​. Non-poisonous.
3. Black Phosphorus:
   • More stable. Two forms: α-black P (opaque monoclinic or rhombohedral crystals) and β-black P (layered structure).

B. Phosphine (PH3​):

• Preparation:
  • By reaction of calcium phosphide with water or dilute HCl. Ca3​P2​+6H2​O→3Ca(OH)2​+2PH3​
  • By disproportionation of white phosphorus with concentrated NaOH. P4​+3NaOH+3H2​O→PH3​+3NaH2​PO2​ (sodium hypophosphite)
• Properties: Colourless gas with a rotten fish smell. Highly poisonous. Less basic than ammonia. Forms complexes with metal ions. Spontaneously flammable in air (due to impurities of P2​H4​).
• Uses: Signal in Holme’s signals (smoke screens – due to spontaneous combustion), fumigant.

C. Phosphorus Halides (PX3​, PX5​):

1. Phosphorus Trichloride (PCl3​):
   • Preparation: By passing dry chlorine over white phosphorus. P4​+6Cl2​→4PCl3​
   • Structure: Pyramidal (sp$^3$ hybridized P with one lone pair).
   • Properties: Hydrolyzes readily in water. PCl3​+3H2​O→H3​PO3​+3HCl

• Phosphorus Pentachloride (PCl5​):
  • Preparation: By reaction of white phosphorus with excess dry chlorine or by reacting PCl3​ with excess chlorine. P4​+10Cl2​→4PCl5​ PCl3​+Cl2​→PCl5​
  • Structure: Gaseous/liquid: Trigonal bipyramidal (sp$^3dhybridized).Solid:Ionic([PCl_4]^+[PCl_6]^-$).
  • Properties: Hydrolyzes readily in water. PCl5​+H2​O→POCl3​+2HCl POCl3​+3H2​O→H3​PO4​+3HCl
  • Reacts with organic compounds containing -OH groups (alcohols, carboxylic acids) to replace -OH with -Cl.

D. Oxoacids of Phosphorus: (Important to remember oxidation state and basicity/acidity)

1. Hypophosphorous Acid (Phosphinic Acid), H3​PO2​:
   • Oxidation State: +1
   • Basicity: Monobasic (contains one P-OH bond).
   • Structure: One P-OH, two P-H bonds.
   • Reducing Agent: Strong reducing agent.
2. Phosphorous Acid (Phosphonic Acid), H3​PO3​:
   • Oxidation State: +3
   • Basicity: Dibasic (contains two P-OH bonds).
   • Structure: Two P-OH, one P-H bond.
   • Reducing Agent: Good reducing agent.
3. Orthophosphoric Acid (Phosphoric Acid), H3​PO4​:
   • Oxidation State: +5
   • Basicity: Tribasic (contains three P-OH bonds).
   • Structure: Three P-OH bonds.
   • Strong acid. Does not act as a reducing agent.

• Key Point: P-H bonds are not ionizable and are responsible for reducing properties. P-OH bonds are ionizable and determine basicity.

NEET Chemistry: The p-Block Elements (Nitrogen Family) – Practice Questions

I. Multiple Choice Questions (MCQs)

1. Question: The general electronic configuration of Group 15 elements is: a) ns2np1 b) ns2np2 c) ns2np3 d) ns2np4

2. Question: Which of the following elements in Group 15 is a metal? a) Nitrogen b) Phosphorus c) Arsenic d) Bismuth

3. Question: The stability of the +5 oxidation state in Group 15 elements: a) Increases down the group b) Decreases down the group c) Remains constant down the group d) First increases, then decreases

4. Question: Which of the following is NOT an allotropic form of phosphorus? a) White phosphorus b) Red phosphorus c) Blue phosphorus d) Black phosphorus

5. Question: Nitrogen shows anomalous behavior from the rest of the group due to: a) Its high atomic mass b) Presence of vacant d-orbitals c) Its ability to form pπ−pπ multiple bonds d) Its low electronegativity

6. Question: In Haber’s process for the synthesis of ammonia, the catalyst used is: a) V2​O5​ b) Fe oxide with K2​O,Al2​O3​ c) Pt/Rh gauze d) Ni

7. Question: Which of the following oxides of nitrogen is neutral? a) N2​O3​ b) NO2​ c) N2​O5​ d) N2​O

8. Question: The basicity of orthophosphoric acid (H3​PO4​) is: a) Monobasic b) Dibasic c) Tribasic d) Non-basic

9. Question: Which of the following forms of phosphorus is highly reactive and stored under water? a) Red phosphorus b) White phosphorus c) Black phosphorus d) Allotropic phosphorus

10. Question: Phosphine (PH3​) is prepared by the reaction of calcium phosphide (Ca3​P2​) with: a) Water b) Concentrated HNO3​ c) Oxygen d) Ammonia

11. Question: In solid state, PCl5​ exists as: a) Simple trigonal bipyramidal molecules b) Ionic solid, [PCl4​]+[PCl6​]− c) Covalent network solid d) Dimer of PCl5​ molecules

12. Question: Which of the following is the correct order of decreasing basicity of the hydrides of Group 15 elements? a) NH3​>PH3​>AsH3​>SbH3​>BiH3​ b) PH3​>NH3​>AsH3​>SbH3​>BiH3​ c) BiH3​>SbH3​>AsH3​>PH3​>NH3​ d) NH3​>BiH3​>PH3​>AsH3​>SbH3​

13. Question: The oxidation state of phosphorus in hypophosphorous acid (H3​PO2​) is: a) +1 b) +3 c) +5 d) -3

14. Question: The bond angle in ammonia (NH3​) is approximately: a) 109.5∘ b) 120∘ c) 107∘ d) 180∘

15. Question: What is the product when phosphorus trichloride (PCl3​) reacts with water? a) H3​PO4​ and HCl b) H3​PO3​ and HCl c) POCl3​ and HCl d) P2​O5​ and HCl

II. Assertion-Reason Type Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option. 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.

16. Assertion (A): Nitrogen exists as a diatomic molecule (N2​) while phosphorus exists as P4​. Reason (R): Nitrogen has a small size and high electronegativity, enabling it to form pπ−pπ multiple bonds.

17. Assertion (A): Bismuth is a stronger oxidizing agent in its +5 oxidation state than antimony. Reason (R): The stability of the +5 oxidation state decreases down Group 15 due to the inert pair effect.

18. Assertion (A): Phosphine is less basic than ammonia. Reason (R): The electron density on the phosphorus atom in phosphine is less available due to larger size and lesser electronegativity than nitrogen.

19. Assertion (A): Nitric acid acts as a strong oxidizing agent. Reason (R): Nitrogen in nitric acid is in its highest oxidation state of +5.

20. Assertion (A): The basicity of oxoacids of phosphorus is determined by the number of P-OH bonds. Reason (R): P-H bonds in oxoacids of phosphorus are not ionizable and do not contribute to basicity.

III. Short Answer / Conceptual Questions

21. Question: Explain why Group 15 elements show −3,+3, and +5 oxidation states.

22. Question: Describe the preparation of dinitrogen in the laboratory and state its two important properties.

23. Question: Draw the structures of white phosphorus and red phosphorus. Briefly mention one key difference in their properties.

24. Question: What is the Ostwald’s process? Write the balanced chemical equations for the steps involved in the industrial preparation of nitric acid.

25. Question: Differentiate between the reaction of primary aliphatic amine and primary aromatic amine with nitrous acid.

26. Question: Identify the type of hybridization and geometry of phosphorus in PCl5​ in gaseous phase. What is its structure in the solid state?

27. Question: Arrange the following oxides of nitrogen in increasing order of oxidation state of nitrogen: NO2​,N2​O,N2​O5​,NO.

28. Question: Explain why HNO3​ is a stronger oxidizing agent than H3​PO4​.

29. Question: How does the reducing character of hydrides of Group 15 elements change down the group? Give a reason.

30. Question: Give two uses of ammonia (NH3​).

Answers and Explanations

I. Multiple Choice Questions (MCQs) – Answers

1. Answer: c) ns2np3 Explanation: Group 15 elements have 5 valence electrons, with 2 in the s-orbital and 3 in the p-orbital.

2. Answer: d) Bismuth Explanation: In Group 15, non-metallic character decreases and metallic character increases down the group. Nitrogen and Phosphorus are non-metals, Arsenic and Antimony are metalloids, and Bismuth is a metal.

3. Answer: b) Decreases down the group Explanation: Due to the inert pair effect, the stability of the +5 oxidation state decreases down Group 15. The ns2 electrons become increasingly reluctant to participate in bonding.

4. Answer: c) Blue phosphorus Explanation: The main allotropic forms of phosphorus are white, red, and black phosphorus. Blue phosphorus is not a commonly recognized allotrope.

5. Answer: c) Its ability to form pπ−pπ multiple bonds Explanation: Nitrogen’s small size and high electronegativity allow it to form stable pπ−pπ multiple bonds (e.g., N≡N). Heavier elements cannot do this effectively due to their larger size and diffused p-orbitals. This is a key reason for nitrogen’s anomalous behavior.

6. Answer: b) Fe oxide with K2​O,Al2​O3​ Explanation: In Haber’s process for ammonia synthesis, finely divided iron oxide, promoted by K2​O and Al2​O3​, is used as a catalyst.

7. Answer: d) N2​O Explanation: N2​O (Nitrous oxide) and NO (Nitric oxide) are neutral oxides of nitrogen. N2​O3​,NO2​, and N2​O5​ are acidic oxides.

8. Answer: c) Tribasic Explanation: Orthophosphoric acid (H3​PO4​) has three P-OH bonds. These three hydrogen atoms are acidic and can be replaced, making it a tribasic acid. Its structure is O=P−(OH)3​.

9. Answer: b) White phosphorus Explanation: White phosphorus exists as discrete P4​ tetrahedral molecules with highly strained bonds, making it extremely reactive. It spontaneously ignites in air and is stored under water to prevent oxidation.

10. Answer: a) Water Explanation: Phosphine (PH3​) can be prepared by the hydrolysis of metal phosphides, such as calcium phosphide (Ca3​P2​) reacting with water or dilute acids. Ca3​P2​+6H2​O→3Ca(OH)2​+2PH3​

11. Answer: b) Ionic solid, [PCl4​]+[PCl6​]− Explanation: In the gaseous and liquid phases, PCl5​ exists as a simple trigonal bipyramidal molecule. However, in the solid state, it exists as an ionic compound, consisting of tetrahedral PCl4+​ cations and octahedral PCl6−​ anions.

12. Answer: a) NH3​>PH3​>AsH3​>SbH3​>BiH3​ Explanation: The basicity of hydrides of Group 15 elements decreases down the group. This is because as atomic size increases, the electron density on the central atom decreases, and the lone pair of electrons becomes less available for donation.

13. Answer: a) +1 Explanation: In hypophosphorous acid (H3​PO2​), the oxidation state of hydrogen is +1 and oxygen is -2. Let the oxidation state of phosphorus be x. 3(+1)+x+2(−2)=0 3+x−4=0 x−1=0⇒x=+1.

14. Answer: c) 107∘ Explanation: Nitrogen in ammonia (NH3​) is sp$^3$ hybridized. It has three bond pairs and one lone pair. The lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, which compresses the ideal tetrahedral angle (109.5∘) to approximately 107∘.

15. Answer: b) H3​PO3​ and HCl Explanation: Phosphorus trichloride (PCl3​) undergoes hydrolysis with water to form phosphorous acid (H3​PO3​) and hydrogen chloride gas. PCl3​+3H2​O→H3​PO3​+3HCl

II. Assertion-Reason Type Questions – Answers

16. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Nitrogen exists as diatomic N2​ because its small size and high electronegativity allow it to form strong pπ−pπ triple bonds. Larger atoms like phosphorus cannot form such strong multiple bonds, hence phosphorus exists as discrete P4​ molecules or polymeric structures.

17. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: The stability of the +5 oxidation state decreases down Group 15, while the stability of the +3 oxidation state increases due to the inert pair effect. Consequently, Bi(V) compounds are less stable and act as stronger oxidizing agents (getting reduced to Bi(III)) compared to Sb(V) compounds.

18. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Phosphine is less basic than ammonia. This is because phosphorus is larger and less electronegative than nitrogen, meaning its valence electrons (including the lone pair) are more diffused and less available for donation to a proton, thus reducing its basicity.

19. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Nitric acid (HNO3​) is a strong oxidizing agent because nitrogen is in its highest possible oxidation state of +5. In this state, nitrogen is electron-deficient and readily accepts electrons (gets reduced) to achieve lower, more stable oxidation states, thus oxidizing other substances.

20. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: The basicity of oxoacids of phosphorus (e.g., H3​PO2​,H3​PO3​,H3​PO4​) is determined by the number of acidic protons, which are the hydrogens attached to oxygen atoms (P-OH bonds). The hydrogens directly attached to phosphorus atoms (P-H bonds) are not acidic and do not ionize, hence they do not contribute to the basicity.

III. Short Answer / Conceptual Questions – Answers

21. Answer: Group 15 elements have a general electronic configuration of ns2np3.

1. −3 Oxidation State: By gaining three electrons, they can achieve a stable noble gas configuration. This tendency is highest for Nitrogen (most electronegative) and decreases down the group as metallic character increases.
2. +3 Oxidation State: By losing the three p-electrons, they can exhibit a +3 oxidation state. The stability of this oxidation state increases down the group due to the inert pair effect, where the ns2 electrons become increasingly reluctant to participate in bonding.
3. +5 Oxidation State: By losing all five valence electrons (both ns2 and np3 electrons), they can exhibit a +5 oxidation state. The stability of this oxidation state decreases down the group due to the inert pair effect.

22. Answer: Preparation of Dinitrogen (N2​) in the laboratory: Dinitrogen is prepared in the laboratory by heating an aqueous solution of ammonium chloride (NH4​Cl) with sodium nitrite (NaNO2​). NH4​Cl(aq)+NaNO2​(aq)Heat​N2​(g)+2H2​O(l)+NaCl(aq) Two important properties of Dinitrogen:

1. Chemical Inertness: Dinitrogen is largely unreactive at room temperature due to the very high bond dissociation enthalpy (941.4kJ/mol) of the N≡N triple bond. This strong triple bond makes it difficult to break and react.
2. Low Solubility: It has very low solubility in water.

23. Answer:

• White Phosphorus (P4​):
  • Structure: It consists of discrete tetrahedral P4​ molecules. Each phosphorus atom is bonded to three other phosphorus atoms. The P-P-P bond angle is 60∘, which is highly strained.
  • Drawing: (Imagine a tetrahedron with P atoms at each vertex) P / \ P---P \ / P
  • Key difference in property: It is highly reactive and spontaneously ignites in air (chemiluminescence), hence stored under water.
• Red Phosphorus:
  • Structure: It has a polymeric structure, formed by linking P4​ tetrahedral units together. This forms a network structure, which is less strained than white phosphorus.
  • Drawing: (Imagine multiple tetrahedrons linked in a chain-like or network fashion)-P-P-P- | | | P P P | | | -P-P-P- (Simplified representation of a polymeric structure)
  • Key difference in property: It is much less reactive than white phosphorus and is stable in air. It is also non-poisonous, unlike white phosphorus.

24. Answer: Ostwald’s Process: It is an industrial method used for the large-scale production of nitric acid (HNO3​). The process involves three main steps:

1. Catalytic Oxidation of Ammonia: Ammonia gas (NH3​) reacts with atmospheric oxygen (O2​) in the presence of a platinum-rhodium gauze catalyst at high temperature and pressure. This step produces nitric oxide (NO). 4NH3​(g)+5O2​(g)Pt/Rh gauze catalyst,500K,9bar​4NO(g)+6H2​O(g)
2. Oxidation of Nitric Oxide: The nitric oxide (NO) produced in the first step reacts with more oxygen (from the air) to form nitrogen dioxide (NO2​). This reaction is exothermic and spontaneous. 2NO(g)+O2​(g)⇌2NO2​(g)
3. Absorption of Nitrogen Dioxide in Water: Nitrogen dioxide (NO2​) is then absorbed in water (usually hot water) in an absorption tower to form nitric acid. The nitric oxide (NO) regenerated in this step is recycled. 3NO2​(g)+H2​O(l)→2HNO3​(aq)+NO(g) The concentrated nitric acid can be obtained by distillation.

25. Answer: The reaction with nitrous acid (HNO2​, prepared in situ from NaNO2​ and dilute HCl) is a key test to distinguish between primary aliphatic and primary aromatic amines.

1. Primary Aliphatic Amines (e.g., Ethylamine, CH3​CH2​NH2​): Primary aliphatic amines react with nitrous acid to form unstable aliphatic diazonium salts, which rapidly decompose even at low temperatures (0−5∘C) to yield primary alcohols with the evolution of nitrogen gas (N2​). The evolution of bubbles of N2​ gas is a characteristic observation. R−NH2​+HNO2​NaNO2​/HCl,0−5∘C​[R−N2+​Cl−] (unstable)→R−OH+N2​↑+HCl
2. Primary Aromatic Amines (e.g., Aniline, C6​H5​NH2​): Primary aromatic amines react with nitrous acid at very low temperatures (0−5∘C) to form stable arenediazonium salts. These diazonium salts are stable only at low temperatures and can be used for further reactions (like coupling reactions to form azo dyes). There is no immediate evolution of N2​ gas. Ar−NH2​+HNO2​+HClNaNO2​/HCl,0−5∘C​Ar−N2+​Cl− (stable below 5∘C)+2H2​O This difference in stability and product formation (alcohol + N2​ gas vs. stable diazonium salt) is used for distinction.

26. Answer:

• In gaseous phase:
  • Hybridization: Phosphorus in PCl5​ is sp$^3$d hybridized.
  • Geometry: It has a trigonal bipyramidal geometry. In this structure, three chlorine atoms are in equatorial positions (lying in a plane with P), and two chlorine atoms are in axial positions (above and below the plane).
• In solid state:
  • In the solid state, PCl5​ exists as an ionic solid composed of two distinct ions:
    • Tetrahedral tetrachlorophosphonium cation ([PCl4​]−).
    • Octahedral hexachlorophosphate anion ([PCl6​]−).
  • The structure can be represented as [PCl4​]+[PCl6​]−.

27. Answer: To arrange the oxides in increasing order of oxidation state of nitrogen, we calculate the oxidation state for each:

• N2​O: Let N be x. 2x+(−2)=0⇒2x=2⇒x=+1.
• NO2​: Let N be x. x+2(−2)=0⇒x−4=0⇒x=+4.
• N2​O5​: Let N be x. 2x+5(−2)=0⇒2x−10=0⇒2x=10⇒x=+5.
• NO: Let N be x. x+(−2)=0⇒x=+2. Increasing order of oxidation state: N2​O(+1)<NO(+2)<NO2​(+4)<N2​O5​(+5)

28. Answer: Nitric acid (HNO3​) is a stronger oxidizing agent than orthophosphoric acid (H3​PO4​) for the following reasons:

1. Oxidation State of Central Atom:
   • In HNO3​, Nitrogen is in its highest oxidation state of +5.
   • In H3​PO4​, Phosphorus is also in its highest oxidation state of +5.
2. Stability of Oxidation State:
   • While both are in their highest oxidation states, nitrogen in +5 state is less stable than phosphorus in +5 state. Nitrogen has no d-orbitals to expand its valency, and its small size makes the highly positive charge less dispersed.
   • Phosphorus, being a larger atom and having vacant d-orbitals, can accommodate the +5 oxidation state more comfortably and form more stable structures.
3. Bond Polarity and Acid Strength:
   • HNO3​ is a strong acid and completely ionizes to give H+ and NO3−​. The NO3−​ ion is a relatively good oxidizing agent.
   • H3​PO4​ is a moderately strong acid. Therefore, the nitrogen in HNO3​ has a greater tendency to accept electrons and get reduced (i.e., oxidize other substances) compared to phosphorus in H3​PO4​.

29. Answer: The reducing character of hydrides of Group 15 elements increases down the group (from NH3​ to BiH3​). Reason: As we move down the group from Nitrogen to Bismuth, the atomic size of the central atom (N, P, As, Sb, Bi) increases. This leads to:

1. Decrease in Bond Dissociation Enthalpy: The M−H bond length (where M is the Group 15 element) increases down the group, making the M−H bond weaker. Consequently, the bond dissociation enthalpy of the M−H bond decreases.
2. Ease of Hydrogen Release: A weaker M−H bond means that hydrogen can be released more easily. The ability to donate hydrogen (or provide electrons to reduce other species) defines reducing character. Therefore, the order of increasing reducing character is: NH3​<PH3​<AsH3​<SbH3​<BiH3​.

30. Answer: Two uses of ammonia (NH3​):

1. Manufacture of Fertilizers: Ammonia is a crucial raw material for the production of nitrogenous fertilizers such as urea (CO(NH2​)2​), ammonium nitrate (NH4​NO3​), ammonium sulphate ((NH4​)2​SO4​), etc. These fertilizers are essential for agriculture to enhance crop yield.
2. Manufacture of Nitric Acid: Ammonia is the primary starting material for the industrial production of nitric acid (HNO3​) via the Ostwald’s process. Nitric acid, in turn, is used in explosives, dyes, and other chemicals. (Other uses include: as a refrigerant, in the manufacture of sodium carbonate (Solvay process), and in the production of synthetic fibers like nylon.)