Group 15 Elements (Nitrogen Family)

• Explain the exceptional inertness of N₂ and contrast it with P₄
• Write the Haber process for ammonia synthesis and state the industrial conditions
• Describe the Ostwald process for HNO₃ manufacture with all three steps
• Compare the allotropes of phosphorus (white, red, black) in terms of structure and reactivity
• Explain the structures and hydrolysis of PCl₃ and PCl₅
• Write the structures and names of oxoacids of nitrogen and phosphorus

Level 1 — Foundations

Group 15 — Overview

Inertness of N₂:

• N≡N bond energy = 945 kJ/mol (very high)
• Small size → strong π overlap in triple bond
• No d-orbitals → cannot expand beyond 4-coordination easily
• N₂ is kinetically inert at room temperature despite being thermodynamically capable of forming compounds

Phosphorus — no P₂ at room temperature because:

• P atoms are large → weak p–p π bonds
• P prefers to form 3 σ bonds using sp³ hybrids → stable P₄ tetrahedron

Ammonia (NH₃)

Haber Process:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol

Properties of NH₃:

• Pyramidal shape; sp³ hybridised; bond angle 107° (lone pair compression)
• Strong H-bonding → high BP (−33°C) for its molar mass
• Weak base: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ (Kb = 1.8 × 10⁻⁵)
• Lewis base: donates lone pair to Lewis acids (e.g., BF₃·NH₃)
• Reducing agent at high T: 4NH₃ + 5O₂ → 4NO + 6H₂O (Ostwald step 1)

Nitric Acid (HNO₃) — Ostwald Process

Step 1: Catalytic oxidation of NH₃
4NH₃ + 5O₂ → 4NO + 6H₂O (Pt/Rh catalyst, 800°C)

Step 2: Oxidation of NO
2NO + O₂ → 2NO₂

Step 3: Absorption in water
3NO₂ + H₂O → 2HNO₃ + NO (NO recycled)

Concentrated HNO₃ reactions:

Brown ring test for NO₃⁻:
FeSO₄ + HNO₃ + H₂SO₄ → [Fe(H₂O)₅NO]²⁺ (brown ring at interface)

Phosphorus Allotropes

White P₄ has P–P–P bond angle = 60° (highly strained → reactive).

PCl₃ and PCl₅:

In PCl₅, the axial P–Cl bonds are longer (219 pm) and weaker than equatorial P–Cl bonds (204 pm).

P₄O₁₀ (phosphorus pentoxide):

• Correct formula P₄O₁₀ (not P₂O₅)
• Strong dehydrating agent: absorbs water to give H₃PO₄
• Reacts with HNO₃: P₄O₁₀ + 4HNO₃ → 4HPO₃ + 2N₂O₅

Oxoacids of Nitrogen

Oxoacids of Phosphorus

Key rule: Only O–H bonds are ionisable; P–H bonds are not. H₃PO₃ is dibasic (not tribasic) because one H is directly bonded to P.

Level 2 — JEE Depth

Why does N not form pentahalides but P does?

• N has no d-orbitals in valence shell (n=2); cannot expand octet beyond 4
• P has available 3d-orbitals; can use sp³d hybridisation to form PCl₅
• This explains NF₃ (maximum) but PCl₃ and PCl₅ both exist

Basicity of hydrides (Group 15):
NH₃ > PH₃ > AsH₃ > SbH₃
NH₃ has highest electron density on N (small, electronegative) → best proton acceptor.
PH₃ is a much weaker base (lone pair on large P; less available).

Reducing power of hydrides:
NH₃ < PH₃ < AsH₃ < SbH₃
Thermal stability decreases down the group → hydrides become better reducing agents.

NO vs NO₂ — odd electron molecules:

• NO: 11 electrons, paramagnetic; HOMO is an antibonding π* orbital
  Bond order = (8−3)/2 = 2.5 (between double and triple bond)
• NO₂: 17 electrons, paramagnetic; bent molecule (134°)

Thermal stability of nitrogen oxides (JEE trap):

Aqua regia = 3 parts conc. HCl + 1 part conc. HNO₃
Dissolves gold and platinum:
Au + HNO₃ + 4HCl → H[AuCl₄] + NO↑ + 2H₂O

Example 1: Predict the hybridisation, shape, and bond angles in PCl₅ and explain why axial bonds are longer.

Step 1: Count electron pairs around P in PCl₅:
        5 bonding pairs, 0 lone pairs → sp³d hybridisation

Step 2: Shape = trigonal bipyramidal
        3 equatorial bonds at 120° (in a plane with P)
        2 axial bonds at 90° to equatorial plane

Step 3: Axial bonds are longer because:
        - Axial positions involve p-orbital contribution (less directional)
        - Axial Cl experiences greater repulsion from 3 equatorial Cl atoms
          (three 90° repulsions vs two 90° for equatorial)
        - Axial P–Cl = 219 pm; equatorial P–Cl = 204 pm

Answer: sp³d, trigonal bipyramidal, equatorial 120°, axial 90°.
        Axial bonds are longer due to greater electron-pair repulsion.

Example 2: In the Haber process, what happens to yield if pressure is doubled but temperature is kept constant? Use Le Chatelier's principle.

Step 1: Write the equilibrium:
        N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
        Moles of gas: reactant side = 4, product side = 2

Step 2: Increasing pressure shifts equilibrium to the side with
        fewer moles of gas → toward products (right)

Step 3: Yield of NH₃ INCREASES when pressure is doubled.

Step 4: In industry, very high pressure (200–300 atm) is used,
        but above ~300 atm the engineering cost outweighs yield gain.

Answer: Yield of NH₃ increases; equilibrium shifts right (fewer gas moles).

• Q1: Why cannot nitrogen form NCl₅ while phosphorus forms PCl₅?
• Q2: State the conditions (temperature, pressure, catalyst) for the Haber process and explain why high temperature is a compromise.
• Q3: How many acidic hydrogen atoms are in H₃PO₂? Draw the structure to justify.
• Q4: Write balanced equations for both dilute and concentrated HNO₃ reacting with copper.
• Stretch: Q5: NO is paramagnetic but N₂ and O₂ are also paramagnetic. Using molecular orbital theory, calculate the bond order of NO and explain why its ionisation energy is higher than that of O₂.

NCERT Chapter 7 link: The p-Block Elements, NCERT Chemistry Part 1, Class 12 (Group 15)

Exam connections: JEE Mains frequently tests Haber/Ostwald conditions, basicity of oxoacids of P (especially H₃PO₃), and shapes of PCl₃/PCl₅. JEE Advanced has asked for MO theory of NO and structures of nitrogen oxides.

Study strategy: Make a reaction map for nitrogen: N₂ → NH₃ → NO → NO₂ → HNO₃ → salts. For phosphorus, build a table of oxoacids with OS, basicity, and whether H is on P or O. Practise drawing trigonal bipyramidal structures with correct bond angles.

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