Physical Properties of Group 15 Elements
Overview of Group 15 (Pnictogens)
Group 15 elements: Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi). All have the outer electron configuration ns²np³ — giving 5 valence electrons and 3 unpaired electrons (by Hund's rule).
The group shows the same non-metal → metalloid → metal progression as Group 14: N and P are non-metals, As and Sb are metalloids, Bi is a metal.
| Element | Z | Config | State | M.p. (°C) | Character | Common OS |
|---|---|---|---|---|---|---|
| Nitrogen (N) | 7 | [He]2s²2p³ | Gas (N₂) | −210 | Non-metal | −3, 0, +1 to +5 |
| Phosphorus (P) | 15 | [Ne]3s²3p³ | Solid (white/red) | 44 (white) | Non-metal | −3, +3, +5 |
| Arsenic (As) | 33 | [Ar]3d¹⁰4s²4p³ | Solid | 817 (sublimes) | Metalloid | −3, +3, +5 |
| Antimony (Sb) | 51 | [Kr]4d¹⁰5s²5p³ | Solid | 631 | Metalloid | +3, +5 |
| Bismuth (Bi) | 83 | [Xe]4f¹⁴5d¹⁰6s²6p³ | Solid | 271 | Metal | +3 (stable), +5 |
Trends Down Group 15
- Atomic radius — increases (more electron shells)
- Ionisation energy — generally decreases, but note N has higher IE₁ than O (half-filled 2p³ = extra stability)
- Electronegativity — decreases: N (3.0) → P (2.1) → As (2.0) → Sb (1.9) → Bi (1.9)
- Metallic character — increases (N = typical non-metal; Bi = metal)
- +5 stability — decreases down group (Bi(V) is rare and strongly oxidising — inert pair effect)
- +3 stability — increases down group (Bi(III) most stable)
- Hydrides MH₃ — boiling point: NH₃ > PH₃ (NH₃ has hydrogen bonds; PH₃ has only van der Waals)
Nitrogen — Occurrence and Inertness
Occurrence of Nitrogen
Nitrogen makes up 78.1% of the atmosphere (by volume) as diatomic N₂. It is also found in proteins, nucleic acids (DNA, RNA), and minerals (saltpetre NaNO₃, Chile saltpetre NaNO₃). The nitrogen cycle describes how nitrogen moves between the atmosphere, soil, and living organisms.
Structure of N₂ — Why is Nitrogen so Inert?
Nitrogen exists as N≡N — a triple bond: one σ bond and two π bonds. The N≡N bond is exceptionally strong: 945 kJ mol⁻¹ (the strongest homonuclear diatomic bond). This makes N₂ extremely unreactive at room temperature.
Reasons for N₂ inertness:
- Very high N≡N bond energy (945 kJ mol⁻¹) — requires enormous activation energy to break
- N₂ is non-polar — no δ⁺/δ⁻ centres for nucleophilic or electrophilic attack
- The π bonds prevent rotation and make the molecule rigid
- N is small — steric hindrance prevents approach of reagents to the strongly bound nitrogen atoms
Reactions of Nitrogen
Despite its general inertness, N₂ reacts under forcing conditions:
Biological nitrogen fixation: certain bacteria (Rhizobium, Azotobacter) use the enzyme nitrogenase to convert N₂ to NH₃ at room temperature and pressure — a remarkable feat that chemistry struggles to replicate industrially.
Ammonia (NH₃)
Structure and Properties of Ammonia
Ammonia has the molecular formula NH₃. Nitrogen is sp³ hybridised with one lone pair and three N–H bonds. The molecular shape is trigonal pyramidal with a bond angle of 107° (lone pair repels bonding pairs → angle less than 109.5°).
Key properties of ammonia:
- Colourless gas with pungent smell; b.p. = −33°C
- Highly soluble in water (700 volumes at STP) — forms alkaline solution: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
- Hydrogen bonding — N–H···N bonds between molecules → higher b.p. than PH₃ (−88°C)
- Acts as a Lewis base (lone pair donor) and a Brønsted-Lowry base (proton acceptor)
- Forms complex ions with transition metals: Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ (deep blue)
Industrial Manufacture — The Haber Process
The Haber Process manufactures ammonia from nitrogen and hydrogen:
Conditions:
- Temperature: ~450°C — compromise between yield (lower T favours product) and rate (higher T gives faster reaction)
- Pressure: ~200 atm — high pressure favours product (4 moles gas → 2 moles gas); limited by engineering/economic constraints
- Catalyst: Finely divided iron (Fe) with promoters K₂O and Al₂O₃
- Recycling: Unreacted N₂ and H₂ are recycled — yield per pass is only ~15%; overall conversion ~97%
Sources of raw materials: N₂ from fractional distillation of liquid air. H₂ from steam reforming of methane: CH₄ + H₂O → CO + 3H₂ (then water-gas shift: CO + H₂O → CO₂ + H₂).
Reactions of Ammonia
Question: In the Haber process, N₂ + 3H₂ ⇌ 2NH₃, if the equilibrium mixture at 450°C contains 15% NH₃ by moles, calculate the mole fraction of N₂ if H₂:N₂ = 3:1 initially.
Start with 1 mol N₂ and 3 mol H₂ (4 mol total). Let x mol N₂ react. Then: N₂ remaining = 1−x; H₂ remaining = 3−3x; NH₃ formed = 2x. Total = (1−x)+(3−3x)+2x = 4−2x.
NH₃ mole fraction = 0.15: 2x/(4−2x) = 0.15 → 2x = 0.15(4−2x) = 0.6−0.3x → 2.3x = 0.6 → x = 0.261 mol.
N₂ remaining = 1−0.261 = 0.739 mol. Total = 4−2(0.261) = 3.478 mol. Mole fraction of N₂ = 0.739/3.478 = 0.212 (21.2%).
Oxides of Nitrogen
| Formula | Name | OS of N | State | Properties / Notes |
|---|---|---|---|---|
| N₂O | Dinitrogen oxide (nitrous oxide) | +1 | Colourless gas | "Laughing gas" — anaesthetic; greenhouse gas; decomposes to N₂+O₂ above 600°C |
| NO | Nitrogen monoxide (nitric oxide) | +2 | Colourless gas | Radical (odd electron); turns brown in air (→NO₂); made in lightning, Haber step 1 |
| N₂O₃ | Dinitrogen trioxide | +3 | Blue liquid/gas | Acid anhydride of HNO₂; unstable above −20°C |
| NO₂ | Nitrogen dioxide | +4 | Brown gas | Radical; toxic; reacts with water → HNO₃ + HNO₂; in equilibrium with N₂O₄ |
| N₂O₄ | Dinitrogen tetroxide | +4 | Colourless gas/liquid | 2NO₂ ⇌ N₂O₄; pale yellow at low conc. |
| N₂O₅ | Dinitrogen pentoxide | +5 | White solid | Acid anhydride of HNO₃; reacts vigorously with water |
Important Reactions of Nitrogen Oxides
Nitric Acid (HNO₃)
Industrial Manufacture — The Ostwald Process
Nitric acid is manufactured from ammonia by the Ostwald Process in three steps:
Concentrated HNO₃ is ~68% HNO₃. It can be made more concentrated by distillation with H₂SO₄.
Properties and Reactions of Nitric Acid
Nitric acid is a strong acid (fully ionises: HNO₃ → H⁺ + NO₃⁻) and a strong oxidising agent. The products of reaction with metals depend on the concentration and the metal.
Reactions with Metals
Reactions with Non-metals and Organic Compounds
Thermal Decomposition of Nitrates
Question: Determine the oxidation state of nitrogen in (a) HNO₃ (b) NH₄⁺ (c) N₂O (d) NO₃⁻.
HNO₃: H(+1) + N + 3O(−2) = 0 → N = +5
NH₄⁺: N + 4H(+1) = +1 → N = +1−4 = −3
N₂O: 2N + O(−2) = 0 → 2N = +2 → N = +1
NO₃⁻: N + 3O(−2) = −1 → N = −1+6 = +5
Phosphorus — Allotropes and Properties
Allotropes of Phosphorus
Phosphorus exists in several allotropic forms. The two most important are white phosphorus and red phosphorus.
| Property | White Phosphorus | Red Phosphorus |
|---|---|---|
| Structure | Discrete P₄ molecules (tetrahedral, P–P–P angle = 60°) | Polymeric — chains of P₄ tetrahedra linked in a network |
| Appearance | White/yellow waxy solid | Dark red powder |
| Melting point | 44°C (simple molecular) | ~600°C (polymeric — giant structure) |
| Reactivity | Very reactive — ignites in air at 34°C; kept under water | Stable in air; ignites only at ~250°C |
| Toxicity | Extremely toxic and flammable | Not toxic at room temp |
| Solubility | Soluble in CS₂ (carbon disulfide) | Insoluble in CS₂ |
| Uses | Historically — incendiary weapons (now banned) | Match heads, fireworks, fertilisers |
P₄ Structure — Why is White Phosphorus so Reactive?
The P₄ molecule has a tetrahedral arrangement with P–P–P bond angles of 60°. This is much smaller than the natural sp³ angle of ~107°. The severe angle strain (electrons repel most when forced to bond at 60°) stores energy in the molecule — making white phosphorus very reactive and self-igniting in air.
Compounds of Phosphorus
Phosphorus Oxides: P₄O₆ and P₄O₁₀
P₄O₆ (phosphorus(III) oxide): reacts slowly with cold water → phosphorous acid H₃PO₃ (diprotic): P₄O₆ + 6H₂O → 4H₃PO₃
P₄O₁₀ (phosphorus(V) oxide): powerful desiccant (drying agent). Reacts vigorously with water → phosphoric acid: P₄O₁₀ + 6H₂O → 4H₃PO₄. Also reacts with alcohols to form esters.
Phosphorus Chlorides: PCl₃ and PCl₅
PCl₅ structure: P is sp³d hybridised → trigonal bipyramidal (5 Cl around P). It can expand its octet because phosphorus (Period 3) has accessible 3d orbitals. This contrasts with NCl₃ — nitrogen cannot form NCl₅ (no accessible d orbitals, limited to octet).
Phosphoric Acid (H₃PO₄) and Its Salts
Phosphoric acid (orthophosphoric acid) is a triprotic (three ionisable protons) weak acid:
Important phosphate salts:
- Ca₃(PO₄)₂ — calcium phosphate, main component of bones and teeth; sparingly soluble
- CaHPO₄ — calcium hydrogen phosphate; used in animal feed supplements
- Na₃PO₄ — trisodium phosphate; used in cleaning agents and water treatment
- NH₄H₂PO₄ — ammonium dihydrogen phosphate; important NPK fertiliser
- Superphosphate: Ca₃(PO₄)₂ + H₂SO₄ → Ca(H₂PO₄)₂ + CaSO₄ (water-soluble fertiliser)
Phosphine (PH₃)
Phosphine is the hydride of phosphorus — analogous to ammonia. However, there are key differences:
| Property | NH₃ | PH₃ |
|---|---|---|
| Boiling point | −33°C | −88°C |
| Reason for b.p. difference | Hydrogen bonding (N–H···N) | Only van der Waals forces (P–H bond too weak/long to H-bond) |
| Bond angle | 107° (sp³-like) | 93° (closer to 90° — P uses pure p orbitals, minimal hybridisation) |
| Basic strength | Moderate base (Kb = 1.8×10⁻⁵) | Very weak base — lone pair less accessible on larger P |
| Stability | Very stable | Unstable — flammable, toxic |
| Smell | Pungent (familiar) | Extremely unpleasant (garlic/decaying fish) |
Trends in Group 15 Properties
Hydrides: NH₃, PH₃, AsH₃, SbH₃
The stability of hydrides decreases down the group — the N–H bond is much stronger than P–H, As–H, Sb–H bonds. Going down, bond strength decreases (larger central atom, weaker overlap) → hydrides decompose more easily at lower temperatures.
| Hydride | B.p. (°C) | Bond angle | Stability | Basic strength |
|---|---|---|---|---|
| NH₃ | −33 | 107° | Very stable | Moderate (Kb = 1.8×10⁻⁵) |
| PH₃ | −88 | 93° | Moderately stable | Very weak |
| AsH₃ | −55 | 92° | Unstable | Negligible |
| SbH₃ | −17 | 91° | Very unstable | None |
Note on boiling points: NH₃ has anomalously HIGH b.p. (−33°C) due to strong N–H···N hydrogen bonds. PH₃ has lower b.p. (−88°C) — only van der Waals. The b.p. then increases from PH₃ → SbH₃ because van der Waals forces increase with larger electron clouds/more electrons.
Oxides: Trend from Acidic to Amphoteric/Basic
| Oxide | Nature | Reaction with water |
|---|---|---|
| N₂O₃ | Acidic | N₂O₃ + H₂O → 2HNO₂ |
| N₂O₅ | Acidic | N₂O₅ + H₂O → 2HNO₃ |
| P₄O₆ | Acidic | P₄O₆ + 6H₂O → 4H₃PO₃ |
| P₄O₁₀ | Acidic | P₄O₁₀ + 6H₂O → 4H₃PO₄ |
| As₂O₃ | Amphoteric | Slight solubility; reacts with acid and base |
| Bi₂O₃ | Basic | Dissolves in acid: Bi₂O₃ + 6HCl → 2BiCl₃ + 3H₂O |
Trend: oxides become more basic going down Group 15 — as metallic character increases.
No videos added yet for this unit.
Exercises
-
Explain why N₂ is so chemically inert at room temperature. Your answer should refer to bond order, bond energy, and the electronic structure of nitrogen.
N₂ contains a triple bond (N≡N) — one σ and two π bonds. The N≡N bond energy is 945 kJ mol⁻¹ — the highest of any homonuclear diatomic bond. Breaking this bond requires enormous activation energy. Additionally: N₂ is non-polar (no δ+/δ−) so electrophiles and nucleophiles cannot easily attack. The small size of N also causes steric resistance. The lone pairs on each N are not easily donated (low-energy, held close to the nucleus). All these factors combine to make N₂ very unreactive at room temperature.
-
State the conditions and write the overall equation for the Haber process. Explain why the temperature used (450°C) is a compromise.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Conditions: ~450°C, ~200 atm, Fe catalyst with K₂O/Al₂O₃ promoters. Temperature compromise: the forward reaction is exothermic — lowering temperature shifts equilibrium right (Le Châtelier), giving better yield. But lower temperature = slower rate of reaction (fewer molecules with Eₐ). At 450°C: rate is acceptably fast AND yield is reasonable (~15% per pass). The catalyst helps achieve this yield at a lower temperature than would otherwise be needed. Higher temperature would increase rate but give very poor yield.
-
Write the three steps of the Ostwald process for manufacturing nitric acid. State the catalyst and conditions for Step 1.
Step 1: 4NH₃ + 5O₂ → 4NO + 6H₂O — Pt/Rh gauze catalyst, ~850°C, 5–10 atm. Step 2: 2NO + O₂ → 2NO₂ — cooling the gas; NO is oxidised in air. Step 3: 3NO₂ + H₂O → 2HNO₃ + NO — NO recycled to Step 2. Overall: NH₃ is converted to HNO₃. The process is linked to the Haber process (NH₃ feedstock).
-
Explain why phosphorus can form PCl₅ but nitrogen cannot form NCl₅. Include reference to electron configuration and orbital availability.
Nitrogen is in Period 2 — its electron configuration is [He]2s²2p³. The 2p subshell is the highest available; there are no empty low-energy d orbitals accessible. Nitrogen cannot exceed an octet → maximum 4 bonds. Phosphorus is in Period 3 — [Ne]3s²3p³. It has empty 3d orbitals at accessible energy. When 5 Cl atoms approach, the 3d orbitals can participate in bonding → sp³d hybridisation → 5 bonds → PCl₅ (trigonal bipyramidal). This "octet expansion" is only possible for elements in Period 3 and below that have available d orbitals.
-
Compare the structures, melting points, and reactivities of white and red phosphorus. Explain the differences in terms of their structures.
White P: discrete P₄ molecules (tetrahedral), simple molecular solid, m.p. = 44°C (only weak van der Waals between P₄ units). Very reactive — P–P–P bond angle is 60° (severe strain); ignites in air at 34°C. Red P: polymeric chain structure (P₄ tetrahedra linked into a network), giant molecular/polymeric solid, m.p. = ~600°C (many covalent bonds must be broken). Much more stable — the strain is relieved in the polymeric structure; only ignites at ~250°C. Reactivity: white >> red, because white P has strained bonds that release energy on reaction, while red P has a more stable structure with less strain energy.
-
Write equations to show: (a) the reaction of concentrated HNO₃ with copper (b) the reaction of dilute HNO₃ with copper. Explain the difference in the nitrogen oxide produced.
(a) Concentrated: Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂↑ + 2H₂O. Brown/reddish-brown gas NO₂ produced. (b) Dilute: 3Cu + 8HNO₃(dil.) → 3Cu(NO₃)₂ + 2NO↑ + 4H₂O. Colourless gas NO produced (turns brown when it meets air: 2NO+O₂→2NO₂). Difference: concentrated HNO₃ is a stronger oxidising agent → N is reduced further from +5 to +4 (NO₂). Dilute HNO₃ → N reduced from +5 to +2 (NO). Both reactions: HNO₃ acts as oxidising acid; H⁺ + NO₃⁻ oxidises the metal.
Multiple Choice Quiz — 25 Questions
Unit 10: Group 15 Elements and Compounds
25 Questions · Select one answer eachWhat is the outer electron configuration of all Group 15 elements?
The bond energy of N≡N is 945 kJ mol⁻¹. This makes N₂:
The conditions for the Haber process are approximately:
Why is a high pressure used in the Haber process?
The shape of NH₃ is trigonal pyramidal with a bond angle of 107°. The angle is less than 109.5° because:
Which reagent produces white fumes when held near a flask of ammonia?
In the Ostwald process, what catalyst is used in Step 1 (oxidation of NH₃)?
What gas is produced when copper reacts with dilute nitric acid?
The oxidation state of nitrogen in HNO₃ is:
White phosphorus ignites spontaneously in air because:
Why does NH₃ have a much higher boiling point than PH₃, despite having a lower molar mass?
Phosphorus pentachloride (PCl₅) has a trigonal bipyramidal shape because:
The reaction of PCl₅ with water produces:
What is the main product of the thermal decomposition of ammonium nitrate (NH₄NO₃) at ~230°C?
Which of the following is the oxidation state of N in NO₂?
Superphosphate fertiliser is made by treating calcium phosphate with:
The reaction 2NO + O₂ → 2NO₂ is responsible for:
P₄O₁₀ is used industrially as a:
Which of the following statements about nitrogen fixation is correct?
The bond angle in PH₃ (93°) is much smaller than in NH₃ (107°) because:
Iron is passivated by concentrated nitric acid. This means:
Phosphoric acid (H₃PO₄) is described as triprotic. What does this mean?
Which oxide of nitrogen is a radical (contains an unpaired electron)?
The thermal decomposition of potassium nitrate (KNO₃) produces:
Biological nitrogen fixation is carried out by:
Unit Test — 50 Marks
Section A — Short Answer
30 marksExplain why N₂ is kinetically inert at room temperature. Describe two conditions under which N₂ does react, giving an equation for each. [5]
Describe the Haber process for the manufacture of ammonia. Include: the equation, conditions, catalyst, reason for the conditions chosen, and how the process achieves a high overall yield despite low conversion per pass. [5]
Describe the Ostwald process. Write the equation for each step. State the catalyst and conditions for Step 1. Write the overall equation (from NH₃ to HNO₃). [5]
Compare white phosphorus and red phosphorus under the headings: (a) structure (b) melting point (c) reactivity with air (d) solubility in CS₂ (e) toxicity. Explain the differences in terms of structure. [5]
Explain why nitrogen cannot form NCl₅ while phosphorus forms PCl₅. State the hybridisation and shape of PCl₅. Write an equation for the hydrolysis of PCl₅. [5]
Give the oxidation state of nitrogen in each species: (a) N₂ (b) NH₃ (c) N₂O (d) NO (e) NO₂ (f) HNO₃ (g) NO₃⁻ (h) NH₄⁺ (i) N₂H₄ (j) NaNO₂. [5 — ½ mark each]
Section B — Extended Answer
20 marks(a) Describe the industrial importance of ammonia and nitric acid, giving three specific uses of each. [4]
(b) Explain the environmental problems associated with nitrogen oxides (NOₓ) in the atmosphere, including acid rain, photochemical smog, and ozone depletion. Write a relevant equation for each. [6]
(a) Compare ammonia (NH₃) and phosphine (PH₃) with respect to: structure and bond angle, boiling point and reason, basic strength in water, stability, and ability to form 5-coordinate compounds. [5]
(b) Describe the nitrogen cycle, explaining how nitrogen moves between the atmosphere and living organisms. Include the roles of nitrogen-fixing bacteria, nitrifying bacteria, denitrifying bacteria, and decomposers. [5]