Chemistry Labs

Grade 10

Halogens, oxygen–sulfur, nitrogen–phosphorus, carbon–silicon

Compare the chemistry of groups 17–14 through their valence patterns, characteristic molecules and industrially important compounds, then relate halogen redox behavior to thermodynamics.

IntuitionA map of the nonmetals

Elements in groups 17, 16, 15 and 14 share outer-electron patterns. Moving down a group adds an electron shell, changing size, bonding and reactivity; neighboring groups still show distinct chemical families.

This article compares their representative molecular forms and compounds: halogens, oxygen and sulfur, nitrogen and phosphorus, then carbon and silicon.

Rotate the side-by-side F₂, Cl₂, Br₂ and I₂ ball-and-stick molecules. Their X–X bond lengths are 142, 199, 228 and 267 pm, respectively.

SchoolValence patterns and group trends

Definition: Valence-electron pattern

Group 17: ns2np5ns^2np^5 (7 valence electrons); group 16: ns2np4ns^2np^4; group 15: ns2np3ns^2np^3; group 14: ns2np2ns^2np^2. These patterns help explain common bonding and oxidation states, but do not by themselves determine every compound.

Representative group trends
Group / familyValence patternTypical examples and trend
17, halogensns²np⁵F₂, Cl₂, Br₂, I₂; oxidising ability generally decreases down the group.
16, chalcogensns²np⁴O₂/O₃; sulfur commonly forms S₈ rings; oxidation states include −2, +4, +6.
15, pnictogensns²np³N₂ is kinetically inert; phosphorus has several allotropes, including P₄ molecular forms.
14, carbon familyns²np²C forms diverse covalent networks; Si is a semiconductor and forms SiO₂ and silicates.

Down a group, atomic radius generally increases and electronegativity decreases. In group 17 the aqueous standard reduction potentials fall from F₂/F⁻ to I₂/I⁻, although phase, solvent and bond-energy terms all contribute.

Common oxidation states (selected examples)
Element/familyExamples
Halogens−1; Cl, Br, I also +1, +3, +5, +7; F is −1 in compounds.
O, SO usually −2 (−1 in peroxides); S: −2, +4, +6.
N, P−3 to +5 are common; N₂ is 0.
C, Si−4 to +4; CO has C +2, CO₂ has C +4.

SchoolRepresentative chemistry across the families

Halogens are diatomic. At room conditions F₂ is a pale-yellow gas, Cl₂ a yellow-green gas, Br₂ a red-brown liquid and I₂ a dark grey-violet solid (violet vapour). A more reactive halogen displaces a less reactive halide from solution.

ClX2+2 KBr→2 KCl+BrX2\ce{Cl2 + 2KBr -> 2KCl + Br2}

Example: Halogen displacement

What forms when chlorine water is added to excess potassium bromide?

Solution

Chlorine is the stronger oxidant, so it oxidises Br⁻ to Br₂. The balanced net ionic equation is ClX2+2 BrX−→2 ClX−+BrX2\ce{Cl2 + 2Br- -> 2Cl- + Br2}; K⁺ is a spectator ion.

Hydrogen-halide acid strength in water increases HF < HCl < HBr < HI as the H–X bond becomes easier to break. As reducing agents, the halide ions strengthen in the order F⁻ < Cl⁻ < Br⁻ < I⁻; do not confuse acid strength with the oxidising strength of X₂.

Silver nitrate gives AgCl white, AgBr cream and AgI yellow precipitates; dilute ammonia dissolves AgCl, concentrated ammonia dissolves AgBr less readily, and AgI is insoluble. Chlorine reacts reversibly with water to form hydrochloric acid and hypochlorous acid; hypochlorous acid provides bleach action.

ClX2+HX2O⇌HCl+HClO\ce{Cl2 + H2O <=> HCl + HClO}

Oxygen has O₂ and O₃ allotropes; ozone is a reactive, protective trace gas in the stratosphere. Elemental sulfur commonly occurs as S₈ rings. Burning sulfur forms pungent SO₂; catalytic oxidation to SO₃ and absorption in concentrated sulfuric acid are central to the contact process. Concentrated H₂SO₄ is a strong acid and, depending on conditions, a dehydrating and oxidising agent. H₂S is a toxic weak acid and reducing agent.

2 SOX2+OX2⇌VX2OX52 SOX3\ce{2SO2 + O2 <=>[V2O5] 2SO3}
SOX3+HX2SOX4→HX2SX2OX7\ce{SO3 + H2SO4 -> H2S2O7}

Nitrogen gas is relatively inert because its N≡N bond is very strong. The Haber–Bosch process combines N₂ and H₂ reversibly to make NH₃; the Ostwald process oxidises NH₃ to nitric acid. NO₂ is brown and dimerises to colourless N₂O₄ more at lower temperature. White phosphorus consists of P₄ molecules and is dangerously reactive; phosphoric acid H₃PO₄ and phosphate salts are important in fertilisers.

NX2+3 HX2⇌2 NHX3\ce{N2 + 3H2 <=> 2NH3}

Carbon forms CO, a colourless toxic gas that binds haemoglobin, and CO₂, a linear greenhouse gas and product of complete combustion. Silicon dioxide is a giant covalent solid; silicates built from linked SiO₄ tetrahedra dominate many rocks. Crystalline silicon is a semiconductor whose conductivity can be tuned by doping.

UndergraduateRedox potentials and bond energetics

The Frost-diagram idea plots oxidation state against standard reduction potential (often through nE∘nE^\circ): the thermodynamically favored redox direction corresponds to a downhill change in free energy. For halogens, the large positive potentials make X₂ strong oxidants, with the trend F₂ > Cl₂ > Br₂ > I₂ in water.

Standard reduction potentials in water at 25 °C
CoupleE° / V
FX2/FX−\ce{F2/F-}+2.87
ClX2/ClX−\ce{Cl2/Cl-}+1.36
BrX2/BrX−\ce{Br2/Br-}+1.07
IX2/IX−\ce{I2/I-}+0.54

Fluorine has an unusually weak F–F bond because the very short bond brings three lone pairs on each small atom into strong repulsion. The net standard potential also includes atomisation, electron-affinity and hydration contributions; the high hydration of F⁻ helps make E∘(FX2/FX−)=+2.87 VE^\circ(\ce{F2/F-})=+2.87\,V exceptionally positive. Thus the weak bond helps explain, but alone does not determine, fluorine’s oxidising strength.

UndergraduateQuantitative stoichiometry in the sulfur cycle

Example: Sulfur converted to sulfuric acid

Assuming complete conversion, how many moles of H₂SO₄ can be made from 3.20 g of sulfur?

Solution

The net atom balance gives 1 mol S per 1 mol H₂SO₄. With M(S)=32.06 g mol−1M(S)=32.06\,g\,mol^{-1}, n(S)=3.20/32.06=0.0998 moln(S)=3.20/32.06=0.0998\,mol, so the theoretical amount is n(H2SO4)=0.0998 moln(H₂SO₄)=0.0998\,mol (about 0.100 mol). Industrial yield and absorption conditions are not included.

UndergraduateOxidation-state bookkeeping

2 NHX3+32 OX2→2 NO+3 HX2O\ce{2NH3 + 3/2 O2 -> 2NO + 3H2O}

Oxidation states are formal electron-bookkeeping assignments, not literal charges in covalent molecules. In the Ostwald first step N changes from −3 in NH₃ to +2 in NO, while O changes from 0 in O₂ to −2; later oxidation of NO gives NO₂, followed by absorption and reactions that produce nitric acid.

References

  • Inorganic Chemistry, 5th edition · Catherine E. Housecroft and Alan G. Sharpe, 2018
  • Chemistry of the Elements, 2nd edition · N. N. Greenwood and A. Earnshaw, 1997