CL

Chapter 5 (Part 2): Binary Ionic Nomenclature

Ionic Compounds: Core Ideas

  • Binary ionic compounds = compounds composed of exactly two elements (one metal, one non-metal).
  • Formation process
    • Metal (electron donor) loses 1⁺, 2⁺, 3⁺… electrons → forms a cation (positive ion).
    • Non-metal (electron acceptor) gains electrons → forms an anion (negative ion).
    • Electrostatic attraction between oppositely charged ions holds the lattice together.
  • Structural nature
    • Do not exist as discrete molecules; instead form an extended 3-D crystal lattice of alternating cations/ anions.
    • Smallest representative unit = formula unit (gives the simplest whole-number ratio of ions; e.g., NaCl means 1 Na⁺ : 1 Cl⁻).
  • Electrical neutrality rule
    • Overall charge of a formula unit is 0, so \sum \text{positive charges} + \sum \text{negative charges}=0.

Predictable Charges from the Periodic Table

  • Metals forming only one common charge (no Roman numeral in names)
    • Group 1: +1 (e.g.
    • \text{Li}^+, \text{Na}^+, \text{K}^+)
    • Group 2: +2 (e.g.
    • \text{Mg}^{2+}, \text{Ca}^{2+})
    • Group 13 elements Al, Ga: +3
    • Transition-metal exceptions (single stable charge)
    • \text{Zn}^{2+}, \text{Cd}^{2+}, \text{Ag}^+, \text{Sc}^{3+}
  • Common non-metal (fixed) charges
    • Halogens (Group 17): -1
    • Chalcogens O, S, Se: -2
    • N, P: -3

Algorithm: Writing a Binary Ionic Formula ("Crossover" approach)

  1. Write element symbols with charges: Metal cation first, non-metal anion second.
  2. Cross over magnitudes of charges → become subscripts for the other ion.
    • Example: Al³⁺ with O²⁻ → \text{Al}2\text{O}3 because 3 → O, 2 → Al.
  3. Reduce subscripts to the smallest integer ratio (divide by their greatest common divisor).
  4. Verify neutrality: multiply each ion’s subscript by its charge; their sum must be 0.
    • Check for \text{Al}2\text{O}3:
      2( +3 ) + 3( -2 ) = +6 -6 = 0 ✔️

Extra Worked Formula Examples from Transcript

  • Aluminum + sulfide
    • Charges: \text{Al}^{3+}, \text{S}^{2-}
    • Swap → \text{Al}2\text{S}3 (can’t reduce).
  • Magnesium + oxide
    • Charges: \text{Mg}^{2+}, \text{O}^{2-}
    • Cross → \text{Mg}2\text{O}2, reduce → \text{MgO}.

Naming Binary Ionic Compounds (No Variable Charge)

  • General form: name of metal + base of non-metal + “-ide”.
  • Never use prefixes (mono-, di-, etc.) for ionic names.
  • Common “-ide” conversions
    • F → fluoride, Cl → chloride, Br → bromide, I → iodide
    • O → oxide, S → sulfide, N → nitride
  • Examples
    • \text{MgCl}_2 → magnesium chloride
    • \text{Na}_2\text{O} → sodium oxide
    • \text{CaBr}_2 → calcium bromide

Metals with Variable Charges: Stock (Roman Numeral) System

  • Many transition metals & a few post-transition metals can form more than one stable ion (e.g., Fe²⁺ / Fe³⁺, Sn²⁺ / Sn⁴⁺).
  • Name structure:
    1. Metal name
    2. Roman numeral in parentheses = magnitude of metal’s charge (not subscript count!)
    3. Non-metal base + “-ide”.
    • \text{Fe}^{2+} = iron(II), \text{Fe}^{3+} = iron(III).
  • Older Latin ous/ic system (NOT tested but may appear)
    • Lower charge → “-ous” (ferrous = Fe²⁺), higher → “-ic” (ferric = Fe³⁺).

Determining Unknown Metal Charge from a Formula

  1. Let x = metal charge.
  2. Multiply non-metal charge by its subscript → total negative charge.
  3. Set x(\text{metal subscript}) + \text{total neg.} = 0 and solve for x.
    • Example: \text{FeCl}_3
      • Cl has -1; three Cl → -3.
      • One Fe must be +3 → iron(III) chloride.

Special & Exception Cases

  • Metals that always keep single charge (no Roman numeral): \text{Ag}^+ (silver), \text{Zn}^{2+}, \text{Cd}^{2+}, \text{Sc}^{3+}, \text{Al}^{3+}, \text{Ga}^{3+}.
  • Mercury(I) ion
    • Diatomic cation \text{Hg}_2^{2+} (each Hg bears +1).
    • Named “mercury(I)” despite 2 Hg atoms because numeral references individual atom charge.

Naming & Formula Exercises from Transcript

  • Silver chloride → \text{Ag}^+ vs \text{Cl}^- → \text{AgCl}.
  • Iron(III) oxide → Fe³⁺, O²⁻ → \text{Fe}2\text{O}3.
  • Tin(IV) oxide → Sn⁴⁺, O²⁻ → initial \text{Sn}2\text{O}4 → reduce → \text{SnO}_2.
  • PbO₂ → oxygen totals -4 (two O²⁻). Lead must be +4 → lead(IV) oxide.
  • \text{Hg}_2\text{O}
    • O²⁻ contributes -2, 2 Hg atoms total +2 → each Hg +1 → mercury(I) oxide.
  • \text{FeCl}_3 analyzed above → iron(III) chloride.

Quick Naming Flowchart

  1. Does the formula start with a metal?
    • If NO → probably molecular or acid; ionic rules don’t apply.
    • If YES → go to 2.
  2. Does the metal form only one charge? (Group 1, 2, Al, Ga, Zn, Cd, Ag, Sc)
    • If YES → metal name + non-metal*ide.
    • If NO → determine charge, insert Roman numeral, add non-metal*ide.
  3. Check charge balance to validate formula.

Key Takeaways & Exam Tips

  • Always respect charge neutrality; most mistakes arise from forgetting to cross-check charges.
  • Prefixes (mono-, di-, …) never appear in ionic names.
  • Roman numerals ≠ subscripts; they show charge magnitude.
  • Reduce subscripts to lowest terms; a correct but unreduced formula loses points.
  • Be cautious of mercury(I): diatomic \text{Hg}_2^{2+}.
  • Memorize fixed-charge transition metals to spare time on Roman numerals.