Explaining Molecular Formulas Section 3.2 (pg. 85-90)

Explaining Molecular Formulas Objectives: Draw electron dot diagrams of atoms and molecules, writing structural formulas for molecular substances using Lewis structures to predict bonding in simple molecules Illustrate, by drawing or building models, the structure of simple molecular substances Explain why the formulas for molecular substances refer to the number of atoms of each constituent element Section 3.2 (pg. 85-90)

Molecular Elements Many molecular elements are diatomic and some are polyatomic You will need to memorize the formulas of the 9 molecular elements as they will not be given to you: Name Symbol hydrogen H2(g) nitrogen N2(g) oxygen O2(g) fluorine F2(g) chlorine Cl2(g) iodine I2(g) bromine Br2(g) phosphorous P4(g) sulfur S8(g)

In structural formulas, lone pairs are not shown Why are they diatomic? Remember fluorine has 7 valence e-’s and needs 1 more e- to be stable? Well 2 fluorine atoms could obtain a stable octet of e-’s if they shared a pair with each other F - F F2 In structural formulas, lone pairs are not shown Remember: This is a simplified 2-D version, not where the electrons actually are

Diatomic Elements O = O N Ξ N What about oxygen and nitrogen? Would sharing only 1 electron each work? What about oxygen and nitrogen? O = O Each oxygen atom only has 6 valence electrons. So by sharing 2 electrons each, the two oxygen atoms can create a full octet. This creates a double bond N Ξ N Each nitrogen atom only has 5 valence electrons. So by sharing 3 electrons each, the two nitrogen atoms can create a full octet. This creates a triple bond

Molecular Compounds Background: Molecular compounds have covalent bonds (shared electrons) between non-metals and non-metals Can be solid, liquid or gas as SATP May or may not be soluble in water (more later) Don’t ever conduct electricity - even when (aq) Generally have lower m.p. and b.p than ionic compounds Review from Section 1.5 Notes

Molecular Compounds Background: Empirical Formulas – show the simplest whole number ratios of atoms in a compound Very useful for IONIC compounds Formula Unit – the ratio of ions that repeats in a pattern within the crystal; the chemical formula of ionic compounds represents the formula unit Not useful for MOLECULAR compounds Na242Cl242  Na16Cl16  NaCl CH  C2H2  acetylene  C6H6  benzene  C8H8  octene All are extremely different compounds but the empirical formula would be the same

Molecular Compounds Background: C2H4O2  CH3COOH Molecular Formulas – a molecular formula shows the actual number of atoms that are covalently bonded to make-up each molecule We use this because chemical formulas for molecular compounds result from sharing electrons, therefore a variety of compounds are possible (which we determine empirically through experiments) Often the symbols are written in a sequence that helps you determine how the atoms are bonded C2H4O2  CH3COOH Empirical formula: CH2O incorrect for molecular compounds

Review of Molecular Compound Formulas See Pg. 88

Q: How do we know how molecular compounds bond? (Aka: How do we draw Lewis Formulas?) Where do these come from???

Determining Lewis Formulas Bonding Capacity: the maximum number of single covalent bonds that an atom can form REMINDER: How many e-’s does an atom want in its valence energy level to be satisfied? H = 2 e- C, N, O, F, P, S, Cl, etc. = 8 e- REMINDER: What types of covalent bonds are possible? F – F single = sharing one e- pair O = O double = sharing two e- pairs N Ξ N triple = sharing three e- pairs

Determining Lewis Formulas So why do we care about bonding capacity? If we know how many bonding e-’s an atom has, we can predict what structure a molecular compound will have Atom Number of valence electrons bonding electrons Bonding capacity carbon 4 nitrogen 5 3 oxygen 6 2 halogens 7 1 hydrogen H I.e. Carbon can form 4 single bonds, 2 double bonds, 1 triple and 1 single, or 1 double and 2 singles

Lewis Formulas- Guided Ex. #1 Determine the Lewis formula and structural formula for sulfur trioxide, SO3(g) 1. Count the number of valence electrons there are in total? (If polyatomic ions are included, subtract or add electrons to account for the net charge) Oxygen = 3 atoms x 6 valence e-’s each = 18 valence e-’s Sulfur = 1 atom x 6 valence e-’s each = 6 valence e-’s 24 valence e-’s

Lewis Formulas – Guided Ex. #1 2. Choose your central atom In our course, we will limit our formulas to ones with one central atom (unless extra info is provided) So how do you know which is the central atom? Usually the one in lesser quantity (SO3(g)) OR The one with the higher bonding capacity Carbon usually – because 4 is the highest bonding capacity So which is the central atom?

Lewis Formulas – Guided Ex. #1 3. Arrange peripheral atoms around central atom and place one pair of valence e-’s between them 4. Place lone pairs on all peripheral atoms to complete their octet 5. Place any remaining valence e-’s on the central atom as lone pairs. For this example, all 24 have been assigned

Lewis Formulas – Guided Ex. #1 6. If the central atom’s octet is not complete, move a lone pair from a peripheral atom to a new position between the peripheral and central atom. 7. Show the structural formula but omit lone pairs and replace every bond with a line (Count the electrons around each atom to confirm the octet rule. Each atom should have 8 e-’s around it; exception: H)

Lewis Formulas – Guided Ex. #2 Determine the Lewis formula & structural formula for the nitrate ion, NO3- 1. Count the valence electrons (*look for a net charge if an ion). nitrogen = 1 x 5 valence e-’s = 5 oxygen = 3 x 6 valence e-’s = 18 23 + 1 (b/c net charge is -1) = 24 2. Which is the central atom? Nitrogen (in lesser quantity) 3. Arrange peripheral atoms around central atom and place 1 pair of valence e-’s between them N

Lewis Formulas – Guided Ex. #2 4. Place lone pairs on all peripheral atoms to complete their octet 5. Place any remaining valence e-’s on the central atom as lone pairs. 6. If the central atom’s octet is not complete, move a lone pair from a peripheral atom to a new position between the peripheral and central atom. 7. If the entity is a polyatomic ion, place square brackets around the entire Lewis formula and then write the net charge outside the bracket on the upper left. N N N

Practice Pg. 89 #5 (a-f), 6 (a-e) Pg. 90 #1-4, 6 Watch 5 (f) there is an exception noted. The central atom does not follow the octet rule. We will go through these answers as a class. Pg. 90 #1-4, 6 The theory presented today is not absolute – there are exceptions. But rather than presenting a more detailed theory, your textbook will always note such exceptions.

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