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Molecular and Ionic Compound Structure and Properties

Molecular and Ionic Compound Structure and Properties. Unit 2. 2.1 Types of Chemical Bonds. Atoms or ions bond due to interactions between them, forming molecules or formula units Ionic bonding occurs when oppositely charged ions are close enough to bond

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Molecular and Ionic Compound Structure and Properties

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  1. Molecular and Ionic Compound Structure and Properties Unit 2

  2. 2.1 Types of Chemical Bonds • Atoms or ions bond due to interactions between them, forming molecules or formula units • Ionic bonding occurs when oppositely charged ions are close enough to bond • They form a crystal lattice structure whose strength is relative to Coulomb’s law

  3. Nonpolar Covalent Bond • Valence electrons shared between atoms of similar electronegativity constitute a nonpolar covalent bond • For example, bonds between C and H are effectively nonpolar even though C is slightly more electronegative than H • Nonpolar bonds are formed in diatomic molecules

  4. Polar Covalent Bond • Valence electrons shared between atoms of unequal electronegativity constitute a polar covalent bond • The atom with the higher electronegativity will develop a partial negative charge • In a single bond, greater differences in electronegativity lead to greater bond dipoles

  5. Bond Comparison • All polar bonds have some ionic character • The difference between ionic and covalent bonding is not distinct, but rather a continuum • Generally bonds between a metal and nonmetal are ionic and bonds between nonmetals are covalent • Examination of the properties of a compound is the best was to characterize the type of bonding

  6. Metallic Bonding • The valence electrons in a metallic solid are considered to be delocalized • The valence electrons are not associated with any individual atom

  7. 2.2 Intramolecular Force and Potential Energy • A graph of potential energy versus the distance between atoms describes the interactions between atoms • The following graph shows the separation between atoms at which the potential energy is lowest (when the bond will form) • It also shows the bond energy (the energy required to separate the atoms

  8. Graph of PE vs. Distance

  9. Bond Formation Example • When two hydrogen atoms approach each other, two “bad” energy things happen • Electron/electron repulsion and proton/proton repulsion • One “good” energy thing happens • Proton/electron attraction • When the attractive forces offset the repulsive forces, the energy of the two atoms decreases and a bond is formed

  10. Before Bond Formation

  11. Strength of Covalent Bonds • The bond length is influenced by both the size of the atom's core and the bond order (single, double, triple) • Bonds with a higher order are shorter and have larger bond energies (are stronger)

  12. Ionic Interaction Strength • Coulomb’s law can be used to understand the strength of interactions between cations and anions • The interaction strength is proportional to the charge on each ion, larger charges lead to stronger interactions

  13. Ionic Interaction Strength • The interaction strength increases as the distance between the centers of the ions decreases • Smaller ions lead to stronger interactions • This is when the ionic size trend becomes helpful in estimating or comparing bond strength

  14. Relative Strength of the Interactions

  15. 2.3 Structure of Ionic Solids • The cations and anions in an ionic crystal are arranged in a systematic, periodic 3-D array • This crystal lattice maximizes the attractive forces among the cations and anions while minimizing the repulsive forces • The energy of the crystal lattice is proportional to the strength of interactions between cations and anions

  16. Typical Ionic Crystal Lattice

  17. Comparative Strength

  18. Practice • Explain why the lattice energy of LiF is so much stronger than the energy of LiI • Answer: this reasoning is totally due to atomic size. Iodine is a much larger atom than fluorine, so fluorine will be closer to Li in the lattice, making a stronger interaction • This should be answered using the formula for Coulomb’s Law

  19. Another Practice • Explain why the lattice energy of NaCl is so much weaker than the energy of MgCl2 • Answer: this reasoning is due to two factors, atomic size and charge size. Mg is a smaller atom than sodium, so this enables atoms to be closer together, strengthening the interaction. Also, Mg has a 2+ charge and requires two Cl- charges to balance it, this gives a stronger interaction. (use the formula for Coulomb’s law)

  20. 2.4 Structure of Metals and Alloys • Metallic bonding can be represented as an array of positive metal ions surrounded by delocalized valence electrons • This is referred to as an “electron sea” • This array allows for alloys to be easily formed • Alloy: a metallic mixture made by combining two or more metallic elements, especially to give greater strength or resistance to corrosion.

  21. Electron Sea Model

  22. Metal Alloys • Alloys typically retain a sea of mobile electrons and so remain conducting • Often the surface of a metal or alloy is changed through a chemical reaction (for example: the formation of a chemically inert oxide layer in stainless steel, through reaction with oxygen in the air • Two main types: substitutional and interstitial

  23. Substitutional Alloy • Substitutional alloys form between atoms of comparable radius, where one atom substitutes for the other in the lattice • Example: brass, in which some Cu atoms are substituted usually with Zn • The density typically lies between those of the component metals and the alloy remains malleable and ductile

  24. Interstitial Alloy • Interstitial alloys are formed when some of the interstices (holes) in the metal structure are occupied by small atoms • The presence of the interstitial atoms changes the properties of the host metal • Frequently the alloy is harder, stronger, and less ductile

  25. Examples of the Two Types of Alloys

  26. 2.5 Lewis Structures • The Lewis structure of a molecule show how the valence electrons are arranged among the atoms in the molecule • Named after G. N. Lewis • From experimentation, chemists learned that the most important requirement for the formation of a stable compound is that the atoms achieve noble gas electron configurations

  27. Steps to Draw a Lewis Structure • Draw the Lewis structure for each atom • Add the total number of valence electrons • Add the total number of e-’s in lone pairs • Subtract the number of electrons in lone pairs from the total number of valence e-’s • Divide this number by 2, and this is usually the number of bonds you have • Draw sticks for the bonds that represent 2 e-’s, draw in lone pairs, check for octets

  28. Helpful Hints • Hydrogen only needs to form a duet; it is stable with 2 valence electrons • Carbon forms 4 bonds (some can be multiple), oxygen forms 2 bonds (can be a double bond), nitrogen forms 3 bonds (can be multiple) • The atom with the lowest electronegativity is usually the central atom • If carbon is present, it is the central atom

  29. Practice • Draw Lewis Structures for the following: • H2O • O2 • CH2O • PCl3 • CHI3 • C2H4 • CO2

  30. Answers

  31. Answers

  32. Answers

  33. Ions and Lewis Structures • You must consider the absence of an electron or the extra electrons • Subtract an absent electron from the total and add the extra electrons to the total (it helps to draw them differently) • Do not change the number of electrons on the central atom • Place the entire structure in brackets and put the charge in the upper right corner outside the brackets

  34. Practice • Draw Lewis Structures for the following: • CN- • NO+

  35. Answers

  36. Exceptions to the Octet Rule • Fewer than 8: hydrogen has at most 2 electrons to form one bond, beryllium will have only 4 valence electrons, boron will have only 6 valence electrons (BF3) • Expanded valence: this can only happen if the central element has d-orbitals, the element is surrounded by more than four valence pairs in certain compounds • Odd-electron compounds: a few stable compounds contain an odd number of valence electrons, for example, NO, NO2, ClO2

  37. Comments About the Octet Rule • The 2nd row elements C, N, O, and F obey the octet rule • 2nd row elements B and Be often have fewer than 8 e-’s. They are very reactive • 2nd row elements never exceed the octet rule • 3rd row and heavier elements often satisfy the octet rule, but can exceed the octet rule by using their empty valence d orbitals

  38. Octet Rule Violations • Write the Lewis structure for the following: • PCl5 • I3- • ClF3 • RnCl2 • BeCl2 • ICl4-

  39. Answers

  40. Answers

  41. 2.6 Resonance and Formal Charge • Sometimes more than one valid Lewis structure is possible for a given molecule • Consider NO3- • The structure implies there are two types of bonds, but experiments clearly show that there is only one type of bond

  42. Resonance • The structure for NO3- is valid, but does not represent the true bonding, so the model must be modified • There is no reason for choosing a particular O atom to have the double bond

  43. Resonance • The nitrate ion doesn’t exist as any of the three structures, but as an average of all three • Resonance: a condition occurring when more than one valid Lewis structure can be written for a molecule

  44. Example • Draw the Lewis structure for the following: • NO2- • CO32- • O3

  45. Answers

  46. Formal Charge • Formal charge: the difference between the number of valence electrons on the free atoms and the number of valence electrons assigned to the atom in the molecule • Easy way to calculate: • # valence e- − # nonbonding e- − ½ #bonding e- • Formal charge is used to determine the correct Lewis structure when more than one is possible, but experimentation is needed to confirm

  47. Example • There are two possible Lewis structures for the sulfate ion • Calculate the formal charge for all atoms on each structure realizing that 0 is ideal

  48. Answer to Previous Slide • For the first structure: • S = 6 – 0 – 4 = 2 • For each O = 6 – 6 – 1 = -1 • For the second structure: • S = 6 – 0 – 6 = 0 • For O with double bond: 6 – 4 – 2 = 0 • For O with single bond: 6 – 6 – 1 = -1

  49. Justification for Choice • Atoms in molecules try to achieve formal charges as close to zero as possible • We can expect resonance structures for the sulfate ion Usually you will see just these two

  50. Examples • Give the possible Lewis structures after considering formal charge. • XeO3 (an explosive compound, strong oxidizer, unstable but slow acting) • SO2 (a reducing agent and is used for bleaching and as a fumigant and food preservative, large quantities of sulfur dioxide are used in the contact process for the manufacture of sulfuric acid)

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