BONDING

1. BONDING song

2. Introduction to Bonding • Atoms are generally found in nature in combination held together by chemical bonds. • A chemical bond is a mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together. • There are two types of chemical bonds: ionic, and covalent. song

3. Introduction to Bonding • What determines the type of bond that forms? • The valence electrons of the two atoms involved are redistributed to the most stable arrangement. • The interaction and rearrangement of the valence electrons determines which type of bond that forms. • Before bonding the atoms are at their highest possible potential energy

4. Introduction to Bonding • There are understandings of bond electron interaction • One understanding of the formation of a chemical bond deals with balancing the opposing forces ofrepulsion and attraction • Repulsion occurs between thenegative e-clouds of each atom • Attraction occurs between thepositive nucleiand the negative electron clouds

5. Introduction to Bonding • When two atoms approach each other closely enough for their electron clouds to begin to overlap • The electrons of one atom begin to repel the electrons of the other atom • And repulsion occurs between the nuclei of the two atoms

6. Introduction to Bonding • As the optimum distance is achieved that balances these forces, there is a release of potential energy • The atoms vibrate within the window of maximum attraction/minimum repulsion • The more energy released the stronger the connecting bond between the atoms

8. Introduction to Bonding • Another understanding of the form-ation of a chemical bond between two atoms centers on achieving the most stable arrangement of the atoms’ valence electrons • By rearranging the electrons so that each atom achieves a noble gas-like arrangement of its electrons creates a pair of stable atoms (only occurs when bonded)

9. Ionic Bonding Covalent Bonding Introduction to Bonding • Sometimes to establish this arrange-ment one or more valence electrons are transferred between two atoms • Basis for ionic bonding • Sometimes valence electrons are shared between two atoms • Basis for covalent bonding

10. Introduction to Bonding • A good predictor for which type of bonding will develop between a set of atoms is the difference in their electronegativities. • Remember, electronegativity is a measure of the attraction an atom has for e-s after developing a bond • The more extreme the difference between the two atoms, the less equal the exchange of electrons

12. Introduction to Bonding • Let’s consider the compound Cesium Fluoride, CsF. • The electronegativity value (EV) for Cs is .70; the EV for F is 4.00. • The difference between the two is 3.30, which falls within the scale of ionic character. • When the electronegativity difference between two atoms is greater than 2.1 the bond is mostly ionic.

14. Introduction to Bonding • The take home lesson on electro-negativity and bonding is this: • The closer together the atoms are on the P.T., the more evenly their e- interact, and are therefore more likely to form a covalent bond • The farther apart they are on the P.T., the less evenly their e- interact, and are therefore more likely to form an ionic bond. metal w/nonmetal = ionic nonmetal w/nonmetal = covalent

15. Introduction to Covalent Bonding • In a co-valent bond: • The electronegativity difference between the atoms involved is not extreme • So the interaction between the involved electrons is more like a sharing relationship • It may not be an equal sharing relationship, but at least the electrons are being “shared”.

16. Cl Cl Cl Cl Shared Electrons Covalent Bonds Lets look at the molecule Cl2 +

17. Shared electrons are counted with both atoms Cl 2 each atom must have 8 valence e's Cl Cl Notice 8 e- in each valence shell!!!

18. H Cl Cl H 2.1 3.0 Covalent Bonds How about the molecule HCl? + (Polar Covalent) shared, but not evenly

19. So what’s the bottom line? To be stable the two atoms involved in the covalent bond share their electrons in order to achieve the arrangement of a noble gas.

20. Introduction to Ionic Bonding • In an ion - ic bond: • The electronegativity difference is extreme, • So the atom with the stronger pull doesn’t really share the electron • Instead the electron is essentially transferred from the atom with the least attraction to the atom with the most attraction

21. + - - - - - - - - - - - - - - - - - - - - - - - - - - - - An electron is transferred from the sodium atom to the chlorine atom + Na Cl

22. + - - - - - - - - - - - - - - - - - + - - - - - - - - - - - Both atoms are happy, they both achieve the electron arrangement of a noble gas. Notice 8 e- in each valence shell!!! -1 +1 Na Cl

23. Very Strong Electrostatic attraction established… IONIC BONDS

25. So what’s the bottom line? To be stable the two atoms involved in the ionic bond will either lose or gain their valence electrons in order to achieve a stable arrangement of electrons.

26. Bond Energies and Bonding • As we’ve learned so far ionic com-pounds are formed by the transfer of electrons from ametal to a nonmetal • The ionic compound is held together by the strong electrostatic attraction between oppositely charged ions. • There is a tremendous amount ofenergystored in thebondsformed in an ionic compound.

27. Bond Energies and Bonding • It takes a lot of energy (A.K.A. bond energy) to pull the two ions apart once they have established their stable arrangement through bonding • Energy can bereleasedorabsorbedwhen ions form • Removing electrons from atoms requires aninputof energy • Remember from last chapter this energy is calledionization energy

28. Energy and Ionic Bonding • On the other hand adding electrons to atomsreleasesenergyinto the environment • Remember this has to do with the atomsaffinityfor electrons • Sometimes this energy is used to help remove the electron from another atom • The ionization energy to remove 1 e- from each atom in a mole of Na atoms is495.8kJ

29. Energy and Ionic Bonding • A mol of Cl atoms releases 348.6 kJ when an e- is added to the atom • Notice that it takes more energy to remove Na’s e- than the amount released from the Cl atoms. • Forming an ionic bond is a multi-step process • The final step releases a substantial amount of energy (a.k.a. the driving force)

30. Cl Na 2 +110.8 kJ/mol Na Crystal Formation At the beginning there is solid sodium and chlorine gas. Na(s) & Cl2(g) Step #1 A mol of sodium is converted from a solid to a gas: Na(s) + energy  Na(g) Step #2 ENERGY IN

31. + Na Na +495.8 kJ/mol e +100.8 kJ/mol Cl 2 One electron is then removed from each sodium atom of form a sodium cation Na(g) + energy  Na+(g) + e- Step #3 ENERGY IN Energy is required to break the bond holding 0.5mol of Cl2 molecules together to form a mole of chlorine atoms Cl2(g) + energy  2Cl(g) Step #4 ENERGY IN

32. Cl - Cl e The next step involves adding an electron to each chlorine atom to form a chloride anion: Cl(g) + e- Cl-(g) + energy Step #5 +348.6 kJ/mol ENERGY OUT The final step provides the driving force for the reaction. Na+(g) + Cl-(g)  NaCl(s) + energy Step #6 +787.5 kJ/mol ENERGY OUT

33. Crystal Formation • Energy released in the final step is called the lattice energy • Energy released when the crystal lattice of an ionic solid is formed • For NaCl, the lattice energy is 787.5 kJ/mol, which is greater than the input of energy in the previous steps • The lattice energy provides enough energy to allow for the formation of the sodium ion

34. Crystal Formation • We can use the lattice energy as a method for measuring the strength of thebond in ionic compounds. • The amount of energy necessary to break a bond is called bond energy. • This energy is equal to the lattice energy, but • Bond energy movesinto the system • Lattice energy movesout of the system

35. Bond energy Lattice energy Compound kJ/mol (out) kJ/mol (in) LiCl -861.3 +861.3 LiBr -817.9 +817.9 LiI -759.0 +759.0 NaCl -787.5 +787.5 NaBr -751.4 +751.4 NaI -700.1 +700.1 CaF2 -2634.7 +2634.7 MgO -3760.2 +3760.2

36. END OF THE GOOD STUFF • ARE YOU READY TO TRY • If you are using this for review of the classroom presentation STOP HERE

37. Hydrate Formation • In the construction of a crystal lattice, depending on the ions involved there can be small “pores” develop between ions in the ionic crystal. • Some ionic compnds have enough space between the ions that water molecules can get trapped in between the ions • Ionic compounds that absorb water into their pores form a special type of ionic compound called a hydrate.

38. Hydrate Formation • Hydrates typically have different properties than their dry versions - A.K.A.anhydrides • Anhydrous CuSO4 is nearlycolorless • CuSO4•5 H2O is a brightblue color • WhenCopper (II) Sulfateis fully hydrated there are5 water moleculespresent for every Copper ion. • The hydrated name would be Copper (II) Sulfate Pentahydrate

39. Hydrate Formation • Have you ever bought a new purse or camera and found a small packet of crystals labeled – do not eat? • These crystals are there to absorb water that might lead to mildew or mold • The formula of a hydrate is XAYB • Z H2O(Z is a coefficient indicating how many waters are present per formula unit)

40. Percent Composition • The FDA requires manufacturers to provide nutritional information on the labels of processed food products. • Dietary guidelines are based on the percent of calories that the average person should consume from fats, carbohydrates, proteins, etc.

41. Percent Composition • percent composition in a compnd can be determined in 2 ways • The 1st is by calculating the percent composition by mass from a chemical formula. • The 2nd is a lab scenario where an unknown compound is chemically broken up into its individual components and percent compo-sition is determined by analyzing the results.

42. 18 g/mol What is the percent composition of Hydrogen & Oxygen in Water (H2O)? 1st Assume you have a mole of the compound in question, and calculate its molar mass (2•1) + (1•16)= 18 g H2O 2nd Use the MM of each component and the MM of the compound to calculate the percent by mass of each component (2•1) = 2 g/mol H: 11.1% x 100 = O: 100% – 11.1% = 88.9%

43. Calculating PC Using Analysis Data • In this method, the mass of the sample is measured, then the sample is decomposed or separated into the component elements • The masses of the component ele-ments are then determined and the percent composition is calculated as before • divide the mass of each element by the total mass of the sample and multiply by 100.

44. Find the percent composition of a compound that contains 1.94g of carbon, 0.48g of Hydrogen, and 2.58g of Sulfur in a 5.0g sample of the compound. • Calculate the percents for each component by the equation: (Component Mass/Total Sample Mass) x 100 C: 1.94g/5.0g x 100 = 38.8% H: 0.48g/5.0g x 100 = 9.6% S: 2.58g/5.0g x 100 = 51.6%

45. Empirical Formulas • Percent compositions can be used to calculate the a simplechemical formulaof a compound, called an empirical formula • Empirical formula is the simplest ratio of the atoms in a compound • Ionic compounds are always written as empirical formulas

46. Empirical Formulas • Procedure for calculating Empirical Formula • convert the percent compositions into moles • compare the mols of each compo-nent to calculate the simplest whole number ratio • divide each amount in moles by the smallest of the mole amounts • This sets up a simple ratio

47. Calculate the empirical formula of a compound that is 80.0% Carbon and 20.0% Hydrogen by mass • We have 80 grams of Carbon and 20 grams of Hydrogen • We need to calculate the number of moles of each element that we have. Since we don’t know the original mass of the sample, we can assume a 100 g sample:

48. 6.66 mol 6.66 mol = = 6.66 mol C 19.8 mol H Calculating Empirical Formulas 1 mole C 80.0g C 12.01 g C = 1 • Now we need to calculate the smallest whole number ratio in order to find the empirical formula. • Divide each component by the smallest number in moles CH3 1 mole H 20.0g H = 2.97 1.008 g H

49. .01345mol .01345mol Calculating Empirical Formulas Determine the empirical formula of a compound containing 2.644g of Au and 0.476g of Cl. 1 mol Au 2.664g Au = .01352mol Au 197 g Au = 1 1 mol Cl .476g Cl = .01345mol Cl = 1 35.4 g Cl AuCl

50. Molecular Formulas • The empirical formula for a compound provides the simplest ratio of the atoms in the compound • However, it does not tell you the actual numbers of atoms in each molecule of the compound • For instance the empirical formula for glucose is CH2O (1:2:1) • While the molecular formula for glucose is C6H12O6