Why do atoms bond?

1. Why do atoms bond?

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 outer electrons of different atoms that binds the atoms together. • There are two types of chemical bonds: ionic, and covalent.

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

5. Introduction to Bonding • There are 2 philosophies of atom to atom interaction • One deals with balancing the opposing forces ofrepulsion and attraction • As the atoms approach repulsion occurs between thenegative e-clouds of each atom • And attraction occurs between thepositive nucleiand the negative electron clouds

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 chemical bond philosophy 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. • The more extreme the difference between the two atoms, the less equal the exchange of electrons • This leaves us with three different levels of interaction: pure covalent, polar covalent, and ionic

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 1.7 the bond is mostly ionic.

13. .3<Polar Covalent <1.7 Ionic 1.7

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 so are more likely to form a pure 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. • In between exists the polar covalent interactions

15. Rule of Thumb metal w/nonmetal = usually ionic nonmetal w/nonmetal = usually covalent

16. 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”.

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

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.

21. 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

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

23. + - - - - - - - - - - - - - - - - - + - - - - - - - - - - - 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

24. Very Strong Electrostatic attraction established… IONIC BONDS

25. Hydrated vs. Anhydrous • 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.

26. Trapped Water Molecules Hydrated Crystal

27. Hydrate Formation • Hydrates typically have somewhat different properties than their “dry” versions - A.K.A.anhydrate or anhydrous • Anhydrous copper sulfate is nearlycolorless • The hydrated version is a brightblue color • WhenCopper (II) Sulfateis fully hydrated there are5 water moleculestrapped for every Copper sulfate.

28. Hydrate Formation • These hydrate-able ionic compounds are sometimes used to indicate the presence of water. • For example, Cobalt Chloride is a compound that is blue in its anhydrous version, and magenta when it is hydrated.

29. 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.

31. Percent Composition • An important quantitative measurement that can be made for any chemical substance is a Percent Composition. • The percent composition of a compound is a relative measure of the mass of each different element present in the compound. • It gives you a rough comparison of the masses of the each component in the total sample

32. 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 composition is determined by analyzing the results.

33. 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•1.008) + (1•15.99)= 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•1.008) = 2 g/mol H: 11.1% x 100 = O: 100% – 11.1% = 88.9%

34. 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 elements 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.

35. 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%

36. Classroom Practice 1: Calculate the percent composition of Mg(NO3)2. % Mg = 16.2% % N = 18.9% % O = 64.0%

37. Empirical Formulas • Scientists communicate the atoms involved in a compound through symbolic formulas. • There are three types of formulas that chemistry use: empirical, molecular, and structural • The simplest formula is called an empirical formula • simplest ratio of the atoms in a compnd • Ionic compounds are always written as empirical formulas

38. Empirical Formulas • Procedure for calculating empirical formulas • 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

39. 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:

40. 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

41. .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

42. Classroom Practice 2: Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H. CaO2H2 or Ca(OH)2

43. 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

44. Molecular Formulas • A molecular formula indicates the numbers of each atom involved in the the compound • The molecular formula is always a multiple of the empirical formula • To calculate the molecular formula you must have 2 pieces of info. • Empirical formula • Molar mass of the unknown compound (always given)

45. 1.763mol 1.763mol Calculating Molecular Formula Find the molecular formula of a compound that contains 56.36 g of O and 54.6 g of P. The molar mass of the compound is 189.5 g/mol. 1st find the Emp. Formula: 1 mol O 56.36 g O = 3.525mol O = 1.99 15.99 g O PO2 1 mol P 54.6g P = 1.763mol P = 1 30.97 g P

46. 62.95g Calculating Molecular Formula Now determine the mass of the empirical formula: PO2: (1•30.97g P)+(2•15.99g O)= 62.95g MM Given in the problem = 189.5 g/mol 189.5 g/mol = 3.01 P3O6 Molecular Formula = 3(PO2)

47. One More: A Good 1 Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the molecular formula for methyl acetate, which has the following chemical analysis: 48.64% C, 8.16% H, and 43.20% O. The Molar Mass of the compound in question is reported as 74g/mol. 1st determine the empirical formula 2nd determine the molecular formula

48. 1 mole C = 4.050mol C 48.64g C = 1.50 12.01 g C 2.702 mol 1 mole H 8.16g H = 8.095 mol H = 2.99 1.008 g H 2.702 mol 1 mole O 43.20g O = 2.702 mol O 15.99 g O = 1.00 2.702 mol =74g C1.5H3O1 2 = C3H6O2 = 36+6+32 C9H18O6 222/74 = 3

49. Classroom Practice 3: A hydrocarbon is 84.25% carbon and 15.75% hydrogen and has a molecular weight of 114. What is its molecular formula? C8H18