2. Covalent Bonds and Compounds Molecules

3. Three Kinds of Bonds 1. Non-metal to non-metal 2. metal to non-metal 3. metal to metal Covalent Covalent Ionic Ionic Metallic Metallic

4. Bonds and Electronegativity Electrons are transferred between atoms when the difference in electronegativity between the atoms is quite high. Electrons are transferred between atoms when the difference in electronegativity between the atoms is quite high. The amount of transfer depends on the electronegativity difference. The amount of transfer depends on the electronegativity difference.

5. Bonds and Electronegativity The number 1.67 seems to be the magic number.(Note:1.67 rounds to 1.7) The number 1.67 seems to be the magic number.(Note:1.67 rounds to 1.7) If the electronegativity difference is less than 1.67, the bond is more covalent than ionic. If the electronegativity difference is less than 1.67, the bond is more covalent than ionic. If the electronegativity difference is greater than 1.67, the bond is more ionic than covalent. If the electronegativity difference is greater than 1.67, the bond is more ionic than covalent.

6. Electronegativity Difference Covalent < 1.67 Covalent < 1.67 However – 0 -.6 is non-polar covalent 0 -.6 is non-polar covalent. 6 – 1.67 polar covalent. 6 – 1.67 polar covalent There are 7 instances of perfectly covalent bonds (electronegativity difference = 0) There are 7 instances of perfectly covalent bonds (electronegativity difference = 0) H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2

8. Nomenclature (naming) Covalent bonds are generally between two non-metals. Covalent bonds are generally between two non-metals. CO CO CO 2 CO 2 - Carbon monoxide - Carbon dioxide

9. Prefixes 1 2 3 4 5 6 7 8 9 - Mon(o) - Di - Tri - Tetr(a) - Pent(a) - Hex(a) - Hept(a) - Oct - Non

10. Nitrogen and oxygen (five molecules) N 2 O N 2 O NO NO N 2 O 3 N 2 O 3 NO 2 NO 2 N 2 O 5 N 2 O 5 - Dinitrogen monoxide - Nitrogen monoxide - Dinitrogen trioxide - Nitrogen dioxide - Dinitrogen pentoxide

11. Organic Molecules (hydrocarbons) Composed primarily of carbons (always) and hydrogens (usually). Composed primarily of carbons (always) and hydrogens (usually). Three primary types Three primary types Alkanes – only single bonds Alkanes – only single bonds Alkenes – at least one double bond Alkenes – at least one double bond Alkynes – at least one triple bond Alkynes – at least one triple bond

12. Root name from # of Carbons 1 – meth 1 – meth 2 – eth 2 – eth 3 – prop 3 – prop 4 – but 4 – but 5 – pent 5 – pent 6 – hex 6 – hex 7 – hept 7 – hept 8 – oct 8 – oct 9 – non 9 – non 10 - dec 10 - dec

13. Some examples H H-C-H H C- - - H-C- C- HH H HH methane ethane Propane butane CH 4 C2H6C2H6 C3H8C3H8 C 4 H 10 Notice: # of H is 2n+2 # of carbons

14. Alkenes H-C- C- HH H HH ethane ethene # of H’s is just 2n H H-C=C-H

15. Saturated vs. Unsaturated H-C-C-C-C-C-C-C-C-C-H ɩ ɩɩ ɩ ɩɩɩɩɩ ɩ ɩ ɩɩɩɩ ɩɩɩ H H H H H H H H H H H H H H H H H-C-C=C-C-C-C-C-C-C-H ɩ ɩ ɩɩɩɩɩɩɩ ɩ ɩ ɩ ɩ H H H H H H H H H ɩ ɩɩ H-C-C-C=C-C-C-C=C-C-H ɩ ɩ ɩɩɩɩɩɩɩ ɩ ɩ ɩ ɩɩ H H H H H H H H H H H H H H nonane 2-nonene 2,6 nonadiene

16. Formula Mass Mass of one formula unit. Mass of one formula unit. Add the mass of all of the elements times their subscripts. Add the mass of all of the elements times their subscripts. Sodium phosphate: Sodium phosphate: Na 3 PO 4 Na 3 PO 4 Na = 22.99 x 3 = 68.97 amu Na = 22.99 x 3 = 68.97 amu P = 30.97 x1 = 30.97 amu P = 30.97 x1 = 30.97 amu O = 16.00 x 4 = 64.00 amu O = 16.00 x 4 = 64.00 amu 68.97 + 30.97 + 64 = 163.94 amu 68.97 + 30.97 + 64 = 163.94 amu

17. Percent Composition What is the percent of sodium in sodium phosphate? What is the percent of sodium in sodium phosphate? What is the mass of sodium in Na 3 PO 4 ? What is the mass of sodium in Na 3 PO 4 ? Na = 22.99 x 3 = 68.97 amu Na = 22.99 x 3 = 68.97 amu What is the mass of Na 3 PO 4 ? What is the mass of Na 3 PO 4 ? 163.94 amu 163.94 amu % of Na is 68.97/163.94 x 100 or % of Na is 68.97/163.94 x 100 or 40.07% 40.07%

18. Empirical Formulas A formula in lowest terms. (Use the GCF) A formula in lowest terms. (Use the GCF) Greatest Common Factor Greatest Common Factor The empirical formula for C 2 H 6 is The empirical formula for C 2 H 6 is CH 3 CH 3 What is the formula mass for C 2 H 6 ? What is the formula mass for C 2 H 6 ? What is the formula mass for CH 3 ? What is the formula mass for CH 3 ? What is 30 ÷ 15? What is 30 ÷ 15?

19. Determining Empirical Formula from Percent composition. A certain compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. A certain compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. We’re going to do percent composition in reverse… We’re going to do percent composition in reverse… Assume a 100g sample – therefore 32.38g of sodium etc. Assume a 100g sample – therefore 32.38g of sodium etc.

20. These are the subscripts Smallest value Divide all values by the smallest of them(this will give whole numbers) Determine the molar ratios So, the empirical formula is Na 2 SO 4 So, the empirical formula is Na 2 SO 4 Na S O 32.28g1 mole 22.99g 22.65g1 mole 32.07g 44.99g1 mole 16.0g = 1.408 moles = 0.7063 moles = 2.812moles 0.7063 = 2 = 1 = 4

21. Actual Formula from Empirical Formula If you know the empirical formula and the formula mass of the actual formula, you can determine actual formula by finding the formula mass of the empirical formula and dividing it into the formula mass of the actual formula. This will give you the GCF. If you know the empirical formula and the formula mass of the actual formula, you can determine actual formula by finding the formula mass of the empirical formula and dividing it into the formula mass of the actual formula. This will give you the GCF. Multiply the subscripts of the empirical formula by the GCF. Multiply the subscripts of the empirical formula by the GCF.

22. Example What is the formula of a compound whose molecular mass is 150.1amu and its empirical formula is CH 2 O? What is the formula of a compound whose molecular mass is 150.1amu and its empirical formula is CH 2 O? Formula mass of CH 2 O is 30.02amu. Formula mass of CH 2 O is 30.02amu. 150.1 ÷ 30.02 = 5 (That’s the GCF) 150.1 ÷ 30.02 = 5 (That’s the GCF) Multiply the subscripts 1,2,1 by the GCF (5) which gives the new subscripts of 5,10,5 or Multiply the subscripts 1,2,1 by the GCF (5) which gives the new subscripts of 5,10,5 or C 5 H 10 O 5 C 5 H 10 O 5

23. Metallic Bonds Bonding in metals is due to delocalized electrons. Bonding in metals is due to delocalized electrons. These often exist in what is called a sea of electrons. These often exist in what is called a sea of electrons. Metal atoms Sea of “delocalized” electrons

24. Metallic Bonds This explains many of the properties of metals: This explains many of the properties of metals: Malleable Malleable Ductile Ductile Conducts electricity well Conducts electricity well

25. Alloys Two Metals (and sometimes other substances) bonded (mixed) together. Two Metals (and sometimes other substances) bonded (mixed) together.

26. Alloys

27. Alloys

28. Alloys