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Chemical Bonds The Formation of Compounds From Atoms Chapter 11

Chemical Bonds The Formation of Compounds From Atoms Chapter 11

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Chemical Bonds The Formation of Compounds From Atoms Chapter 11

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  1. Chemical BondsThe Formation ofCompounds From AtomsChapter 11 Hein and Arena

  2. 11.2Lewis Structures of Atoms

  3. Metals form cations and nonmetals form anions to attain a stable valence electron structure.

  4. These rearrangements occur by losing, gaining, or sharing electrons. This stable structure often consists of two s and six p electrons.

  5. The Lewis structure of an atom is a representation that shows the valenceelectrons for that atom. • Na with the electron structure 1s22s22p63s1 has 1 valence electron. • Fluorine with the electron structure 1s22s22p5 has 7 valence electrons

  6. The Lewis structure of an atom uses dots to show the valence electrons of atoms. Unpaired electron B Paired electrons Symbol of the element 2s22p1 The number of dots equals the number of s and p electrons in the atom’s outermost shell.

  7. The Lewis structure of an atom uses dots to show the valence electrons of atoms. S 2s22p4 The number of dots equals the number of s and p electrons in the atom’s outermost shell.

  8. Lewis Structures of the first 20 elements. 11.4

  9. 11.3The Ionic BondTransfer of Electrons FromOne Atom to Another

  10. The chemistry of many elements, especially the representative ones, is to attain the same outer electron structure as one of the noble gases.

  11. With the exception of helium, this structure consists of eight electrons in the outermost energy level.

  12. After sodium loses its 3s electron, it has attained the same electronic structure as neon.

  13. After chlorine gains a 3p electron, it has attained the same electronic structure as argon.

  14. Formation of NaCl

  15. A sodium ion (Na+) and a chloride ion (Cl-) are formed. The 3s electron of sodium transfers to the 3p orbital of chlorine. The force holding Na+ and Cl- together is an ionic bond. Lewis representation of sodium chloride formation.

  16. Formation of MgCl2

  17. Two 3s electrons of magnesium transfer to the half-filled 3p orbitals of two chlorine atoms. A magnesium ion (Mg2+) and two chloride ions (Cl-) are formed. The forces holding Mg2+ and two Cl- together are ionic bonds.

  18. In the crystal each sodium ion is surrounded by six chloride ions. NaCl is made up of cubic crystals.

  19. In the crystal each chloride ion is surrounded by six sodium ions. 11.5

  20. The ratio of Na+ to Cl- is 1:1 There is no molecule of NaCl 11.5

  21. 11.6Electronegativity

  22. electronegativity: The relative attraction that an atom has for a pair of shared electrons in a covalent bond.

  23. If the two atoms that constitute a covalent bond are identical, then there is equal sharing of electrons. • This is called nonpolar covalent bonding. • Ionic bonding and nonpolar covalent bonding represent two extremes.

  24. If the two atoms that constitute a covalent bond are not identical, then there is unequal sharing of electrons. • This is called polar covalent bonding. • One atom assumes a partial positive charge and the other atom assumes a partial negative charge. • This charge difference is a result of the unequal attractions the atoms have for their shared electron pair.

  25. + - H Cl Chlorine has a greater attraction for the shared electron pair than hydrogen. Partial positive charge on hydrogen. Partial negative charge on chlorine. Polar Covalent Bonding in HCl : : The attractive force that an atom of an element has for shared electrons in a molecule or a polyatomic ion is known as its electronegativity. Shared electron pair. The shared electron pair is closer to chlorine than to hydrogen.

  26. A scale of relative electronegativities was developed by Linus Pauling.

  27. Electronegativity decreases down a group for representative elements. Electronegativity generally increases left to right across a period.

  28. The electronegativities of the metals are low. The electronegativities of the nonmetals are high. 11.1

  29. The polarity of a bond is determined by the difference in electronegativity values of the atoms forming the bond.

  30. If the electronegativity difference between two bonded atoms is greater than 1.7-1.9, the bond will be more ionic than covalent. • If the electronegativity difference is greater than 2, the bond is strongly ionic. • If the electronegativity difference is less than 1.5, the bond is strongly covalent.

  31. H H Electronegativity 2.1 Electronegativity 2.1 Hydrogen Molecule If the electronegativities are the same, the bond is nonpolar covalent and the electrons are shared equally. The molecule is nonpolar covalent. Electronegativity Difference = 0.0 11.10

  32. Cl Cl Electronegativity 3.0 Electronegativity 3.0 Chlorine Molecule If the electronegativities are the same, the bond is nonpolar covalent and the electrons are shared equally. The molecule is nonpolar covalent. Electronegativity Difference = 0.0 11.10

  33. + - H Cl Electronegativity 2.1 Electronegativity 3.0 If the electronegativities are not the same, the bond is polar covalent and the electrons are shared unequally. The molecule is polar covalent. Electronegativity Difference = 0.9 Hydrogen Chloride Molecule 11.10

  34. Na+ Cl- Electronegativity 0.9 Electronegativity 3.0 Sodium Chloride If the electronegativities are very different, the bond is ionic and the electrons are transferred to the more electronegative atom. No molecule exists. The bond is ionic. Electronegativity Difference = 2.1 11.10

  35. + - A dipole can be written as A dipole is a molecule that is electrically asymmetrical, causing it to be oppositely charged at two points.

  36. O H Br H Cl H H An arrow can be used to indicate a dipole. The arrow points to the negative end of the dipole. Molecules of HCl, HBr and H2O are polar .

  37. A molecule containing different kinds of atoms may or may not be polar depending on its shape. The carbon dioxide molecule is nonpolar because its carbon-oxygen dipoles cancel each other by acting in opposite directions.

  38. Relating Bond Type to Electronegativity Difference. 11.11

  39. 11.7Lewis Structuresof Compounds

  40. In writing Lewis structures, the most important consideration for forming a stable compound is that the atoms attain a noble gas configuration.

  41. The most difficult part of writing Lewis structures is determining the arrangement of the atoms in a molecule or an ion. • In simple molecules with more than two atoms, one atom will be the central atom surrounded by the other atoms.

  42. Cl2O has two possible arrangements. The two chlorines can be bonded to each other. Cl-Cl-O The two chlorines can be bonded to oxygen. Cl-O-Cl Usually the single atom will be the central atom.

  43. Procedures for WritingLewis Structures

  44. Valence Electrons of Group A Elements

  45. Step 1. Obtain the total number of valence electrons to be used in the structure by adding the number of valence electrons in all the atoms in the molecule or ion. • If you are writing the structure of an ion, add one electron for each negative charge or subtract one electron for each positive charge on the ion.

  46. Write the Lewis structure for H2O. Step 1. The total number of valence electrons is eight, two from the two hydrogen atoms and six from the oxygen atom.

  47. Step 2. Write the skeletal arrangement of the atoms and connect them with a single covalent bond (two dots or one dash). • Hydrogen, which contains only one bonding electron, can form only one covalent bond. • Oxygen atoms normally have a maximum of two covalent bonds (two single bonds, or one double bond).

  48. : : : : H O or H O H H Write the Lewis structure for H2O. Step 2. The two hydrogen atoms are connected to the oxygen atom. Write the skeletal structure: Place two dots between the hydrogen and oxygen atoms to form the covalent bonds.

  49. Step 3. Subtract two electrons for each single bond you used in Step 2 from the total number of electrons calculated in Step 1. • This gives you the net number of electrons available for completing the structure.

  50. : : H O H Write the Lewis structure for H2O. Step 3. Subtract the four electrons used in Step 2 from eight to obtain four electrons yet to be used.