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October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 material  Electron configuration  Bonding IonicIonic CovalentCovalent MetallicMetallic.

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Presentation on theme: "October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 material  Electron configuration  Bonding IonicIonic CovalentCovalent MetallicMetallic."— Presentation transcript:

1 October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 material  Electron configuration  Bonding IonicIonic CovalentCovalent MetallicMetallic Hydrogen (next)Hydrogen (next)  Apply Lewis diagram to get chemical formulas (at least a good guess- some formulas are exceptions to rules)

2 October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 Lewis Rules   Arrange the atoms (central atom lowest electornegativity)   Count up total valence electronsDraw single bonds between central atom and surrounding atoms   Place remaining electrons, in pairs, around appropriate atoms; start with outer atoms.   Make sure all atoms that need octets have octets

3 October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 Example nitrite ion is NO2− nitrite  Nitrogen is the least electronegative atom, so it is the central atom by multiple criteria.  Nitrogen has 5 valence electrons; each oxygen has 6, for a total of (6 × 2) + 5 = 17. The ion has a charge of −1, which indicates an extra electron, so the total number of electrons is 18.  Each oxygen must be bonded to the nitrogen, which uses four electrons — two in each bond. The 14 remaining electrons should initially be placed as 7 lone pairs. Each oxygen may take a maximum of 3 lone pairs, giving each oxygen 8 electrons including the bonding pair. The seventh lone pair must be placed on the nitrogen atom.  Both oxygen atoms currently have 8 electrons assigned to them. The nitrogen atom has only 6 electrons assigned to it. One of the lone pairs on an oxygen atom must form a double bond, but either atom will work equally well. We therefore must have a resonance structure.  Two Lewis structures must be drawn: one with each oxygen atom double-bonded to the nitrogen atom. The second oxygen atom in each structure will be single-bonded to the nitrogen atom. Place brackets around each structure, and add the charge (−) to the upper right outside the brackets. Draw a double-headed arrow between the two resonance forms. O6N5 charge total18

4 October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 For ionic bonds sometimes the elctron is shown as completely transferred and the ion charges shown.

5 October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010 Reactions  Oxidation (burn) H2+O2  H2O unbalancedH2+O2  H2O unbalanced H2+O2  2H2OH2+O2  2H2O 2H2+O2  2H2O2H2+O2  2H2O  2 moles of hydrogen gas (diatomic) and one mole of oxygen gas (diatomic) reacts to create two moles of water.

6 October 6, 2015October 6, 2015October 6, 2015 GSCI 163 Spring 2010


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