logging in or signing up Oxidative states rbartelt Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1706 Category: Education License: All Rights Reserved Like it (1) Dislike it (0) Added: November 18, 2007 This Presentation is Public Favorites: 0 Presentation Description A simple step by step method for determining the Comments Posting comment... By: vinbhatt (43 month(s) ago) vvv Saving..... Post Reply Close Saving..... Edit Comment Close Premium member Presentation Transcript Oxidation numbers: Oxidation numbers How to assign oxidation numbers: How to assign oxidation numbers Electronegativity is important when assigning oxidation numbers. When assigning oxidation numbers to atoms in a covalently bonded molecule, you must first look up the electronegativity of all atoms bonding in the molecule. We’ll look at the example of carbon dioxideAssigning oxidation numbers for CO2: Assigning oxidation numbers for CO2 Look up the electronegativity of carbon and oxygen (pg 161 in the text, or online) Carbon = 2.5 Oxygen = 3.5 Give the more electronegative atom its “preferred oxidative state”Assigning oxidation numbers for CO2 (continued…): Assigning oxidation numbers for CO2 (continued…) In the example of CO2 oxygen has the higher electronegativity (3.5). Therefore oxygen gets its “preferred oxidative state”. What is oxygen’s “preferred oxidative state” and how do you determine what it would be? The peferred oxidative state can be looked up on the periodic table.Preferred oxidative states by column in the periodic table: Preferred oxidative states by column in the periodic table +1 +2 +3 -3 -2 -1 0Still on CO2: Still on CO2 As can be seen on the periodic table from the previous slide, oxygen has a preferred oxidative state of -2, therefore the oxidative state of oxygen is -2. Carbon needs to have an oxidative state that balances out the two oxygens. Since each oxygen has a charge of -2, the total negaive charge from the oxygens is -4. Therefore carbon’s charge must be equal and opposite, +4.Another example, NH3: Another example, NH3 Look up the electronegativity of nitrogen and hydrogen N = 3.0 H = 2.1 Nitrogen has the greater electronegativity, thus its oxidative state is -3. Now solve for hydrogen.Solving for H in NH3: Solving for H in NH3 Since nitrogen has an oxidative state of -3 you set up the equation below. 3(H) -3 = 0 H stand for the oxidative state of hydrogen. You must muliply that number by three because there are three hydrogens in NH3. You set the equation equal to zero because NH3 is a neutral molecule with no charge. NH3 where N= -3 Solving for H in NH3: Solving for H in NH3 Solve the equation: 3(H) -3 = 0 3(H)=3 H = 1 Therefore the oxidative state of hydrogen is +1. You would report the your answer as follows: H = +1 N = -3The hydrogen rule, example H2SO4: The hydrogen rule, example H2SO4 Hydrogen rule states: Whenever hydrogen is present in a molecule that contains any atom that has a higher electronegativity than itself its oxidative state is always +1. Armed with that knowledge we can solve for hydrogen.The hydrogen rule, example H2SO4: The hydrogen rule, example H2SO4 Look up all electronegativities H = 2.1 S = 2.6 O = 3.5 Oxygen has the highest electronegativity and thus takes the charge of -2. Since oxygen has a higher electronegativity than hydrogen, hydrogen takes the charge of +1The hydrogen rule, example H2SO4: The hydrogen rule, example H2SO4 Set up the equation: 2(1)+S+4(-2) = 0 2 + S – 8 = 0 2 + S = 8 S = 6 Sulfur’s oxidative state is +6. H2SO4 The equality rule, example H2, O2, N2, F2, anything2: The equality rule, example H2, O2, N2, F2, anything2 The equality rule states: All atoms in an elemental molecule have an oxidative state of 0. Therefore the oxidative state of fluorine in F2 is 0. It makes no difference that fluorine has the highest electronegativity on the periodic table, if it’s bound to itself the electrons show no favor between ether atom.The equality rule continued: The equality rule continued This goes for all elemental substances. In pure iron, Fe has an oxidative state of 0. S8, sulfur’s oxidative state is 0 P4, phosphorus's oxidative state is 0 03, oxygen’s oxidative state is 0 C60, or C540 carbon’s oxidative state is 0 I think you get the point.Polyatomic ions, example SO4-2: Polyatomic ions, example SO4-2 The steps don’t change. First look up both electronegativities S = 2.6 O = 3.5 Oxygen wins, it takes the state of -2 Set up the equation When you set up the equation you set it equal to the charge on of the ion S + 4(-2) = -2 SO4-2 Polyatomic ions, example SO4-2: Polyatomic ions, example SO4-2 Solve the equation S + 4(-2) = -2 S – 8 = -2 S = 6 Again, sulfur’s oxidative state is +6Polyatomic ions, example ClO4-: Polyatomic ions, example ClO4- Look up both electronegativities Cl = 3.0 O = 3.5 Oxygen wins, it takes the state of -2 Set up the equation Cl + 4(-2) = -1 Cl – 8 = -1 Cl = +7 ClO4- Polyatomic ions, example ClO-: Polyatomic ions, example ClO- Look up both electronegativities Cl = 3.0 O = 3.5 Oxygen wins, it takes the state of -2 Set up the equation Cl + 1(-2) = -1 Cl – 2 = -1 Cl = +1 ClO- That’s all: That’s all You are now ready to do some example problems of your own. Once you’re confident that you can determine the oxidative state of any atom in any simple molecule or polyatomic ion you can move onto the next topic. You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
Oxidative states rbartelt Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1706 Category: Education License: All Rights Reserved Like it (1) Dislike it (0) Added: November 18, 2007 This Presentation is Public Favorites: 0 Presentation Description A simple step by step method for determining the Comments Posting comment... By: vinbhatt (43 month(s) ago) vvv Saving..... Post Reply Close Saving..... Edit Comment Close Premium member Presentation Transcript Oxidation numbers: Oxidation numbers How to assign oxidation numbers: How to assign oxidation numbers Electronegativity is important when assigning oxidation numbers. When assigning oxidation numbers to atoms in a covalently bonded molecule, you must first look up the electronegativity of all atoms bonding in the molecule. We’ll look at the example of carbon dioxideAssigning oxidation numbers for CO2: Assigning oxidation numbers for CO2 Look up the electronegativity of carbon and oxygen (pg 161 in the text, or online) Carbon = 2.5 Oxygen = 3.5 Give the more electronegative atom its “preferred oxidative state”Assigning oxidation numbers for CO2 (continued…): Assigning oxidation numbers for CO2 (continued…) In the example of CO2 oxygen has the higher electronegativity (3.5). Therefore oxygen gets its “preferred oxidative state”. What is oxygen’s “preferred oxidative state” and how do you determine what it would be? The peferred oxidative state can be looked up on the periodic table.Preferred oxidative states by column in the periodic table: Preferred oxidative states by column in the periodic table +1 +2 +3 -3 -2 -1 0Still on CO2: Still on CO2 As can be seen on the periodic table from the previous slide, oxygen has a preferred oxidative state of -2, therefore the oxidative state of oxygen is -2. Carbon needs to have an oxidative state that balances out the two oxygens. Since each oxygen has a charge of -2, the total negaive charge from the oxygens is -4. Therefore carbon’s charge must be equal and opposite, +4.Another example, NH3: Another example, NH3 Look up the electronegativity of nitrogen and hydrogen N = 3.0 H = 2.1 Nitrogen has the greater electronegativity, thus its oxidative state is -3. Now solve for hydrogen.Solving for H in NH3: Solving for H in NH3 Since nitrogen has an oxidative state of -3 you set up the equation below. 3(H) -3 = 0 H stand for the oxidative state of hydrogen. You must muliply that number by three because there are three hydrogens in NH3. You set the equation equal to zero because NH3 is a neutral molecule with no charge. NH3 where N= -3 Solving for H in NH3: Solving for H in NH3 Solve the equation: 3(H) -3 = 0 3(H)=3 H = 1 Therefore the oxidative state of hydrogen is +1. You would report the your answer as follows: H = +1 N = -3The hydrogen rule, example H2SO4: The hydrogen rule, example H2SO4 Hydrogen rule states: Whenever hydrogen is present in a molecule that contains any atom that has a higher electronegativity than itself its oxidative state is always +1. Armed with that knowledge we can solve for hydrogen.The hydrogen rule, example H2SO4: The hydrogen rule, example H2SO4 Look up all electronegativities H = 2.1 S = 2.6 O = 3.5 Oxygen has the highest electronegativity and thus takes the charge of -2. Since oxygen has a higher electronegativity than hydrogen, hydrogen takes the charge of +1The hydrogen rule, example H2SO4: The hydrogen rule, example H2SO4 Set up the equation: 2(1)+S+4(-2) = 0 2 + S – 8 = 0 2 + S = 8 S = 6 Sulfur’s oxidative state is +6. H2SO4 The equality rule, example H2, O2, N2, F2, anything2: The equality rule, example H2, O2, N2, F2, anything2 The equality rule states: All atoms in an elemental molecule have an oxidative state of 0. Therefore the oxidative state of fluorine in F2 is 0. It makes no difference that fluorine has the highest electronegativity on the periodic table, if it’s bound to itself the electrons show no favor between ether atom.The equality rule continued: The equality rule continued This goes for all elemental substances. In pure iron, Fe has an oxidative state of 0. S8, sulfur’s oxidative state is 0 P4, phosphorus's oxidative state is 0 03, oxygen’s oxidative state is 0 C60, or C540 carbon’s oxidative state is 0 I think you get the point.Polyatomic ions, example SO4-2: Polyatomic ions, example SO4-2 The steps don’t change. First look up both electronegativities S = 2.6 O = 3.5 Oxygen wins, it takes the state of -2 Set up the equation When you set up the equation you set it equal to the charge on of the ion S + 4(-2) = -2 SO4-2 Polyatomic ions, example SO4-2: Polyatomic ions, example SO4-2 Solve the equation S + 4(-2) = -2 S – 8 = -2 S = 6 Again, sulfur’s oxidative state is +6Polyatomic ions, example ClO4-: Polyatomic ions, example ClO4- Look up both electronegativities Cl = 3.0 O = 3.5 Oxygen wins, it takes the state of -2 Set up the equation Cl + 4(-2) = -1 Cl – 8 = -1 Cl = +7 ClO4- Polyatomic ions, example ClO-: Polyatomic ions, example ClO- Look up both electronegativities Cl = 3.0 O = 3.5 Oxygen wins, it takes the state of -2 Set up the equation Cl + 1(-2) = -1 Cl – 2 = -1 Cl = +1 ClO- That’s all: That’s all You are now ready to do some example problems of your own. Once you’re confident that you can determine the oxidative state of any atom in any simple molecule or polyatomic ion you can move onto the next topic.