Chapter 9 Chemical Bonds
Chapter 9 Chemical Bonds Chem 111-003
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This 3 page Class Notes was uploaded by Sharneece Gary on Thursday November 5, 2015. The Class Notes belongs to Chem 111-003 at University of South Carolina taught by Dmitry V Peryshkov in Fall 2015. Since its upload, it has received 28 views.
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Date Created: 11/05/15
Chapter 9 Chemical Bonds Chemical bonds are the forces that hold the atoms together in substances A Lewis electrondot symbol consists of the symbol for the element surrounded by dots one for each valence electron Cations of most representative elements have no valence shell electrons shown in the Lewis symbol Ionic Bonding results from the electrostatic attraction between cations and anions 0 Formation of an ionic bond can be viewed as a transfer of electrons Lattice energy is the energy required to separate one mole of ionic solid into its gaseous ions 0 When arranging ionic compounds by lattice energy you first go off of their charges and if two or more have the same charge then you go off of the sizedistance I Note if the distance is short the energy is large 0 Test Your Skill Arrange the following ionic compounds in order of increasing lattice energy MgO NaCl MgClZ CaO NaCl lt MgClZ lt CaOlt MgO Because NaCL charge is 11 MgClZ is 21 CaO is 22 MgO is 22 and MgO is bigger than CaO because Mg is bigger In size than Ca Note How I determine how one is bigger than the other is that I have a concept that the the closer to group 1 and period 1 the bigger it is In this case Mg is closer to Period one than Ca A covalent bond result from atoms sharing electron pairs Lewis structures represent covalent bonding by showing how the valence electrons are present in a molecule 0 Bonding pairs are shared between two atoms and are represented by lines 0 Lone pairs are entirely on one atom and are represented by two dots 0 They can share electrons to form bond pairs Octet Rule atoms share electrons until each atom is surrounded by eight 0 Single Bond sharing one pair of electrons H 0 Double Bond sharing two pairs of electrons 0 Triple Bond sharing three pairs of electrons The bond order is the number of electron pairs shared between two atoms The skeleton structure shows which atoms are bonded to each other A central atom is bonded to two or more other atoms A terminal atom is bonded to only one other atom Writing Lewis Structures 0 Write the Lewis structure of CN Step One find the sum of valence electrons C4 N6 10 Step two subtract two electrons from each bond 1 bonds x 2 2 102 8 Step three Count the number of electrons needed to make each element have 8 C has a total of 2needs 6 more N has a total of 2 needs 6 more they both have two because of the single bond you initial place to connect the two Total need of 12 12 need 8 Have 4 electrons short share pairs Step 4 Add the correct number of paired bonds indicated by the answer from step 3 C triple bond N because you have 4 electrons which equals a double bond plus the single bone that is already there to give you a triple bond Step 5 Add in the rest as lone pairs to make each element full with 8 electrons each C triple bond N quotDipole moment is a measure of the unequal sharing of electrons quotElectronegativity is a measure of the ability of an atom to attract the shared electrons in a chemical bond quot Example Electronegativity Select the most polar bond ClF OF PF PF is the most polar bond Think of it as the farthest to the bottom and right of the periodic table is the more electronegative quot39 Formal charge is a charge assigned to atoms in Lewis structures by assuming the shared electrons are divided equally between the bonded atoms quotHow to determine formal charge First you must figure out the complete Lewis Structure Once you have completed that step you must look at each element individually comparing the number of electrons it originally has from the periodic table to the number of electrons it has now Note if the elements share paired bonds split the bonds evenly to where each elements has the same amount of electrons from that paired bond If the element has more in the end than the beginning then it is because it gained electrons If the element has less in the end than the beginning then it is because it lost electrons quotStructure Stability Lewis structures that show the smallest formal charges are favored Lewis structures that have adjacent atoms with formal charges of the same sign are much less favorable Lewis structures that place negative formal charges on more electronegative atoms are favored quot39 Resonance structures differ only in the distribution of the valence electrons are indicated by a double headed arrow Basically they are the Lewis structures just with the electrons distributed in a different fashion in which it still gives each element 8 electrons each quotCentral atoms from Group 2A and 3A do not have enough valence electrons to complete an octet the Lewis structures are electrondeficient quotExpanded valence shell molecules have more than 8 electrons about an atom in a Lewis Structure Only atoms in the third period or higher can accommodate more than 8 electrons For example in the lewis structure for SF4 each of the 4 F atoms has 6 lone pairs and the S atoms has 2 lone pairs Each F atom has a total of 8 electrons and the S atom has a total of 10 atoms This is because total there are 34 electrons from the beginning Once you subtract 2 electrons from each bond you are left with 26 From the structure there is a need for 24 With a need of 24 and we have 26 there will be a total of 2 left over which are then placed to the S atom quot The bond dissociation energy or bond energy D is the energy required to break one mole of bonds in a gaseous species Enthalpy change Calculate an approximate enthalpy change for the reaction C2H4g H2g 9 C2H6 g Bond Energies CH 414 kJmol HH 436 kJmol CC 348 kJmol CC 611 kJmol Step 1 Make a lewis structure for each part of the equation With C2H4 there should be a CC bond with 2 H on each C bond With H2 it should be HH With C2H6 there should be a CC bond with 3 H on each C bond Step 2 indicate which bond energies will be going with each equation and their amount of moles C2H4 4mols of CH and 1 mol of CC H2 1 mol of HH C2H6 6 mol of CH and 1 mol of CC Step 3 Reactant Reactant Product 4mol x 414 1 x 6111mo436 6mox 414 1mol x 348 27032832 129 kJmol
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