CHAPTER 8 Bonding General Concepts Faculty Web

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Chapter 8 Bonding General Concepts, unequal attraction results in a polar covalent bond The stronger atom has an overabundance of electron. density lending it a small partial negative charge while the weaker atom with less than its share of. electron density acquires a partial positive charge The bond is now a dipole the magnitude of which. depends on the size of the partial charges and their separation. 8 2 Electronegativity, Electronegativity is the attraction an atom has for a shared pair of electrons The polarity of a bond can be. judged from the difference in electronegativity between the two bonded atoms The values of. electronegativity range from a high of 4 0 for F to a low of 0 7 for Cs The trend of variation is the same as. for electron affinity electronegativity increases from lower left to upper right. When two elements have the same electronegativity the shared pair is equally attracted to both nuclei and. the resulting bond is nonpolar When one atom is more electronegative the shared pair is more strongly. attracted to the more electronegative atom and the resulting bond is polar As the difference in. electronegativity increases so do the partial positive and negative charges on the atoms until the charges. approach unity When this condition occurs the bond is more appropriately described as an ionic bond. Since the transition from nonpolar to ionic is gradual a difference in electronegativity of 2 0 is arbitrarily. chosen to be the minimum difference responsible for an ionic bond A bond between nonmetals or between. metalloids and nonmetals is covalent regardless of the electronegativity difference. EXAMPLE Classify the following bonds as nonpolar polar or ionic. a H F b Na F c C S d C H, SOLUTION a The difference in electronegativity between F and H is 4 0 2 1 1 9 so the bond is polar. b The difference is 4 0 0 9 3 1 so the bond is ionic. c Both electronegativities are 2 5 so the bond is nonpolar. d The difference is 2 5 2 1 0 4 so the bond is very slightly polar. Electronegativity increases across a row and up a column see fig 8 3 pg 353. 8 3 Bond Polarity and Dipole Moments, The polarity of a bond is measured by its dipole moment which is a vector quantity with a magnitude equal. to the charge times the distance of charge separation and a direction pointing from the partially positive end. the less electronegative atom to the partially negative end the more electronegative atom The dipole. moment of a molecule is the vector sum of the dipole moments of the individual bonds If a molecule has a. nonzero dipole moment the molecule is polar otherwise it is nonpolar The polarity or nonpolarity of a. molecule can be determined by examining the polarity of each bond and the geometry of the molecule If. none of the bonds in the molecule is polar then the molecule is nonpolar If any of the bonds are polar then. the polarity depends on the geometry of the molecule. EXAMPLE Which of the following molecules is polar H C N O S O O C O. Chapter 8 Bonding General Concepts, SOLUTION a HCN is linear The difference in electronegativity between H and C is 0 4 pointing toward C.
toward the N end of the molecule The C N dipole vector also points toward the N end of the molecule. These vectors add to yield a nonzero dipole moment so the molecule is polar. b Each S O bond is polar with the dipole moment vector pointing toward O The molecule. is bent so the vectors add to yield a smaller vector pointing between the S O bonds The O. molecule is polar, c The C O bonds are polar but they point 180 apart because the molecule is bent The d d. vector sum is 0 so the molecule is nonpolar O C O,8 4 Ions Electron Configurations and Sizes. Metal atoms tend to lose all valence electrons to form cations. Nonmetal atoms tend to gain enough electrons to fill the outer s and p subshells. cations and anions combine in a ratio that allows for the positive and negative charges to cancel. Ionic radius cations are smaller than parent atom see fig 8 7 pg 361. anions are larger than parent atom,in an isoelectronic series size decreases with Z. Mg Na Ne F O,8 5 Formation of Binary Ionic Compounds. Lattice energy E when separated gaseous ions are packed together to form an ionic solid. Li s Li g E sublimation 161 kJ,Li g Li g e E ionization energy 520 kJ.
1 2 F2 g F g E dissociation 77 kJ,F g e F g E electron affinity 328 kJ. Li g F g LiF s E lattice energy 1047 kJ,Li s 1 2 F2 g LiF s E energy of formation 617 kJ. ionic bonds form because of the lattice energy released when oppositely charged ions form a 3. d crystal lattice,Chapter 8 Bonding General Concepts. 8 6 Partial Ionic Character of Covalent Bonds,100 EN 1 7 corresponds to 50 ionic character. 75 For a bond A B as EN varies from 0 to 3 5,the bond goes from nonpolar covalent EN.
0 to polar covalent EN 0 1 7 to ionic,50 50 ionic character EN 1 7. Generally can assume that compounds,between metals and nonmetals or compounds. which include polyatomic ions are ionic and,0 compounds between nonmetals are covalent. Electronegativity difference,8 7 The Covalent Chemical Bond A Model. 8 8 Covalent Bond Energies and Chemical Reactions,8 9 The Localized Electron Bonding Model.
Localized Electron bonding model LE covalent bonds are described as the sharing of electron pairs through. overlap of atomic orbitals, not allowed Valence Bond theory atoms form bonds by overlapping. orbitals to share electrons,not allowed,there is a maximum of 2 electrons per overlap. Pauli Exclusion Principle,can form bond, either way can also form a bond by overlapping p orbitals. According to the model there are two ways that orbitals can overlap to form bonds. sigma s bond a head on overlap of orbitals Electron density is concentrated along the axis of. the bond an imaginary line connecting the two bonded atoms and is greatest. directly between the two bonded nuclei resulting in the lowest possible electron. energy Because sigma bonds are symmetrical around the bond axis they are free. to rotate about this axis Only one sigma bond can exist between any two atoms. pi p bond a side to side overlap of p orbitals Electron density along the bond axis is zero. with twin maxima of electron density above and below the bond axis This is a. weaker bond because the electron density is zero directly between the two. bonded nuclei A pi bond cannot occur unless a sigma bond is already present. Chapter 8 Bonding General Concepts, Pi bonding allows for the existence of multiple bonds in which more than one pair of electrons are shared. between the same two atoms In a Lewis structure shared pairs are represented by lines connecting the. bonded atoms, single bond a lone sigma bond 1 shared pair of electrons.
double bond one sigma bond and one pi bond 2 shared pairs. triple bond,one sigma bond and two pi bonds 3 shared pairs. 8 10 Lewis Structures,The Octet Rule, The driving force behind the formation of covalent bonds is the lowering of electronic energy that occurs. when a shared pair of electrons can get close to two nuclei at the same time Breaking this bond involves. pulling electrons away from nuclei so it involves the expenditure of an amount of energy on the order of the. ionization energy for the elements involved When we recall that the nonmetals have the highest ionization. energies it is no surprise that the nonmetals form the strongest covalent bonds. In order to form a bond overlap must occur between the orbitals of the bonding atoms Just as for atoms. the Pauli exclusion principle demands that there be no more than two electrons per orbital which since the. shared electrons belong to both overlapped orbitals means that there can be no more than two electrons. total in the two overlapping orbitals This arrangement can happen by. a overlapping a full orbital with an empty orbital or. b overlapping a half full orbital with a half full orbital. With four valence orbitals one s and three p each holding two electrons an upper limit of eight valence. electrons is set It is observed that in the vast majority of covalent compounds this upper limit is met that. is that all valence orbitals are filled This result is to be expected if we view valence vacancies as low energy. holes that electrons can fall into, Octet rule when forming covalent compounds atoms tend to gain electrons through sharing until they. completely occupy the outer s and p subshells, Hydrogen of course has no outer p subshell so it obeys the octet rule by gaining one electron through. sharing the duet rule All other elements except He which is inert have both s and p subshells so for. them we could restate the octet rule all elements but hydrogen tend to form bonds until they are surrounded. by eight valence electrons, Element symbol represents nucleus and all inner shell electrons.
Valence electrons are represented as follows,shared pair lone pair. single bond s bond 1 shared pair double bond s bond bond 2 shared pairs. triple bond s bond 2 bond 3 shared pairs,Chapter 8 Bonding General Concepts. Rules for Drawing Lewis Structures, 1 Draw a skeleton of the molecule The likelihood of an atom being in the center of the molecule increases. a if the atom is further to the left on the chart or. b if the atom is further down a group on the chart. 2 Count the number of valence electrons and subtract 2 for each bond needed to connect the atoms in the. 3 Count the number of electrons each atom in the skeleton still needs to acquire an octet remember that H. only needs two so H will always be finished after step 1. 4 If the number from step 3 is greater than the number from step two for each pair short add one bond to. the skeleton thus forming a multiple bond and subtract two electrons from the total from step 1. 5 Distribute the remaining electrons to give each atom an octet. When drawing the skeleton the element furthest from the F corner of the chart is most likely to be central. For molecules with more than one central atom derive clues from the formula. N2H2 CO2 HCN CH3COOH HF H2O NH3,EXAMPLE Draw the Lewis structure for SO2. SOLUTION S and O are both in the same group but S is further down in the group so S must be the. central atom and the skeleton is, The total number of valence electrons is 6 for O and 6 each for the two sulfur atoms for a total of 18.
Remember that the number of valence electrons for a main group element is equal to the group number. Subtracting two electrons for each bond in the skeleton leaves 14. O S O total 14, Now count the number of electrons each atom in the skeleton needs to reach 8 Remember that a shared pair. counts twice once for each atom sharing it,O has one pair needs 6 more electrons. S has two pair needs 4 more electrons,O has one pair needs 6 more electrons. total needed 16 more electrons, We need 16 more electrons to give each atom an octet yet we have only 14 electrons left so we are one pair. short We make up this deficit by adding another bond thereby counting another pair twice. O S O total 14 2 12,Chapter 8 Bonding General Concepts.
Now distribute the remaining twelve electrons to give each atom an octet O has two pair and needs 4 more. O S O total 12 4 8,S has three pair and needs 2 more electrons. O S O total 8 2 6,O has one pair and needs 6 more electrons. O S O total 6 6 0, We now have a valid Lewis structure We have accounted for all 18 valence electrons and each atom has an. EXAMPLE Draw the Lewis structure of CO, SOLUTION The skeleton is simply C O The total number of valence electrons is 4 from C 6 from O. 10 minus 2 for the bond in the skeleton leaves 8,C O total 8.
Both C and O have one pair and need 6 more electrons each or 12 electrons needed which is 4 short of the. total so 2 bonds must be added,O total 8 4 4, Now C and O each have three pair and need 2 electrons each. O total 4 4 0, The following rule is often useful in drawing molecular skeletons or determining where to place extra bonds. The number of bonds a nonmetal is likely to form is equal to 8 the group number Atoms with low. electronegativities are more likely to form more than the expected number of bonds atoms with high. electronegativities are more likely to form less than the expected number of bonds. In the example above C which is expected to form 8 4 4 bonds forms 3 bonds instead and O which is. expected to form 8 6 2 bonds forms 3 bonds instead in contradiction to the above rule. EXAMPLE Draw the Lewis structure of HSCN, SOLUTION The skeleton as the formula implies is H S C N The total number of valence electrons is. 1 6 4 5 16 The skeleton contains 6 electrons leaving only 10 The number of electrons needed to give. each atom except H an octet is 4 for S 4 for C 6 for N 14 electrons needed We are short 2 pair. so 2 more bonds must be added To determine where to place them consider that S would be expected to. Chapter 8 Bonding General Concepts, have 2 bonds C to have 4 and N to have 3 The best Lewis structure should come as close as possible to. satisfying these expectations This can be done by placing both extra bonds between the C and N. Lewis Structures for Polyatomic Ions, For polyatomic ions follow rules as above except rule 2.
2 add up total of valence electrons subtract 2 for each bond in the skeleton subtract charge from total. CO3 SO4 H3O NH3,8 11 Exceptions to the Octet Rule, Electron Deficient Molecules The elements in groups I II and III are mostly metals and form ionic. 8 1 CHAPTER 8 Bonding General Concepts 8 1 Types of Chemical Bonds Ionic Bonding Oppositely charged ions are attracted to each other by a strong electrostatic force E 2 31 x 10 19 J nm x Q1Q2 r where Q is the ionic charge in atomic units and r is the distance between ions in nm Covalent Bonding When two atoms approach each other the result of the interaction depends on the relative

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