Name AP CHEM / / Chapter 8 Outline Bnding: General Cncepts Types f Chemical Bnds Infrmatin abut the strength f a bnding interactin is btained by measuring the bnd energy, which is the energy required t break the bnd. Inic bnding ccurs between an element that lses electrns easily reacts with an atm that has a high affinity fr electrns. Inic cmpunds are frmed between a metal and a nn-metal. The energy f interactin between a pair f ins can be calculated using Culmb s law in the frm: Q Q E = (2.31 x 10-19 J * nm)( 1 2 ); E has units f jules, r is the distance between the in centers in r nanmeters, and Q 1 and Q 2 are the numerical in charges. Fr example: in slid NaCl, the distance between Na + and Cl - ins is 2.76 Å (0.276 nm), and the inic ( + 1)( 1) energy per pair f ins is: E = (2.31 x 10-19 J * nm)( ) = -8.37 x 10-19 J where the negative sign 0.276 equals an attractive frce. The in pair has lwer energy than the separated ins. A bnd will frm if the system can lwer its ttal energy in the prcess. In a bnd between tw hydrgen atms, the system will act t minimize the sum f the psitive (repulsive) energy terms and the negative (attractive) energy term. The distance where the energy is minimal is called the bnd length. The zer pint f energy is defined with atms at infinite separatin. At very shrt distances the energy rises steeply because f the imprtance f the repulsive frces when atms are very clse tgether. The bnd length is the distance at which the system has minimum energy. H 2 is a nn-plar cvalent bnd where the tw hydrgen nuclei share the valence electrns equally. HF exhibits a plar-cvalent bnd where the electrns are shared unequally, shwing an attractin t the mre electrnegative element. Electrnegativity Electrnegativity is the ability f an atm in a mlecule t attract shared electrn. The widely accepted methd fr determining electrnegativity values is that f Linus Pauling (1901-1995), an American scientist wh wn the Nbel Prizes fr bth chemistry and peace. 1
Electrnegativity values f the elements are shwn belw. Nte that electrnegativity tends t increase acrss a perid. Electrnegativity tends t decrease dwn a grup. Flurine has the highest electrnegativity (4.0) and cesium has the lwest electrnegativity (0.7). The relatinship between electrnegativity difference and bnd type is shwn t the right. Practice Prblem: Order the fllwing bnds accrding t plarity: H-H, O-H, Cl-H, S-H and F-H. Bnd Plarity and Diple Mments When HF is placed in an electrical field, the mlecule rients itself s that its electrpsitive end is facing the negative plate and its electrnegative end is facing the psitive plate. In diagram (a) belw, the plates are nt charged. In diagram (b) belw, the plates are charged and HF rients itself. A mlecule such as HF that has a center f psitive charge and a center f negative charge is said t be diplar, r t have a diplar mment. A diple mment is shwn using an arrw pinting t the mre electrnegative element. The electrpsitive end f the arrw has a crss thrugh it. Cpy here: 2
Any diatmic (tw-atm, but nt necessarily the same atms) mlecule that has a plar bnd als will shw a diple mment. Plyatmic mlecules can als exhibit diplar behavir. Fr example, in water, xygen has a higher electrnegativity than hydrgen. The mlecular charge distributin is shwn belw. Sme mlecules have plar bnds but d nt have a diple mment. This ccurs when the individual bnd plarities are arranged in such a way that they cancel each ther ut. CO 2 is an example f this behavir. The charge distributin fr CO 2 is shwn belw. SO 3 is als a mlecule with plar bnds but verall n diple mment. This is due t its mlecular shape. Mre will be cvered n mlecular shapes later this chapter. Ins: Electrn Cnfiguratin and Sizes In virtually every case, the atms in a stable cmpund have a nble gas arrangement f electrns. When tw nnmetals react t frm a cvalent bnd, they share electrns in a way that cmpletes the valence electrn cnfiguratins f bth atms. Bth nnmetals attain nble gas electrn cnfiguratins. When a nnmetal and a representative-grup metal react t frm a binary inic cmpund, the ins frm s that the valence electrn cnfiguratin f the nnmetal achieves the electrn cnfiguratin f the next nble gas atm and the valence rbitals f the metal are emptied. There are sme imprtant exceptins t the rules abve. Sn frms Sn 2+ and Sn 4+ Pb frms Pb 2+ and Pb 4+ Bi frms Bi 3+ and Bi 5+ Tl frms Tl + and Tl 3+ In size plays an imprtant rle in determining the structure and stability f inic slids, the prperties f ins in aqueus slutin, and the bilgical effects f ins. Since a psitive in is frmed by remving ne r mre electrns frm a neutral atm, the resulting catin is smaller than its parent atm. The ppsite is true fr negative ins; the additin f electrns t a neutral atm prduces an anin significantly larger than its parent atm. In size increases ging dwn a grup. Hrizntally, the trend is cmplicated because metals n the left side f the peridic table frm catins, and nnmetals n the right side f the peridic table frm anins. One trend wrth nting invlves relative sizes f a set f iselectrnic ins ins cntaining the same number f electrns. Cnsider the ins O 2-, F -, Na +, Mg 2+ and Al 3+. 3
Since all species have the same number f electrns, it can be assumed that as the number f prtns increases, the attractin between prtns and electrns will increase and make the inic size smaller. In summary, fr iselectrnic ins, the greater the atmic number, the smaller the inic radius. Yu try: Arrange the ins Se 2-, Br -, Rb + and Sr 2+ in rder f decreasing size. Frmatin f Binary Inic Cmpunds Inic cmpunds frm because the aggregated ppsitely charged ins have a lwer energy that the riginal elements. Just hw strngly the ins attract each ther in the slid state is indicated by the lattice energy the change in energy that takes place when separated gaseus ins are packet tgether t frm an inic slid: M + (g) + X - (g) MX (s) Lattice energy is ften defined as the amunt f energy released when an inic slid frms frm its ins, s lattice energy will have a negative value. Cnsider the energy change assciated with the fllwing reactin: Li(s) + ½ F 2 (g) LiF(s) Step 1: Sublimatin f slid lithium. Li(s) Li(g) ΔE = 161 kj/ml Step 2: Inizatin f lithium atm. Li(g) Li + (g) + e - ΔE = 520 kj/ml Step 3: Dissciatin f flurine mlecules. ½ F 2 (g) F(g) ΔE = 77 kj/ml Step 4: Frmatin f fluride ins. F(g) + e - F - (g) ΔE = -328 kj/ml Step 5: Frmatin f LiF frm Li + (g) and F - (g) ins. Li + (g) + F - (g) LiF(s) ΔE = -1047 kj/ml By adding the ΔE values, the ΔE verall = -617 kj/ml f LiF. This is an exthermic reactin. The lattice structure f lithium fluride is shwn abve. All the binary inic cmpunds frmed by an alkali metal(except cesium) and a halgen have the same lattice structure as lithium fluride. It is smetimes called the sdium chlride structure. 4
Partial Inic Character f Cvalent Bnds When atms with different electrnegativities react t frm mlecules, the electrns are nt shared equally. The pssible result is a plar cvalent bnd r, if the electrnegativity difference is great enugh, a cmplete transfer f ne r mre electrns t frm ins. Percent inic character f a bnd = (measured diple mment f X Y) (calculated diple mment f X + Y - ) x 100% Generally cmpunds with mre than 50% inic character are nrmally cnsidered t be inic slids. Since cmpunds such as ammnium chlride and sdium sulfate cntain bth inic and cvalent bnds, inic cmpunds will be defined as any cmpund that cnducts an electric current when melted. Cvalent Bnd Energies and Chemical Reactins Cnsider the stepwise decmpsitin f methane: CH 4 (g) CH 3 (g) + H(g) energy required: 435 kj/ml CH 3 (g) CH 2 (g) + H(g) energy required: 453 kj/ml CH 2 (g) CH(g) + H(g) energy required: 425 kj/ml CH(g) C (g) + H(g) energy required: 339 kj/ml Ttal = 1652 4 = 413 kj/ml Nte that the C H bnd is smewhat sensitive t its envirnment Cnsider the fllwing mlecules and the measured C H bnd energy (kj/ml) Mlecule Measure C-H bnd energy (kj/ml) HCBr 3 380 HCCl 3 380 HCF 3 430 C 2 H 6 410 Again, C H bnd strength varies significantly with its envirnment, but the cncept f an average bnd energy is helpful t chemists. The average bnd energies fr varius types f bnds are listed belw. 5
Nte that single, duble and triple bnds exist. In a single bnd, ne pair f electrns is shared, in a duble bnd, tw pairs f electrns are shared and in a triple bnd, three pairs f electrns are shared. Single bnds are the lngest and weakest f the bnds. Triple bnds are the shrtest and the strngest f the bnds. Fr bnds t be brken, energy must be added t a system an endthermic prcess. ΔH = ΣD (bnds brken) - ΣD (bnds frmed), Σ represents the sum f terms and D represents the bnd energy per mle f bnds Example: Using the bnd energies frm abve, calculate the ΔH fr the reactin f methane with chlrine and flurine t give Fren-12 (CF 2 Cl 2 ) CH 4 (g) + 2Cl 2 (g) + 2F 2 (g) CF 2 Cl 2 (g) + 2HF(g) + 2HCl(g) Bnds brken: C H: 4 x 413 = 1652 kj Cl Cl : 2 x 239 = 478 kj F F: 2 x 154 kj = 308 kj Ttal = 2438 kj Bnds frmed: C F: 2 x 485 = 970 kj C Cl: 2 x 339 = 678 kj H F: 2 x 565 = 1130 kj H Cl: 2 x 427 = 854 kj Ttal energy released = 3632 kj ΔH = ΣD (bnds brken) - ΣD (bnds frmed) ΔH = 2438 kj 3632 kj = -1194 kj Yu try: Calculate the ΔH fr the reactin belw: NN(g) + 3H 2 (g) 2NH 3 (g) The Lcalized Electrn Bnding Mdel The lcalized electrn (LE) mdel assumes that a mlecule is cmpsed f atms that are bund tgether by sharing pairs f electrns using atmic rbitals f the bund atms. Electrn pairs in the mlecule are assumed t be lcalized n a particular atm r in the space between tw atms. Thse pairs f electrns lcalized n an atm are called lne pairs, and thse fund in the space between the atms are called bnding pairs. Lewis Structures The Lewis structure f a mlecule shws hw the valence electrns are arranged amng the atms in the mlecule. Elements tend t fllw the ctet rule where there are surrunded by eight electrns. Hydrgen fllws the duet rule and nly has tw electrns in a cvalent bnd. When writing Lewis structures, d nt wrry abut which electrns cme frm which atms in a mlecule. The best way t lk at a mlecule is t regard it as a new entity that uses all available valence electrns n the atms t achieve the lwest pssible energy. 6
Hw t Write Lewis Structures Example: Write the structural frmula fr diatmic nitrgen, N 2 Descriptin f Actin 1. Determine hw many electrns yu have t wrk with; this is dne by multiplying the number f valence electrns fr that element by the number f thse atms. Label yur ttal as HAVE. Actin 1. N 2 nitrgen s valence # f nitrgen HAVE 5 x 2 = 10 S, we have 10 electrns with which t wrk. 2. If the given mlecule is an in, add r subtract electrns frm yur ttal, depending n the charge f the mlecule. If the mlecule has a psitive charge, subtract electrns frm yur ttal. If the mlecule has a negative charge, add electrns t yur ttal. 3. Write the symbls fr the given frmula with sme space between them. If there are mre than tw atms, put the ne with the lwest electrnegativity in the center and the thers n each f the fur sides. Exceptin: Hydrgen can NEVER be in the center. An element s electrnegativity can be fund n the frnt f yur peridic table. 4. Draw ne dash between the symbls t represent a bnd. Each dash drawn represents 2 electrns. Subtract 2 frm yur HAVE amunt fr each bnd yu draw. 5. Each element in a cvalent cmpund wants t share 8 electrns. We have t check t see hw many electrns each element needs t reach 8. Every bnd that is drawn t a symbl shuld be cunted as 2 electrns fr that element. Add these numbers tgether and label this number as NEED. 6. Cmpare the amunt f electrns yu HAVE with the number f electrns yu NEED. If the values match, draw the crrect amunt f dts n each symbl. Dts must be drawn in pairs. If the number f electrns yu HAVE and the number f electrns yu NEED d nt match, draw anther bnd between the symbls and subtract 2 mre electrns frm yur ttal. 2. HAVE N 2 5 x 2 = 10 N 2 is nt an in s we d nt have t add r subtract electrns t r frm ur ttal. 3. HAVE N 2 5 x 2 = 10 4. HAVE N 2 5 x 2 = 10-2 8 5. HAVE N 2 5 x 2 = 10 NEED - 2 8 6 6 = 12 Each nitrgen atm has 2 electrns because f the bnd between them. That means each nitrgen still needs 6 mre electrns. 6 + 6 = 12, s we NEED 12 electrns. 6. HAVE N 2 5 x 2 = 10 NEED - 2 8 6 6 = 12-2 6-2 4 4 = 8 4 2 2 = 4 7. Once the numbers match, draw in yur dts. The number f dts t be drawn n each symbl is indicated by the number belw it nce the number f electrns yu NEED and the number f electrns yu HAVE match. All dts must be written in pairs. 8. If the mlecule is an in and has an verall charge, put brackets arund the structural frmula and write the charge utside the brackets in the upper right crner. (See the next example t see what I mean.) 7. Because we have a 2 under each nitrgen, each nitrgen gets 2 dts. Write them as a pair. 8. Nitrgen is nt an in. Thus, ur structural frmula des nt need brackets. We are dne. 7
Yu Try: Write Lewis Structures fr the fllwing: HF, CH 4, NO +, CO 2, CO 3 2- Exceptins t the Octet Rule The lcalized electrn mdel is successful but there are sme exceptins that must be nted. The secnd-rw elements C,N,O and F shuld always be assumed t bey the ctet rule. The secnd rw elements B and Be ften have fewer than eight electrns arund them in their cmpunds (knw BF 3 ). These electrn deficient cmpunds are very reactive. Secnd rw elements never exceed the ctet rule, since their valence rbitals (2s and 2p) can accmmdate nly 8 electrns. Third rw and heavier elements ften satisfy the ctet rule but can exceed the ctet rule by using their empty valence d rbitals. (PCl 5, SF 6 ). When writing the Lewis structure fr a mlecule, satisfy the ctet rule fr all atms first. If electrns remain after the ctet rule has been satisfied, then place them n the elements having available d rbitals. When it is necessary t exceed the ctet rule fr ne f several third rw (r higher) elements, assume that the extra electrns shuld be placed n the central atm. Yu try: Write the structural frmula fr the fllwing: SF 6, ClF 3, RnCl 2 and BeCl 2 8
Resnance Smetimes mre than ne valid Lewis structure is pssible fr a given mlecule. Observe the Lewis structure fr nitrate t the right. It shws ne duble bnd and tw single bnds. But, experiments clearly shw that nly ne type f N O bnd ccurs with length and strength between thse expected fr a single and duble bnd. The structure t the right is a valid Lewis Structure but it des nt crrectly represent the bnding in NO 3 -. Resnance ccurs when mre than ne valid Lewis structure can be written fr a particular mlecule. The resulting electrn structure f the mlecule is the average f the resnance structures. The three resnance structures fr nitrate are shwn belw. ad ad Resnance shws that electrns are nt lcalized t ne atm but instead travel thrughut the mlecule. Odd-Electrn Mlecules Few mlecular cmpunds cntain dd numbers f electrns. Tw exceptins are nitric xide (NO) and nitrgen dixide (NO 2 ). The lcalized electrn mdel is based n pairs f electrns; it des nt handle dd-electrn cases in a natural way. A mre sphisticated mdel is needed. Frmal Charge The frmal charge f an atm in a mlecule is the difference between the number f valence electrns n the free atm and the number f valence electrns assigned t the atm in the mlecule. Frmal Charge = (valence e - n free atm) (valence e - assigned t atm in mlecule) Valence e - assigned = (number f lne pair e - ) + ½ (number f shared e - ) Fr Example: Calculate the frmal charge fr each element in the structural frmula shwn t the right. Oxygen: Valence e - assigned = (6) + ½ (2) = 7 Frmal Charge = (6) (7) = -1 Sulfur: Valence e - assigned = (0) + ½ (8) = 4 Frmal Charge = (6) (4) = +2 Atms in mlecules try t achieve frmal charges as clse t zer as pssible. Any negative frmal charges are expected t reside n the mst electrnegative atms. Yu try: Abve it was mentined that NO and NO 2 have an dd number f electrns. Use yur knwledge f frmal charge t determine the crrect Lewis structures fr each f the mlecules. Yu must shw a frmal charge calculatin fr each atm in each pssible Lewis structure. 9
Mlecular Structure: The VSEPR Mdel The term trignal can be used instead f triangular in shapes such as triangular planar. Central Atms with Three Electrn Pairs # f lne pair General Frmula shape Example Lewis Structure 0 AX 3 E 0 triangular planar BCl 3 1 AX 2 E 1 bent (angular) NO 2 - Central Atms with Fur Electrn Pairs # f lne pair General Frmula shape Example Lewis Structure 0 AX 4 E 0 tetrahedral CCl 4 1 AX 3 E 1 triangular pyramidal NH 3 2 AX 2 E 2 bent (angular) H 2 O 10
Central Atms with Five Electrn Pairs # f lne pair General Frmula shape Example Lewis Structure 0 AX 5 E 0 triangular bipyramidal PF 5 1 AX 4 E 1 seesaw SF 4 2 AX 3 E 2 T-shaped ClF 3 3 AX 2 E 3 linear XeF 2 Central Atms with Six Electrn Pairs # f lne pair General Frmula shape Example Lewis Structure 0 AX 6 E 0 ctahedral SF 6 1 AX 5 E 1 square pyramidal BrF 5 2 AX 4 E 2 square planar XeF 4 11
Nte in the chart belw that even thugh CH 4, NH 3 and H 2 O are cmpsed f central atms with 4 electrn pair, the bnd angles between atms varies in the different mlecules. These bservatins prvide evidence that lne pairs f electrns need mre rm than bnding pairs and tend t cmpress the angles between the bnding pairs. 12
Name AP CHEM / / Chapter 9 Outline Cvalent Bnding: Orbitals Hybridizatin and the Lcalized Electrn Mdel sp 3 Hybridizatin In general we assume that bnding nly invlves valence electrns. Cnsider methane (CH 4 ). The valence electrns f hydrgen use 1s rbitals. The 2p and 2s atmic rbitals f carbn will lead t tw different type f C H bnds: thse frm the verlap f the carbn 2p rbital with the 1s rbital f hydrgen and thse frm the verlap f a carbn 2s rbital with the 1s rbital f hydrgen. This is nt the case. The methane mlecule is tetrahedral with bnd angles f 109.5 S either the lcalized electrn mdel is wrng r carbn adpts a set f rbitals ther than its native 2s and 2p rbitals t bnd t the hydrgen atms in frming the methane mlecule. The 2s and 2p rbitals present n an islated carbn atm may nt be the best set f rbitals fr bnding. It makes sense t assume that the carbn atm has fur equivalent atmic rbitals arranged tetrahedrally. Such a set f rbitals can be btained by cmbining the carbn 2s and 2p rbitals as shwn belw. This mixing f the native atmic rbitals t frm special rbitals fr bnding is called hybridizatin. The fur new rbitals are called sp 3 rbitals because they are frmed frm ne 2s rbital and three 2p rbitals(s 1 p 3 ). We say that the carbn atm underges sp 3 hybridizatin r is sp 3 hybridized. Whenever a set f equivalent tetrahedral atmic rbitals is required by an atm, this mdel assumes that the atm adpts a set f sp 3 rbitals; the atm becmes sp 3 hybridized. What the atms in a mlecule were like befre the mlecule was frmed is nt as imprtant as hw the electrns are best arranged in the mlecule. This mdel assumes that the individual atms respnd as needed t achieve the minimum energy fr the mlecule. 13
sp 2 hybridizatin Ethylene (C 2 H 4 ) is an imprtant material in the manufacture f plastics. A duble bnd acts as ne effective pair, s the ethylene is surrunded by three effective pair. This requires a trignal planar arrangement with bnd angles f 120. Since ne 2s and tw 2p rbitals are used t frm these hybrid rbitals, this is called sp 2 hybridizatin. In frming the sp 2 rbitals, ne 2p rbital n carbn has nt been used. This remaining p rbital (p z ) is riented perpendicular t the plane f the sp 2 rbitals. The three sp 2 rbitals n each carbn can be used t share electrns, as shwn belw. In each f these bnds, the electrn pair is shared in an area centered n a line running between the atms. This type f cvalent bnd is called a sigma(σ) bnd. In the ethylene mlecule, the σ bnds are frmed using sp 2 rbitals n each carbn atm and the 1s rbital n each hydrgen atm. The secnd bnd must therefre result frm sharing an electrn pair in the space abve and belw the σ bnd. The parallel p rbitals can share an electrn pair, t frm a pi(π) bnd. σ bnds are frmed frm rbitals whse lbes pint tward each ther, but π bnds result frm parallel rbitals. A duble bnd always cnsists f ne σ bnd where the electrns are lcated between the atms and ne π bnd which ccupies the space abve and belw the σ bnd. 14
sp hybridizatin In the CO 2 mlecule the carbn has tw effective pair that will be arranged at an angle f 180. T btain tw hybrid rbitals arranged at 180 requires sp hybridizatin, invlving ne s rbital and ne p rbital. Tw effective pairs arund an atm will always require sp hybridizatin f that atm. In CO 2 the sp rbitals n carbn frm σ bnds with the sp 2 rbitals n the tw xygen atms. The remaining sp 2 rbitals n the xygen atms hld lne pair. The π bnds between the carbn atm and each xygen atm are frmed by the verlap f parallel 2p rbitals. dsp 3 hybridizatin Since 5 pairs f electrns are needed fr atms such as phsphrus in PCl 5, a dsp 3 hybridizatin is necessary. The dsp 3 hybridized phsphrus atm in PCl 5 mlecule uses its 5 dsp 3 rbitals t share electrns with five chlrine atms. A set f five effective pairs arund a given atm always requires a trignal bipyramidal arrangement which in turn requires dsp 3 hybridizatin f that atm. d 2 sp 3 hybridizatin Since 6 pairs f electrns are needed fr atms such as sulfur in SF 6, a d 2 sp 3 hybridizatin is necessary. A set f six effective pairs arund a given atm always requires an ctahedral arrangement which in turn requires d 2 sp 3 hybridizatin f that atm. Paramagnetism Mst materials have n magnetism until they are placed in a magnetic field. Hwever, in the presence f such a field, magnetism f tw types can be induced. Paramagnetism causes a substance t be attracted t the inducing magnetic field. Diamagnetism causes the substance t be repelled frm inducing magnetic field. Studies have shwn that paramagnetism is assciated with unpaired electrns and diamagnetism is assciated with paired electrns. The O 2 mlecule is knwn t be paramagnetic. 15
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