CHAPTER 10 CHEMICAL BONDING II: MOLECULAR GEOMETRY AND HYBRIDIZATION OF ATOMIC ORBITALS

Transcription

1 APTER 10 EMIAL BNDING II: MLEULAR GEMETRY AND YBRIDIZATIN ATMI RBITALS 10.7 (a) The Lewi tructure of P 3 i hown below. Since in the VSEPR method the number of bonding pair and lone pair of electron around the central atom (phophoru, in thi cae) i important in determining the tructure, the lone pair of electron around the chlorine atom have been omitted for implicity. There are three bond and one lone electron pair around the central atom, phophoru, which make thi an AB 3 E cae. The information in Table 10.2 how that the tructure i a trigonal pyramid like ammonia. P What would be the tructure of the molecule if there were no lone pair and only three bond? The Lewi tructure of 3 i hown below. There are four bond and no lone pair around carbon which make thi an AB 4 cae. The molecule hould be tetrahedral like methane (Table 10.1). The Lewi tructure of Si 4 i hown below. Like part, it i a tetrahedral AB 4 molecule. Si (d) The Lewi tructure of Te 4 i hown below. There are four bond and one lone pair which make thi an AB 4 E cae. onulting Table 10.2 how that the tructure hould be that of a ditorted tetrahedron like S 4. Te Are Te 4 and S 4 ioelectronic? Should ioelectronic molecule have imilar VSEPR tructure?

2 APTER 10: EMIAL BNDING II Strategy: The equence of tep in determining molecular geometry i a follow: draw Lewi find arrangement of find arrangement determine geometry tructure electron pair of bonding pair baed on bonding pair Solution: Lewi tructure Electron pair Electron Lone pair Geometry on central atom arrangement (a) Al 3 trigonal planar 0 trigonal planar, AB 3 Zn 2 linear 0 linear, AB 2 2 Zn 4 tetrahedral 0 tetrahedral, AB Lewi Structure e pair arrangement geometry (a) Br Br Br tetrahedral tetrahedral Br B trigonal planar trigonal planar N tetrahedral trigonal pyramidal (d) Se tetrahedral bent (e) N trigonal planar bent We ue the following equence of tep to determine the geometry of the molecule. draw Lewi find arrangement of find arrangement determine geometry tructure electron pair of bonding pair baed on bonding pair

3 APTER 10: EMIAL BNDING II The Lewi tructure i: AB 4 tetrahedral AB 3 trigonal planar AB 2 E 2 bent nly molecule with four bond to the central atom and no lone pair are tetrahedral (AB 4 ). 2 I Si I I d I What are the Lewi tructure and hape for Xe 4 and Se 4? All four molecule have two bond and two lone pair (AB 2 E 2 ) and therefore the bond angle are not linear. Since electronegativity decreae going down a column (group) in the periodic table, the electronegativity difference between hydrogen and the other Group 6 element will increae in the order Te < Se < S <. The dipole moment will increae in the ame order. Would thi concluion be a eay if the element were in different group? The electronegativity of the halogen decreae from to I. Thu, the polarity of the X bond (where X denote a halogen atom) alo decreae from to I. Thi difference in electronegativity account for the decreae in dipole moment = Br 4 (μ = 0 for both) < 2 S < N 3 < 2 < Draw the Lewi tructure. Both molecule are linear (AB 2 ). In S 2, the two S bond moment are equal in magnitude and oppoite in direction. The um or reultant dipole moment will be zero. ence, S 2 i a nonpolar molecule. Even though S i linear, the and S bond moment are not exactly equal, and there will be a mall net dipole moment. ence, S ha a larger dipole moment than S 2 (zero) Molecule will have a higher dipole moment. In molecule (a), the tran arrangement cancel the bond dipole and the molecule i nonpolar.

4 266 APTER 10: EMIAL BNDING II Strategy: Keep in mind that the dipole moment of a molecule depend on both the difference in electronegativitie of the element preent and it geometry. A molecule can have polar bond (if the bonded atom have different electronegativitie), but it may not poe a dipole moment if it ha a highly ymmetrical geometry. Solution: Each vertex of the hexagonal tructure of benzene repreent the location of a atom. Around the ring, there i no difference in electronegativity between atom, o the only bond we need to conider are the polar bond. The molecule hown in and (d) are nonpolar. Due to the high ymmetry of the molecule and the equal magnitude of the bond moment, the bond moment in each molecule cancel one another. The reultant dipole moment will be zero. or the molecule hown in (a) and, the bond moment do not cancel and there will be net dipole moment. The dipole moment of the molecule in (a) i larger than that in, becaue in (a) all the bond moment point in the ame relative direction, reinforcing each other (ee Lewi tructure below). Therefore, the order of increaing dipole moment i: = (d) < < (a). (a) A 3 ha the Lewi tructure hown below. There are three bond pair and one lone pair. The four electron pair have a tetrahedral arrangement, and the molecular geometry i trigonal pyramidal (AB 3 E) like ammonia (See Table 10.2). The A (arenic) atom i in an p 3 hybridization tate. A Three of the p 3 hybrid orbital form bond to the hydrogen atom by overlapping with the hydrogen 1 orbital. The fourth p 3 hybrid orbital hold the lone pair Strategy: The tep for determining the hybridization of the central atom in a molecule are: draw Lewi Structure ue VSEPR to determine the ue Table 10.4 of of the molecule electron pair arrangement the text to determine urrounding the central the hybridization tate atom (Table 10.1 of the text) of the central atom Solution: (a) Write the Lewi tructure of the molecule. Si

5 APTER 10: EMIAL BNDING II 267 ount the number of electron pair around the central atom. Since there are four electron pair around Si, the electron arrangement that minimize electron-pair repulion i tetrahedral. We conclude that Si i p 3 hybridized becaue it ha the electron arrangement of four p 3 hybrid orbital. Write the Lewi tructure of the molecule. Si Si ount the number of electron pair around the central atom. Since there are four electron pair around each Si, the electron arrangement that minimize electron-pair repulion for each Si i tetrahedral. We conclude that each Si i p 3 hybridized becaue it ha the electron arrangement of four p 3 hybrid orbital The Lewi tructure of Al 3 and Al 4 are hown below. By the reaoning of the two problem above, the hybridization change from p 2 to p 3. Al What are the geometrie of thee molecule? Al Draw the Lewi tructure. Before the reaction, boron i p 2 hybridized (trigonal planar electron arrangement) in B 3 and nitrogen i p 3 hybridized (tetrahedral electron arrangement) in N 3. After the reaction, boron and nitrogen are both p 3 hybridized (tetrahedral electron arrangement) (a) N 3 i an AB 3 E type molecule jut a A 3 in Problem Referring to Table 10.4 of the text, the nitrogen i p 3 hybridized. N 2 4 ha two equivalent nitrogen atom. entering attention on jut one nitrogen atom how that it i an AB 3 E molecule, o the nitrogen atom are p 3 hybridized. rom tructural conideration, how can N 2 4 be conidered to be a derivative of N 3? The nitrate anion N 3 i ioelectronic and iotructural with the carbonate anion 3 2 that i dicued in Example 9.5 of the text. There are three reonance tructure, and the ion i of type AB 3 ; thu, the nitrogen i p 2 hybridized.

6 268 APTER 10: EMIAL BNDING II (a) Each carbon ha four bond pair and no lone pair and therefore ha a tetrahedral electron pair arrangement. Thi implie p 3 hybrid orbital. The left-mot carbon i tetrahedral and therefore ha p 3 hybrid orbital. The two carbon atom connected by the double bond are trigonal planar with p 2 hybrid orbital. arbon 1 and 4 have p 3 hybrid orbital. arbon 2 and 3 have p hybrid orbital (d) The left-mot carbon i tetrahedral (p 3 hybrid orbital). The carbon connected to oxygen i trigonal planar (why?) and ha p 2 hybrid orbital. (e) The left-mot carbon i tetrahedral (p 3 hybrid orbital). The other carbon i trigonal planar with p 2 hybridized orbital (a) p p p Strategy: The tep for determining the hybridization of the central atom in a molecule are: draw Lewi Structure ue VSEPR to determine the ue Table 10.4 of of the molecule electron pair arrangement the text to determine urrounding the central the hybridization tate atom (Table 10.1 of the text) of the central atom

7 APTER 10: EMIAL BNDING II 269 Solution: Write the Lewi tructure of the molecule. Several reonance form with formal charge are hown. 2 2 N N N N N N N N N ount the number of electron pair around the central atom. Since there are two electron pair around N, the electron arrangement that minimize electron-pair repulion i linear (AB 2 ). Remember, for VSEPR purpoe a multiple bond count the ame a a ingle bond. We conclude that N i p hybridized becaue it ha the electron arrangement of two p hybrid orbital The Lewi tructure i hown below. The two end carbon are trigonal planar and therefore ue p 2 hybrid orbital. The central carbon i linear and mut ue p hybrid orbital. A Lewi drawing doe not necearily how actual molecular geometry. Notice that the two 2 group at the end of the molecule mut be perpendicular. Thi i becaue the two double bond mut ue different 2p orbital on the middle carbon, and thee two 2p orbital are perpendicular. The overlap of the 2p orbital on each carbon i hown below. I the allene molecule polar? Strategy: The tep for determining the hybridization of the central atom in a molecule are: draw Lewi Structure ue VSEPR to determine the ue Table 10.4 of of the molecule electron pair arrangement the text to determine urrounding the central the hybridization tate atom (Table 10.1 of the text) of the central atom Solution: Write the Lewi tructure of the molecule. P ount the number of electron pair around the central atom. Since there are five electron pair around P, the electron arrangement that minimize electron-pair repulion i trigonal bipyramidal (AB 5 ). We conclude that P i p 3 d hybridized becaue it ha the electron arrangement of five p 3 d hybrid orbital It i almot alway true that a ingle bond i a igma bond, that a double bond i a igma bond and a pi bond, and that a triple bond i alway a igma bond and two pi bond. (a) igma bond: 4; pi bond: 0 igma bond: 5; pi bond: 1 igma bond: 10; pi bond: 3

8 270 APTER 10: EMIAL BNDING II A ingle bond i uually a igma bond, a double bond i uually a igma bond and a pi bond, and a triple bond i alway a igma bond and two pi bond. Therefore, there are nine pi bond and nine igma bond in the molecule The molecular orbital electron configuration and bond order of each pecie i hown below σ 2 2 bond order = 1 bond order 1 = bond order = 0 2 The internuclear ditance in the 1 ion hould be greater than that in the neutral hydrogen molecule. The ditance in the 2 ion will be arbitrarly large becaue there i no bond (bond order zero) In order for the two hydrogen atom to combine to form a 2 molecule, the electron mut have oppoite pin. urthermore, the combined energy of the two atom mut not be too great. therwie, the 2 molecule will poe too much energy and will break apart into two hydrogen atom The energy level diagram are hown below. e 2 e e 2 1 σ 1 σ bond order = 0 bond order = 1 2 bond order = 1 2 e 2 ha a bond order of zero; the other two have bond order of 1/2. Baed on bond order alone, e 2 ha no tability, while the other two have roughly equal tabilitie The electron configuration are lited. Refer to Table 10.5 of the text for the molecular orbital diagram. Li 2 : σ2 ( ) ( ) ( ) bond order = 1 Li 2 : Li2 : σ2 ( ) ( ) ( ) σ2 σ2 ( ) ( ) ( ) ( ) bond order = bond order = rder of increaing tability: Li2 = Li 2 < Li 2 In reality, Li 2 i more table than Li2 becaue there i le electrotatic repulion in Li 2.

9 APTER 10: EMIAL BNDING II The Be 2 molecule doe not exit becaue there are equal number of electron in bonding and antibonding molecular orbital, making the bond order zero. σ2 σ2 bond order = See Table 10.5 of the text. Removing an electron from B 2 (bond order = 1) give B 2, which ha a bond order of (1/2). Therefore, B 2 ha a weaker and longer bond than B The energy level diagram are hown below. σ2 p x 2p, π2p π y z 2 2 σ2 p x 2p, π2p σ2 p σ x 2 p x π2p, π y 2p π z 2p, π y 2p z σ2 σ2 σ2 σ2 π y z 2 The bond order of the carbide ion i 3 and that of 2 i only 2. With what homonuclear diatomic molecule i the carbide ion ioelectronic? In both the Lewi tructure and the molecular orbital energy level diagram (Table 10.5 of the text), the oxygen molecule ha a double bond (bond order = 2). The principal difference i that the molecular orbital treatment predict that the molecule will have two unpaired electron (paramagnetic). Experimentally thi i found to be true In forming the N 2 from N2, an electron i removed from the igma bonding molecular orbital. onequently, the bond order decreae to 2.5 from 3.0. In forming the 2 ion from 2, an electron i removed from the pi antibonding molecular orbital. onequently, the bond order increae to 2.5 from We refer to Table 10.5 of the text. 2 ha a bond order of 2 and i paramagnetic (two unpaired electron). 2 ha a bond order of 2.5 and i paramagnetic (one unpaired electron). 2 ha a bond order of 1.5 and i paramagnetic (one unpaired electron). 2 2 ha a bond order of 1 and i diamagnetic.

10 272 APTER 10: EMIAL BNDING II Baed on molecular orbital theory, the tability of thee molecule increae a follow: 2 2 < 2 < 2 < rom Table 10.5 of the text, we ee that the bond order of 2 i 1.5 compared to 1 for 2. Therefore, 2 hould be more table than 2 (tronger bond) and hould alo have a horter bond length A dicued in the text (ee Table 10.5), the ingle bond in B 2 i a pi bond (the electron are in a pi bonding molecular orbital) and the double bond in 2 i made up of two pi bond (the electron are in the pi bonding molecular orbital) Benzene i tabilized by delocalized molecular orbital. The bond are equivalent, rather than alternating ingle and double bond. The additional tabilization make the bond in benzene much le reactive chemically than iolated double bond uch a thoe in ethylene The ymbol on the left how the pi bond delocalized over the entire molecule. The ymbol on the right how only one of the two reonance tructure of benzene; it i an incomplete repreentation If the two ring happen to be perpendicular in biphenyl, the pi molecular orbital are le delocalized. In naphthalene the pi molecular orbital i alway delocalized over the entire molecule. What do you think i the mot table tructure for biphenyl: both ring in the ame plane or both ring perpendicular? (a) Two Lewi reonance form are hown below. ormal charge different than zero are indicated. N N There are no lone pair on the nitrogen atom; it hould have a trigonal planar electron pair arrangement and therefore ue p 2 hybrid orbital. The bonding conit of igma bond joining the nitrogen atom to the fluorine and oxygen atom. In addition there i a pi molecular orbital delocalized over the N and atom. I nitryl fluoride ioelectronic with the carbonate ion? The ion contain 24 valence electron. f thee, ix are involved in three igma bond between the nitrogen and oxygen atom. The hybridization of the nitrogen atom i p 2. There are 16 non-bonding electron on the oxygen atom. The remaining two electron are in a delocalized pi molecular orbital which reult from the overlap of the p z orbital of nitrogen and the p z orbital of the three oxygen atom. The molecular orbital are imilar to thoe of the carbonate ion (See Section 10.8 of the text) The Lewi tructure of ozone are: The central oxygen atom i p 2 hybridized (AB 2 E). The unhybridized 2p z orbital on the central oxygen overlap with the 2p z orbital on the two end atom.

11 274 APTER 10: EMIAL BNDING II S AB 6 octahedral μ = 0 Why do the bond dipole add to zero in P 5? According to valence bond theory, a pi bond i formed through the ide-to-ide overlap of a pair of p orbital. A atomic ize increae, the ditance between atom i too large for p orbital to overlap effectively in a ide-to-ide fahion. If two orbital overlap poorly, that i, they hare very little pace in common, then the reulting bond will be very weak. Thi ituation applie in the cae of pi bond between ilicon atom a well a between any other element not found in the econd period. It i uually far more energetically favorable for ilicon, or any other heavy element, to form two ingle (igma) bond to two other atom than to form a double bond (igma pi) to only one other atom Geometry: bent; hybridization: p The Lewi tructure and VSEPR geometrie of thee pecie are hown below. The three nonbonding pair of electron on each fluorine atom have been omitted for implicity. Xe Xe Sb AB 3 E 2 AB 5 E AB 6 T-haped Square Pyramidal ctahedral (a) The Lewi tructure i: B The hape will be trigonal planar (AB 3 ) The Lewi tructure i: - The molecule will be a trigonal pyramid (nonplanar). (d) The Lewi tructure and the dipole moment for 2 i preented in Problem The dipole moment i directed from the poitive hydrogen end to the more negative oxygen. The Lewi tructure i: The molecule i bent and therefore polar.

12 APTER 10: EMIAL BNDING II 275 (e) The Lewi tructure i: N The nitrogen atom i of the AB 2 E type, but there i only one unhared electron rather than the uual pair. A a reult, the repulion will not be a great and the N angle will be greater than 120 expected for AB 2 E geometry. Experiment how the angle to be around 135. Which of the pecie in thi problem ha reonance tructure? To predict the bond angle for the molecule, you would have to draw the Lewi tructure and determine the geometry uing the VSEPR model. rom the geometry, you can predict the bond angle. (a) Be 2 : AB 2 type, 180 (linear). B 3 : AB 3 type, 120 (trigonal planar). 4 : AB 4 type, (tetrahedral). (d) 3 : AB 4 type, (tetrahedral with a poible light ditortion reulting from the different ize of the chlorine and hydrogen atom). (e) g 2 2 : Each mercury atom i of the AB 2 type. The entire molecule i linear, 180 bond angle. (f) Sn 2 : AB 2 E type, roughly 120 (bent). (g) 2 2 : The atom arrangement i. Each oxygen atom i of the AB 2 E 2 type and the angle will be roughly (h) Sn 4 : AB 4 type, (tetrahedral) The two approache are dicued in Section 10.1, 10.3, and 10.4 of the textbook Since arenic and phophoru are both in the ame group of the periodic table, thi problem i exactly like Problem A 5 i an AB 5 type molecule, o the geometry i trigonal bipyramidal. We conclude that A i p 3 d hybridized becaue it ha the electron arrangement of five p 3 d hybrid orbital (a) The Lewi tructure i: S The geometry i trigonal planar; the molecule i nonpolar. The Lewi tructure i: P The molecule ha trigonal pyramidal geometry. It i polar.

13 276 APTER 10: EMIAL BNDING II The Lewi tructure i: Si The molecule will be tetrahedral (AB 4 ). Both fluorine and hydrogen are more electronegative than ilicon, but fluorine i the mot electronegative element, o the molecule will be polar (fluorine ide negative). (d) The Lewi tructure i: Si The ion ha a trigonal pyramidal geometry (AB 3 E). (e) The Lewi tructure i: Br Br The molecule will be tetrahedral (AB 4 ) but till polar. The negative end of the dipole moment will be on the ide with the two bromine atom; the poitive end will be on the hydrogen ide nly I 2 and dbr2 will be linear. The ret are bent The Lewi tructure i hown below. 2 Be The molecule i of the AB 4 type and hould therefore be tetrahedral. The hybridization of the Be atom hould be p (a) Strategy: The tep for determining the hybridization of the central atom in a molecule are: draw Lewi Structure ue VSEPR to determine the ue Table 10.4 of of the molecule electron pair arrangement the text to determine urrounding the central the hybridization tate atom (Table 10.1 of the text) of the central atom

14 APTER 10: EMIAL BNDING II 277 Solution: The geometry around each nitrogen i identical. To complete an octet of electron around N, you mut add a lone pair of electron. ount the number of electron pair around N. There are three electron pair around each N. Since there are three electron pair around N, the electron-pair arrangement that minimize electron-pair repulion i trigonal planar. We conclude that each N i p 2 hybridized becaue it ha the electron arrangement of three p 2 hybrid orbital. Strategy: Keep in mind that the dipole moment of a molecule depend on both the difference in electronegativitie of the element preent and it geometry. A molecule can have polar bond (if the bonded atom have different electronegativitie), but it may not poe a dipole moment if it ha a highly ymmetrical geometry. Solution: An N bond i polar becaue i more electronegative than N. The tructure on the right ha a dipole moment becaue the two N bond moment do not cancel each other out and o the molecule ha a net dipole moment. n the other hand, the two N bond moment in the left-hand tructure cancel. The um or reultant dipole moment will be zero (a) The tructure for cyclopropane and cubane are yclopropane ubane The bond in cyclopropane i 60 and in cubane i 90. Both are maller than the expected for p 3 hybridized carbon. onequently, there i coniderable train on the molecule. They would be more difficult to make than an p 3 hybridized carbon in an uncontrained (that i, not in a mall ring) ytem in which the carbon can adopt bond angle cloer to In 1,2-dichloroethane, the two atom are joined by a igma bond. Rotation about a igma bond doe not detroy the bond, and the bond i therefore free (or relatively free) to rotate. Thu, all angle are permitted and the molecule i nonpolar becaue the bond moment cancel each other becaue of the averaging effect brought about by rotation. In ci-dichloroethylene the two bond are locked in poition. The π bond between the atom prevent rotation (in order to rotate, the π bond mut be broken, uing an energy ource uch a light or heat). Therefore, there i no rotation about the = in ci-dichloroethylene, and the molecule i polar.

15 278 APTER 10: EMIAL BNDING II onider the overlap of the 2p orbital on each carbon atom. The geometric plane containing the group at each end of the molecule are mutually perpendicular. Thi i becaue the two carbon-carbon double bond mut ue different 2p orbital on the middle carbon, and thee two 2p orbital are perpendicular. Thi mean that the two chlorine atom can be conidered to be on one ide of the molecule and the two hydrogen atom on the other. The molecule ha a dipole moment. Draw the end-on view if you aren't convinced ,, 2, N 2, N 2, 4, and 3 are greenhoue gae Both ulfur to oxygen bond are double bond which occupy more pace than ingle bond and the repulion between the double bond pread out the angle cloe to The Lewi tructure i: N * * 3 * 2 N * * N N # N * The carbon atom and nitrogen atom marked with an aterik ( and N ) are p 2 hybridized; unmarked carbon atom and nitrogen atom are p 3 hybridized; and the nitrogen atom marked with (#) i p hybridized or an octahedral AX 4 Y 2 molecule only two different tructure are poible: one with the two Y next to each other like and (d), and one with the two Y on oppoite ide of the molecule like (a) and. The different looking drawing imply depict the ame molecule een from a different angle or ide. It would help to develop your power of patial viualization to make ome imple model and convince yourelf of the validity of thee anwer. ow many different tructure are poible for octahedral AX 5 Y or AX 3 Y 3 molecule? Would an octahedral AX 2 Y 4 molecule have a different number of tructure from AX 4 Y 2? Ak your intructor if you aren t ure ha no d orbital but Si doe (3d). Thu, 2 molecule can add to Si in hydrolyi (valence-hell expanion).

16 APTER 10: EMIAL BNDING II (a) Although the atom are p 3 hybridized, they are locked in a planar tructure by the benzene ring. The molecule i ymmetrical and therefore doe not poe a dipole moment. 20 σ bond and 6 π bond Referring to Table 10.5 of the text, we ee that 2 and N 2 have a imilar energy order for their molecular orbital. Therefore, we expect N to have the ame order py 2pz 2p ( σ ) ( σ ) ( σ ) ( σ ) ( π ) ( π ) ( σ x ) i ioelectronic with N (both contain 14 electron) (a) Thi i the only reaonable Lewi tructure for. The electronegativity difference ugget that electron denity hould concentrate on the atom, but aigning formal charge place a negative charge on the atom. Therefore, we expect to have a mall dipole moment. The Lewi tructure how a triple bond. The molecular orbital decription give a bond order of 3, jut like N 2 (ee Table 10.5 of the text). Normally, we would expect the more electronegative atom to bond with the metal ion. In thi cae, however, the mall dipole moment ugget that the atom may form a tronger bond with e 2 than. More elaborate analyi of the orbital involved how that thi i indeed the cae can be repreented by: θ Let ρ be a bond moment. Thu, ρ = 3ρ co θ co θ = 1 3 θ = 70.5 Tetrahedral angle = = 109.5