is an equal mixture of all three resonance structures; i.e. a hybrid: each NO bond is times the strength of a normal NO single bond.

Transcription

1 1. Let's illustrate the approach in detail for NO 3 only (i) 1 O NO 3 N O O This just indicates the skeletal structure i.e. which atoms are connected. Total Number of Valence Electrons = 5 (from N) + 3(6 from each O) + 1 (from the net negative charge) = 24 Trial Structure This structure uses 26 electrons, so we must eliminate the two-electron surplus by introducing one double bond. Acceptable resonance structures are: The actual structure of NO 3 is an equal mixture of all three resonance structures; i.e. a hybrid: each NO bond is times the strength of a normal NO single bond. For the following examples, the valence electron count is placed in parentheses after the empirical formula and only the resonance structures that satisfy the octet rule are given. (iv) 2 N (12) N

2 2 (v) 6 6 (30) 2) NO 3 There are 3 possible resonance structures that satisfy the octet rule. In each resonance structure, the formal charge on N = +1; the formal charge on the N = O oxygen = 0; and the formal charge on each of the N O oxygens = 1. The actual structure is an equal mixture of the 3 resonance structures. 2 N Only one resonance structure: the formal charges on all the atoms = Two possible resonance structures: the formal charges on all the atoms = 0. The actual structure is an equal mixture of the two resonance structures.

3 3 Since structure (c) involves (i) separating larger formal charges of opposite sign and (ii) placing like formal charges at a separation corresponding to the N-N bond length, it will make a much smaller contribution than either (a) or (b). Structure (a) involves separating +1 and 1 formal charges to an N=N distance; structure (b) involves separating +1 and 1 formal charges to an N N distance. Since the N=N distance is considerably shorter than the N N distance, we expect resonance structure (a) to make a larger contribution to the actual structure of N 3 than does resonance structure (b).

4 4 4a) Be 4 2 tetrahedral b) 2 O bent; ( ) O < ) c) NO 3 trigonal planar d) NF 3 pyramidal; ) FNF < ) e) Pl 4 + tetrahedral f) XeF 4 square planar g) N linear h) S 3 2 bent; ) SSS < ) 5. Sl 2 ; The central atom is S. The number of valence Electrons = 6(from S) +2(7 from each l) = Trial structure: :l.. Ṣ. l.. : -- uses 20 electrons so we're O.K. Thus, there are 4 Electron Pairs around S. Since there are two ligands there must be two bonding pairs of electrons and two non-bonding pairs. By VSEPR theory, the electron pair geometry will be tetrahedral. owever ) l S l will be somewhat less than because of the larger size of the non-bonding pairs. 6. NO 2 + : Number of valence of electrons = 2 (6) + 5 1(for the positive charge) = 16.

5 .. Trial structure: :Ọ... Ṇ... Ọ.: uses 20 electrons. Introduce one triple bond or two double bonds. Possible resonance structures are: 5 In (a) there are 4 pairs of valence electrons around the central atom N. Since the N=O bonds are double bonds there are only two electron groups around the central atom. The electron groups geometry around N is linear. In NO 2, there are 9 valence electrons around N. The main resonance structure is.... :Ọ. = N = Ọ. There are 4 pairs of electrons and one odd electron around N, but only three groups. The idealized electron group geometry is trigonal planar O. N >120 O Since the size of a 4e group > the size of a 1e group, the O N O angle will increase somewhat to a value >120 o In NO 2, there are 18 valence electrons: the actual structure is a resonance hybrid (equal mixture of (a) and (b) taken together). For a discussion of the angle consider (a). There are 8 valence electrons around N: There are 4 electron pairs but since there is one double bond there are only 3 electron groups The idealized electron group geometry is trigonal planar. Strong repulsion between the nonbonding electron pair and the bonding pairs will result in a decrease of the O N O angle to a value somewhat smaller than 120 o.

6 7. The first step is to write down all the Lewis structures for 3OO and 3OO which satisfy the octet rule: (i) 3OO (24 valence electrons). Possible Lewis structures are: 6 The relative importance of these structures is (a) >> (b). The Oa bond is much closer to a O double bond than it is to a O single bond. In contrast, the Ob bond is much closer to a O single bond than it is to a O double bond. The two O bonds in 3OO differ in length. (ii) 3OO (24 valence electrons). Possible Lewis structures are: The relative importance of these structures is (a) = (b). Each O bond in 3OO is 1.5 times the strength of a single O bond. We expect the two O bonds to be of the same length and to be shorter than the length of a O single bond, but longer than the length of a O double bond. 8. (a) NO (16 valence electrons). Possible Lewis structures which obey the octet rule are: To compute the formal charges we follow the procedure presented in and the text. This procedure leads to the following formal charges in each of the above structures: N O N O N O From these formal charges, we see that only structure (c) involves any separation of formal charge; structures (a) and (b) simply involve placing the overall 1 charge of the ion on different atoms. Thus, from formal charge considerations, (c) will be the least important of the three resonance structures. From electronegativity considerations, it is energetically preferable to place a negative charge on the more electronegative atom. In this case, this means that a structure in which the 1 charge is placed on O will be preferred over that which

7 has the 1 charge on N. On the basis of formal charge and electronegativity ideas, the relative importance of these structures is: (a) > (b) >> (c). (b) NO (16 valence electrons). Lewis structures which obey the octet rule are: 7 The formal charges on the atoms in each of these three structures are: In structure (c) we separate a 3 charge from a +1 charge. Since the energy required to separate formal charges of opposite sign increases as the magnitude of the charges increases and as the separation increases, the energy of resonance structure (c) will be greater than that of either (a) or (b). Similarly, since the energy required to bring formal charges of the same sign together increases as the magnitude of the charges increases and as the separation decreases, the energy of resonance structure (c) will be greater than that of either (a) or (b) and structure (c) willmake a smaller contribution to the actual structure. Structure (a) which places a 1 charge on the most electronegative atom (i.e O) (and a +1 on N and a 1 on ) will be more important than structure (b) which has zero formal charge on O and a 2 charge on which is not a very electronegative atom. On the basis of formal charge and electronegativity ideas, the relative importance of these structures is: (a) > (b) >> (c). 9. Pbl 4 Number. of valence electrons = 4 (for Pb) + 4(7 for l) = 32 and the Lewis electron dot structure satisfying the octet rule is: Pb is surrounded by 4 pairs of electrons. Since there are no multiple bonds, Pb is surrounded by 4 electron. The idealized electron group geometry is tetrahedral. Since all the electron groups are bonding, the idealized molelcular geometry is tetrahedral

8 Sb 3 Number of valence electrons = 5 (for Sb) + 3(1 for ) = 8 and the Lewis electron dot structure satisfying the octet rule is: 8 Sb is surrounded by 4 pairs of electrons. Since there are no multiple bonds, Sb is surrounded by 4 electron groups. The idelaized electron group geometry is tetrahedral. The geometry of the molecule is determined by the positions of the atoms. Since the 4 electron groups are not equivalent there will be deviations from this ideal geometry. Such effects will increase the angle between the lone pair and the bonding pair and will, thereby, decrease the angle between two different bonding pairs. These effects will result in < Sb < SO 2 Number of valence electrons = 6 (for S) + 2(6 for O) = 18. There are two Lewis electron dot structures which satisfy the octet rule and the actual structure of SO 2 will be an equal mixture of the two resonance structures shown below: S is surrounded by 3 groups of electrons; 2 of these are bonding 3- electron groups and the third is a non-bonding pair of electrons. The idealized electron group geometry is trigonal planar. Since the 3 electron groups are not equivalent there will be deviations from this ideal geometry. Such effects will increase the angle between the lone pair and the bonding 3-electron group and willdecrease the angle between two different bonding 3- electron groups. These effects will result in < O S O < Recall that 1 Debye = 1.0 x esu-cm; it would be preferable to convert this to SI units (i.e. m). In SI units, 1 Debye = x m. For sf, µ = 7.88 D = 7.88 x x m = x m. Since µ = δ + d, then δ + = µ d = x m 2.55 x m = x For a bond that is 100% ionic, δ + = +1e = x % ionic character = x x x 100% = 64.4 %

9 11. ere we must examine the electronegativity difference, χ, between the bonded atoms. As χ increases the degree of ionic character increases. 9 l ( χ = 0.9); lf ( χ = 1.0); l 4 ( χ = 0.5); Kl ( χ = 2.2); l 2 ( χ =0) Thus, the order of decreasing ionic character is: Kl > lf > l > l 4 > l To determine the dipole moment for the whole molecule we must examine the magnitude of the bond moments and the geometry of the bond moments. The magnitude of the bond moment is a measure of the degree of polarity of the bond. To determine the polarity of the bond we examine the electronegativity difference of the bonded atoms. For the molecules considered here the order of decreasing polarity of the bonds is: F > l > > Br Even though each bond in 4 and each l bond in l 4 is polar, the tetrahedral structure of 4 and l 4 results in a complete cancellation of the bond moments. 4 and l 4 have zero net dipole moment. For the remaining molecules the bond moments will not cancel. A consideration of the bond moments presented above leads to the following order of increasing dipole meoments: l 4 = 4 < 3 Br < 3 l < 3 F 13. Recognize that the ordering of the MO's for O and F is different from that for B,, and N. For O and F the ordering is: σ2s < σ 2s < σ2p z < π2p x, π2p y < π 2p x, π 2p y < σ 2p z To answer this question, we determine the ground state electron configuration, calculate the bond order, and then recognize that a molecule with a larger Bond Order will have a greater bond energy. (a) B 2 (6 valence electrons); electron config. (σ2s) 2 (σ 2s) 2 (π2p x ) 1 (π2p y ) 1 Bond Order = 1 B 2 + (5 valence electrons); electron config. (σ2s) 2 (σ 2s) 2 (π2p x ) 1 Bond Order = 0.5 B 2 will have the greater bond energy (b) O 2 + (11 valence electrons): electron configuration: (σ2s) 2 (σ 2s) 2 (σ2p z ) 2 (π2p x ) 2 (π2p y ) 2 (π 2p x ) 1 Bond Order = 2.5 O 2 (13 valence electrons): electron configuration: (σ2s) 2 (σ 2s) 2 (σ2p z ) 2 (π2p x ) 2 (π2p y ) 2 (π 2p x ) 2 (π 2p y ) 1 Bond Order = 1.5 O 2 + will have the greater bond energy

10 10 (c) Be 2 (4 valence electrons): electron config.: (σ2s) 2 (σ 2s) 2 Bond Order = 0 Be 2 + (3 valence electrons): electron config.: (σ2s) 2 (σ 2s) 1 Bond Order = 0.5 Be 2 + will have the greater bond energy (d) & (e): F 2 + (13 valence electrons): electron configuration: (σ2s) 2 (σ 2s) 2 (σ2p z ) 2 (π2p x ) 2 (π2p y ) 2 (π 2p x ) 2 (π 2p y ) 1 Bond Order = 1.5 F 2 (14 valence electrons): electron configuration: (σ2s) 2 (σ 2s) 2 (σ2p z ) 2 (π2p x ) 2 (π2p y ) 2 (π 2p x ) 2 (π 2p y ) 2 Bond Order = 1.0 F 2 (15 valence electrons): electron config.: (σ2s) 2 (σ 2s) 2 (σ2p z ) 2 (π2p x ) 2 (π2p y ) 2 (π 2p x ) 2 (π 2p y ) 2 (σ 2p z ) 1 Bond Order = 0.5 F 2 + will have a greater bond energy than F 2 F 2 will have a greater bond energy than F Extending our ideas to the third period homonuclear diatomic molecules, we recognize that the MO's in these cases will be formed by taking linear combinations of 3s and 3p valence atomic orbitals. The orderings of the resultant MO's are analogous to those found in the second period homonuclear diatomic molecules. That is: For Al, Si, and P: σ3s < σ 3s < π3p x, π3p y < σ3p z < π 3p x, π 3p y < σ 3p z For S and l: σ3s < σ 3s < σ3p z < π3p x, π3p y < π 3p x, π 3p y < σ 3p z P 2 (10 valence electrons): electron config.: (σ3s) 2 (σ 3s) 2 (π3p x ) 2 (π3p y ) 2 (σ3p z ) 2 Bond Order = 3 S 2 (12 valence electrons): electron config. : (σ3s) 2 (σ 3s) 2 (σ3p z ) 2 (π3p x ) 2 (π3p y ) 2 (π 3p x ) 1 (π 3p y ) 1 Bond Order = 2 Thus, P 2 should have the greater bond energy, but S 2 should have the greater bond length.

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