Constructing a MO of NH 3. Nitrogen AO symmetries are

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Oswald Blake
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1 Constructing a MO of NH 3 Nitrogen AO symmetries are

2 To develop a MO scheme for NH 3 assume that only the 2s and2p orbitals of nitrogen interact with the hydrogen 1s orbitals (i.e., the nitrogen 1s orbital is nonbonding). All AOs match with SALCs, so there are no nonbonding levels. Both s and p z AOs match the A 1 SALC, so s p mixing is likely. If bonding and antibonding combinations were formed for both s and p z AOs, we would end up with eight MOs, but only seven AOs on N and H are available (disregarding the 1s AO on N). We must make only three MOs from the s and p z AOs and the A 1 SALC. For simplicity, we will assume that the s and p z AOs each form essentially separate bonding MOs, but that together they form a single mixed antibonding MO. The resulting MO scheme is shown below, where hashed lines indicate lesser contributions from s p mixing. Two AOs of nitrogen match symmetry with one of the SALCs, thereby forming three MOs. Of these, the lowest energy MO is essentially a bonding combination formed between the hydrogen SALC and the 2s AO on nitrogen, and the highest MO is an antibonding combination formed from a mixture of the two AOs from nitrogen and the same symmetry SALC.

3 Symmetry SALCs AOs E' (1 node) p x p y A 1 ' (0 nodes) s p z

4 Symmetry MOs E' 2e' 2e ' * e' e ' * A 1 ' 2a 1 ' 2a 1 ' * a 1 ' a 1 ' *

5 Lewis base characterof NH 3, because the σ(z) MOhas considerable electron density above the nitrogen, not unlike the customary picture of the VB model's lone pair sp 3 hybrid.

6 MOs of Benzene, C 6 H 6 C 6 H 6, has three pairs of electrons delocalized in a system extending around the hexagonal ring. The six 2p orbitals perpendicular to the ring on the six carbon atoms combine to form three bonding ( 1, 2, 3 ) and three antibonding ( 4 *, 5 *, 6 *) MOs. The symmetries and forms of these MOs can be deduced by applying the operations of the point group D 6h to a set of six vectors perpendicular to the ring, one at each carbon, to generate a reducible representation. = B 2g + E 1g + A 2u + E 2u

7 MOs of Benzene, C 6 H 6 h = 24 D 6h E 2C 6 2C 3 C 2 3C 2 ' 3C 2 '' i 2S 3 2S 6 h 3 d 3 v /h A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u B 2g E 1g A 2u +E 2u d

8 = B 2g + E 1g + A 2u + E 2u MOs of Benzene, C 6 H 6 (B 2g ) = (1/ 6) ( a b + c d + e f ) (E 2u ) = (1/2 3) (2 a b c + 2 d e f ) (E 2u ) = (1/2) (- b + c e + f ) (E 1g ) = (1/2 3) (2 a + b c 2 d e f ) (E 1g ) = (1/2) ( b + c e f ) (A 2u ) = (1/ 6) ( a + b + c + d + e + f )

9 = B 2g + E 1g + A 2u + E 2u MOs of Benzene, C 6 H 6 B 2g (6 nodes) anti-bonding (B 2g ) = (1/ 6) ( a b + c d + e f ) E 2u (4nodes) anti-bonding (E 2u ) = (1/2 3) (2 a b c + 2 d e f ) (E 2u ) = (1/2) (- b + c e + f ) E 1g (2 nodes) bonding (E 1g ) = (1/2 3) (2 a + b c 2 d e + f ) (E 1g ) = (1/2) ( b + c e f ) A 2u (0 nodes) bonding (A 2u ) = (1/ 6) ( a + b + c + d + e + f )

10 MO Energy Level Scheme for Benzene, C 6 H 6 By the shadow method we can anticipate that the MO scheme will look like the following: Three pairs in bonding MOs add a total of three bond orders over six C C linkages, or 0.5 for each. When this is added to the sigma bond between each carbon pair, the C C bond order becomes 1.5.

11 MOs of Cyclopentadienyl, C 5 H 5 C 5 H 5, has three pairs of electrons delocalized in a system extending around the pentagonal ring. Thefive2p orbitals perpendicular to the ring on the five carbon atoms combine to form three bonding ( 1, 2, 3 ) and three antibonding ( 4 *, 5 *, 6 *) MOs. The symmetries and forms of these MOs can be deduced by applying the operations of the point group D 5h to a set of five vectors perpendicular to the ring, one at each carbon, to generateareduciblerepresentation tti.

12 MOs of Cyclopentadienyl, C 5 H 5 D 5h E 2C 5 2C 5 2 5C 2 h 2S 5 2S v /h h = A ' A ' E ' E ' A '' A '' E '' E '' d

13 Thefivep orbitals on the planar Cp ring (D 5h symmetry) can be combined to produce five molecular orbitals according to the reducible representation: One combination has the full symmetry of the ring (a 2 ) There are two doubly degenerate combinations (e 1 and e 2 ) having one and two planar nodes at right angles to the plane of the ring. The relative energies of these orbitals increase as the number of nodes increases. The a 2 and e 1 orbitals are both fully occupied in the electronic configuration of the Cp anion whereas the e 2 orbitals are net anti bonding and are unfilled.

14 E 2 '' (4 nodes) anti-bonding E 1 '' (2 nodes) weakly bonding A 2 '' (0 nodes) bonding The molecular orbitals of the cyclopentadienyl ring (D 5h )

15 MOs of the Allyl anion, C 3 H 5 The alkyl anion has a delocalized, open, three center system. Although it is customary to assume that p z orbitals are involved in forming orbitals, in this case for constructing the MO diagram we will assume that p x orbitals are used, in keeping with the standard character table and conventions of defining z as the principal axis and the yz plane as the plane of the C C C chain.

17 B 1 (2 nodes) anti-bonding A 2 (1 node) weakly bonding B 1 (0 nodes) bonding

18 and MOs of Boron Trifluoride, BF 3 Although BH 3 is unstable, the BX 3 trihalides (X = F, Cl, Br) are stable but reactive compounds that have been well characterized. A significant advantage of the BX 3 compounds is the potential for bonding between the 2p z orbital of boron and the np z orbitals of the halogens (n =2,3,4). Although fluorine is the most electronegative of the halides this bonding is most significant for BF 3 due to closely matched energies with the central boron atom.

19 MOs of BF 3 h =12 D 3h E 2C 3 3C 2 h 2S 3 3 v /h A 1 ' A 2 ' E' A 1'' A 2 '' E'' A 1 ' E ' d

20 A 1 ' E '

21 Based upon the symmetry of the 3F SALCs derived from the F 1s orbitals symmetry allowed overlap with the B 2s 2p x and 2p y orbitals is predicted. Due to energy considerations however this interaction is very weak and generally ignored. A 1 ' E ' E' (1 node) A 1 ' (0 nodes) 1 1

22 Overlap of the B 2s 2p x and 2p y orbitals with the 3F SALCs derived from the F 2p x orbitals is more prominent. The 2p y orbitals of the 3F atoms are considered non bonding and are not considered here. A 1 ' E ' Symmetry SALCs AOs E' (1 node) p x p y A 1 ' (0 nodes) s

23 Symmetry allowed sigma MOs derived from the 3F (2p x ) SALCs and the B 2s 2p x and 2p y orbitals. A 1 ' E ' E' (1 node) A 1 ' 1 (0 nodes) 1 1

24 B BF 3 3F 2 1 The non bonding p orbital is responsible for the Lewis acid properties of BF 3 2p x, p y (e') 2p z (a 2 '') n a 1 ' e' 2s (a 1 ') n e' 1 n a 1 '

25 MOs of BF 3 d

26 Symmetry SALCs AOs E'' (1 node) 2 n 3 n A 2 '' (0 nodes) n p z MOs

27 Non bonding SALCs of BF 3 The 2p y orbitals of the 3F atoms are considered non bonding and are considered separately when deriving their irreducible representation. h =12 D 3h E 2C 3 3C 2 h 2S 3 3 v /h n A 1 ' A 2 ' E' A 1 '' A 2 '' E'' d

28 Symmetry y SALCs E' A 2 '

29 B BF 3 3F 2 1 * 2p x, p y (e') 2p z (a 2 '') 1 n 2 n 3 n a 2 '' e'' a 2 ' e' a 1 ' e' 2s (a 1 ') n e' 1 n a 1 '

30 B BF 3 3F 2 1 * 2p x, p y (e') 2p z (a 2 '') a 2 '' e'' a 2 ' e' ' a 1 ' e' 1 n 2 n 3 n 2s (a 1 ') 2 1 Consider the MO diagrams for both NH 3 and BF 3 and how the mechanism for the Lewis acid base reaction may proceed.

Chem Spring, 2017 Assignment 5 - Solutions

Chem Spring, 2017 Assignment 5 - Solutions Page 1 of 10 Chem 370 - Spring, 2017 Assignment 5 - Solutions 5.1 Additional combinations are p z ± d z 2, p x ±d xz, and p y ±d yz. p z ± d z 2 p x ±d xz or p y ±d yz 5.2 a. Li 2 has the configuration