Red veze za benzen. Slika 1.
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1 Red veze za benzen Benzen C 6 H 6 je aromatično ciklično jedinjenje. Njegove dve rezonantne forme (ili Kekuléove structure), prema teoriji valentne veze (VB) prikazuju se uobičajeno kao na slici 1 a), gde su jednostruka i dvostruka veza naizmenično raspoređene u prstenu, a takođe koristi se i prikaz na slici 1b) koji proizilazi iz MO teorije i označava da je 6 π elektrona delokalizovano po celom prstenu Slika 1. Dve rezonantne forme benzena prema VB teoriji su takođe prikazane na slici 2. Prema VB opisu, svaki C atom u benzenu je sp 2 hibridizovan, a ceo molekul je planaran. Svaki C atom povezan je sigma vezom sa dva susedna C atoma i sa jednim H atomom. Na svakom C atomu ostaje po jedna nehibridizovana 2p z orbitala i po jedan zaostali valentni elektron u svakoj orbitali (slika 2). Prema VB teoriji, parovi susednih 2p z orbitala i 6 preostalih elektrona okupiraju regione preklapanja formirajući tri π veze na jedan ili drugi način prikazan na slici 2 (odnosno šematski na slici 1 a) Slika 2. VB prezentacija benzena.
2 Eksperimentalne studije strukture benzena pokazale su da on ne sadrži naizmenične proste i dvogube CC veze. Uobičajena dužina C C jednostruke veze je 1.54 Å, a dužina C=C dvostruke veze je 1.34 Å. Svih šest CC veza u benzenu su iste dužine, 1.39 Å, između dužine jednostruke i dvostruke veze. Ovo je dobro objašnjeno MO teorijom, koja predviđa da se šest 2p z orbitala C atoma preklapaju i mešaju formirajući tri π-vezivne i tri π-antivezivne molekulske orbitale. Na primer, najjače vezujuća π molekulska orbitala u benzenovom π MO sistemu je ona prikazana na slici 3. Šest π elektrona okupiraju tri vezivne molekulske orbitale ovog delokalizovanog sistema, odnosno, ovi elektroni su raspoređeni kroz molekul kao celinu, iznad i ispod ravni koju formiraju σ veze. Ovo rezultuje u identičnom karakteru svih CC veza u benzenu. Svaka CC veza ima red veze 1,5. MO prezentacija proširenog (delokalizovanog) π sistema (Slika 3) je ista kao ona koja se dobija usrednjavanjem dve rezonantne strukture iz teorije valentne veze. Slika 3. MO prezentacija benzena. U benzenu, 6 π elektrona dele 6 ugljenikovih atoma, tako da svaki ugljenikov atom ima polovinu π veze i jednu σ vezu. Odatle je red veze benzena 1,5 za svaku vezu.
3 Slika 3. Kompjuterska prezentacija jedne od molekulskih orbitala benzena. Benzen ima 6 π molekulskih orbitala, (koje sadrže 6 π-elektrona), tri vezivne i tri antivezivne, koje nastaju iz 6 p z orbitala atoma ugljenika. Oblik ovih 6 π-molekulskih orbitala i relativan raspored njihovih energija prikazan je na slikama 4 i 5. Gornji par vezivnih degenerisanih orbitala predstavljaju najviše popunjene molekulske orbitale (HOMO) benzena. π elektroni su delokalizovani kroz ceo prsten ugljenikovih atoma. Slika 4. Šest π-molekulskih orbitala benzena
4 Slika 5. Šest π-molekulskih orbitala benzena. Sa slike 5 može se videti da za energetski najnižu orbitalu, π 1, kooeficijenti orbitale su takvi da je vezivni karakter između svaka dva susedna C atoma podjednak. U π 2 orbitali vezivanje se javlja samo između atoma C 2 and C 3 i između atoma C 5 and C 6, pošto su koeficijenti za atome C 1 i C 4 nula. U orbitali π 3, C 1 je vezan sa C 2 i C 6, odnosno C 4 je vezan sa C 3 i C 5, a postoje anti-vezivne interakcije između C 2 and C 3 i između C 5 i C 6. Otuda, ako razmatramo par orbitala π 2 i π 3 njihov ukupan doprinos CC π vezivanju je podjednak za svaku vezu. Dakle, pošto ima u benzenu tri zauzete vezivne orbitale i 6 CC veza - π red veze je 1 / 2.
5 Bond order is a term used to determine the number of bonds in a covalent bond formation. For example, when Nitrogen forms a triple bond with another Nitrogen atom the bond order of the molecule is 3. Similarly, the bond order between a carbon and hydrogen atom (C-H) is 1. Bond order however, is not always a whole number, it can be a fraction as well. The bond order of resonating structures such as benzene is a fraction. The bond order also determines the stability of a bond. Bond order is directly proportional to the energy required to break the bond, that is the more the bond order higher is the energy required to break it. Bond Order Formula Bond order is a concept very commonly used in molecular orbital theory. According to this theory, bond order is half of the difference between the number of bonding electrons and the number of antibonding electrons. Bond Order = ½ (number of bonding electrons - number of antibonding electrons) Before calculating bond order, you must know what these bonding and antibonding electrons are. Bonding and antibonding orbitals are formed when two atoms combine to form a molecule. The bonding is governed by Pauli's Exclusion Principle, which states that no two electrons can have the same quantum numbers that is n, l, m, s and they must be different for the electrons to have opposite spins. The antibonding orbitals, as the name suggests, are at a higher energy level and are unstable and do not support bond formation, electrons in this orbital are the antibonding electrons. Similarly, bonding electrons, in very simple words are electrons in the binding orbitals, which are at a lower and stable energy state and therefore, support binding. Knowing the structures and placing the electrons in correct orbitals can help determine the electrons in the bonding or antibonding orbitals. How to Calculate Bond Order Calculating bond order can be understood only when you are clear with molecular orbital theory and the orbital structures of atoms. With the help of following examples, we can learn how to find bond order. Bond Order of C 2 - You must know the number of electrons in the bonding and antibonding orbitals. A clue to find this out is to draw an orbital structure and determine the number of electrons in each. In the case of C 2 - there is an extra electron in the bonding orbital. Thus, after applying the formula given above, Bond order = ½ (9-4) = 2.5 Bond Order of N 2 + and N 2 - The bond order for both these is 2.5 because, in both cases there is an unpaired electron either in the bonding or antibonding orbital, reducing the bond order by half. Bond Order for Benzene Benzene is an aromatic cyclic compound and therefore has resonance structures. In benzene, 6 pi electrons are shared by 6 carbon atoms, thus every carbon has half a pi bond and one
6 sigma bond. The bond order of benzene is therefore, 1.5 for each bond. Bond Order for Diatomic Molecule Linus Pauling, derived a formula to determine the bond order, which was later realized, can only be applied to diatomic molecules. It is given as, Bond order(s ij ) = exp[r ij - d ij / b] Here, R ij stands for experimentally measured bond length d ij stands for the single bond length and 'b' is a constant, calculated by Pauling as
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