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9:32 O©& Oe 6 OP #4640, , fe Chemical Bonding and Molecular, , Structure, , , , Question 4.40:, , What is meant by the term bond order? Calculate the bond order of: No, Oo, 0; and 0;, , Answer:, , Bond order is defined as one half of the difference between the number of electrons, present in the bonding and anti-bonding orbitals of a molecule., , |f Ng is equal to the number of electrons in an anti-bonding orbital, then Np is equal to the, , number of electrons in a bonding orbital., , 1, 5M Ny, Bond order= 2.” ), , If Np > Na, then the molecule is said be stable. However, if Ny < Na, then the molecule is, , considered to be unstable., , Bond order of Np can be calculated from its electronic configuration as:, (o(ls)Plo Us)Flo2s) Flo’ 2s)P a2 pla? p,)PLa2 pP, , Number of bonding electrons, N, = 10, , Number of anti-bonding electrons, N, = 4, , (10-4), Bond order of nitrogen molecule 2, , =3, , There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic, configuration of oxygen molecule can be written as:, , [o -(s)F[o' Us) F[o29)F lo 29)F lop. Fa p Va p, Pla 2p 1b (2 pT, , Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding, electrons = 8 = N, and the number of anti-bonding electrons = 4=N,, , =+(N,-N,, Bond order 2! oM), 1, =1(s-4, L(§-4), =2, , Hence, the bond order of oxygen molecule is 2., , Similarly, the electronic configuration of 0; can be written as:, , KKlo@2s)F lo" 2s)F (o(2p¥ [a2 pF [a2 p, VE (2 pT, , There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic, configuration of oxygen molecule can be written as:, , [o = (1s) Po Us) F [629 Plo (28)P [op Pla pV 1a p, Pit (2p Ti (2p, )]!, , Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding, electrons = 8 = N, and the number of anti-bonding electrons = 4=N,, , =1(N,-N,, Bond order 2! oM), 1, =1(s-4, Lg), =2, , Hence, the bond order of oxygen molecule is 2., Similarly, the electronic configuration of 0; can be written as:, KK[o(2s)F lo (25)F lo 2p. FLX pF La pL 2 pT, Np =8, Na=3, +1, , 0; =1(8-3), , Bond order of 2, , =2.5, Thus, the bond order of oO; is 2.5., , The electronic configuration of 0; ion will be:, KK[o(2s)P Lo’ (28) Plo2p Pep Pin pF Le 2p Pe 2p,, Np=8, , Na=5, , 1, - —(8-5, Bond order of ©: = 2! ), , =15, , Thus, the bond order of 0; ion is 1.5., , a