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Chemistry · Ch 5 — Chemical Bonding

MO description of simple diatomic Molecules

5.5.6

MO description of simple diatomic Molecules

Applying the molecular-orbital energy-level ordering (section 5.5.4) and the bond-order formula to five real second-row homonuclear diatomic molecules gives a consistent, testable picture of each one's bonding, stability and magnetism. H2 (2 electrons): configuration σ1s2\sigma1s^2, bond order =(2−0)/2=1=(2-0)/2=1 (a single bond, matching its 74 pm bond length and 438 kJ mol⁻¹ dissociation energy), diamagnetic (no unpaired electrons). Li2 (6 electrons): configuration (σ1s)2(σ∗1s)2(σ2s)2(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2, bond order =(4−2)/2=1=(4-2)/2=1 (one single bond; Li2 molecules occur in the vapour state), diamagnetic. N2 (14 electrons, using the NON-O2/F2 energy order so the degenerate π2p pair fills BEFORE σ2pz): configuration (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2(\pi2p_y)^2(\sigma2p_z)^2, bond order =(10−4)/2=3=(10-4)/2=3 (a triple bond, N≡N — the highest bond order, and correspondingly the highest bond enthalpy, 946 kJ mol⁻¹, of any diatomic molecule covered in this chapter), diamagnetic. O2 (16 electrons, using the O2/F2 energy order, σ2pz BELOW the π2p pair): configuration (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)1(π∗2py)1(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_z)^2(\pi2p_x)^2(\pi2p_y)^2(\pi^*2p_x)^1(\pi^*2p_y)^1 — the final two electrons occupy the two degenerate π* orbitals SINGLY rather than pairing into just one — bond order =(10−6)/2=2=(10-6)/2=2 (a double bond, O=O), and PARAMAGNETIC because of its two unpaired π*-orbital electrons: this is exactly the correct experimental result that valence bond theory (section 5.4. …

Figure 5.8MO diagram for the H2 molecule

What this figure shows. Each H atom contributes one electron from its 1s orbital; combining the two 1s orbitals gives σ1s (bonding, lower energy) and σ1s (antibonding, higher energy). Both of the molecule's two electrons occupy the lower-energy σ1s orbital (paired, opposite spins), leaving σ1s empty. Electronic configuration: σ1s2\sigma1s^2. Bond order =2−02=1=\frac{2-0}{2}=1, i.e. a single covalent bond, consistent with H2's observed bond length of 74 pm and bond dissociation energy of 438 kJ mol⁻¹. Since …

Figure 5.9MO diagram for the Li2 molecule

What this figure shows. Each Li atom (1s2 2s11s^2\,2s^1) contributes 3 electrons, so Li2 as a whole has 6 electrons, filling four molecular orbitals: σ1s, σ1s, σ2s (and leaving σ2s empty). Electronic configuration: (σ1s)2(σ∗1s)2(σ2s)2(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2. Bond order =4−22=1=\frac{4-2}{2}=1, i.e. one single bond between the two lithium atoms (Li2 molecules are observed in the vapour state). With no unpaired …

Figure 5.10MO diagram for the N2 molecule

What this figure shows. Each N atom (1s2 2s2 2p31s^2\,2s^2\,2p^3) contributes 7 electrons, so N2 has 14 electrons in all. Following the non-O2/F2 energy order (πpx, πpy below σ2pz), the configuration is (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2(\pi2p_y)^2(\sigma2p_z)^2, with all higher antibonding orbitals empty. Bond order =10−42=3=\frac{10-4}{2}=3, i.e. a triple bond between the two nitrogen atoms — matching N≡N, and consistent with N2 having, among diatomic molecules, both the highest bond order and the highest bond enthalpy of any (946 kJ mol⁻¹, per the source's own 'Do …

Figure 5.11MO diagram for the O2 molecule

What this figure shows. Each O atom (1s2 2s2 2p41s^2\,2s^2\,2p^4) contributes 8 electrons, so O2 has 16 electrons in all. This time the O2/F2 energy order applies (σ2pz BELOW the degenerate π2p pair), giving configuration (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)1(π∗2py)1(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_z)^2(\pi2p_x)^2(\pi2p_y)^2(\pi^*2p_x)^1(\pi^*2p_y)^1 — note the LAST two electrons go one each into the two degenerate π* orbitals (Hund's-rule-style single occupancy) rather than pairing up in just one. Bond order =10−62=2=\frac{10-6}{2}=2, i.e. an O=O double bond. Because the two π* orbitals each hold one unpaired electron, O2 is paramagnetic — the correct experimental behaviour that valence bond theory (section 5.4.7) co …

Figure 5.12MO diagram for the F2 molecule

What this figure shows. Each F atom (1s2 2s2 2p51s^2\,2s^2\,2p^5) contributes 9 electrons, so F2 has 18 electrons in all. Using the O2/F2 energy order, the configuration is (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)2(π∗2py)2(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_z)^2(\pi2p_x)^2(\pi2p_y)^2(\pi^*2p_x)^2(\pi^*2p_y)^2 — every orbital up to and including both π* orbitals is now completely filled. Bond order =10−82=1=\frac{10-8}{2}=1, i.e. a single F–F bond. With every …

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