Bonding In Diatomic Molecules
Bonding In Some Homonuclear Diatomic Molecules
Homonuclear diatomic molecules are molecules composed of two atoms of the same element. The bonding in these molecules can be effectively explained using Molecular Orbital (MO) Theory. Let's examine the bonding in some common homonuclear diatomic molecules of the second period elements:
1. Lithium Molecule ($$Li_2$$):
- Electronic configuration of Li: $1s^2 2s^1$.
- Atomic orbitals involved: 1s and 2s orbitals from each Li atom.
- Molecular Orbital Configuration: $$(\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^0$$
- Bond Order (BO): $$(2+2 - 2+0)/2 = 4/2 = 2$$. Wait, lithium forms a single bond. Recheck.
Correction: Only valence electrons are usually considered for simple MO diagrams relevant to bonding in typical contexts, but a full MO diagram considers all electrons.
Full MO configuration for $$Li_2$$:
$$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^0 $$Electrons in BMOs ($$\sigma_{1s}, \sigma_{2s}$$): 2 + 2 = 4
Electrons in ABMOs ($$\sigma^{*}_{1s}$$): 2
Bond Order = $$(4 - 2) / 2 = 2 / 2 = 1$$.
Nature: $$Li_2$$ exists as a stable molecule with a single bond. It is diamagnetic because all electrons are paired.
2. Boron Molecule ($$B_2$$):
- Electronic configuration of B: $1s^2 2s^2 2p^1$.
- Atomic orbitals involved: 1s, 2s, and 2p orbitals from each B atom.
- MO Configuration (using the order for $$Li_2$$ to $$N_2$$): $$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\pi_{2p})^2 (\sigma_{2p})^0 $$
Electrons in BMOs ($$\sigma_{1s}, \sigma_{2s}, \pi_{2p}$$): 2 + 2 + 2 = 6
Electrons in ABMOs ($$\sigma^{*}_{1s}, \sigma^{*}_{2s}$$): 2 + 2 = 4
Bond Order = $$(6 - 4) / 2 = 2 / 2 = 1$$.
Nature: $$B_2$$ has a bond order of 1. Crucially, the two electrons in the $$(\pi_{2p})$$ MOs are unpaired (Hund's rule). Therefore, $$B_2$$ is paramagnetic, which is a key experimental observation explained by MO theory and not by simple Lewis structures or VBT.
3. Carbon Molecule ($$C_2$$):
- Electronic configuration of C: $1s^2 2s^2 2p^2$.
- MO Configuration: $$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^0 $$
Electrons in BMOs ($$\sigma_{1s}, \sigma_{2s}, \pi_{2p}$$): 2 + 2 + 4 = 8
Electrons in ABMOs ($$\sigma^{*}_{1s}, \sigma^{*}_{2s}$$): 2 + 2 = 4
Bond Order = $$(8 - 4) / 2 = 4 / 2 = 2$$.
Nature: $$C_2$$ has a bond order of 2. All electrons are paired, so it is diamagnetic. It's interesting to note that carbon primarily forms compounds with single, double, and triple bonds with other elements (like in organic chemistry), rather than forming $$C_2$$ molecules in bulk, but $$C_2$$ is observed in the gaseous state at very high temperatures.
4. Nitrogen Molecule ($$N_2$$):
- Electronic configuration of N: $1s^2 2s^2 2p^3$.
- MO Configuration (order for $$Li_2$$ to $$N_2$$): $$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\pi_{2p})^4 (\sigma_{2p})^2 $$
Electrons in BMOs ($$\sigma_{1s}, \sigma_{2s}, \pi_{2p}, \sigma_{2p}$$): 2 + 2 + 4 + 2 = 10
Electrons in ABMOs ($$\sigma^{*}_{1s}, \sigma^{*}_{2s}$$): 2 + 2 = 4
Bond Order = $$(10 - 4) / 2 = 6 / 2 = 3$$.
Nature: $$N_2$$ has a bond order of 3, corresponding to a strong triple bond (N≡N). All electrons are paired, making $$N_2$$ diamagnetic. Its high bond strength explains why nitrogen gas is relatively inert.
5. Oxygen Molecule ($$O_2$$):
- Electronic configuration of O: $1s^2 2s^2 2p^4$.
- MO Configuration (using the order for $$O_2$$ and $$F_2$$): $$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^{*}_{2p})^2 $$
Electrons in BMOs ($$\sigma_{1s}, \sigma_{2s}, \sigma_{2p}, \pi_{2p}$$): 2 + 2 + 2 + 4 = 10
Electrons in ABMOs ($$\sigma^{*}_{1s}, \sigma^{*}_{2s}, \pi^{*}_{2p}$$): 2 + 2 + 2 = 6
Bond Order = $$(10 - 6) / 2 = 4 / 2 = 2$$.
Nature: $$O_2$$ has a bond order of 2, corresponding to a double bond (O=O). Crucially, the two electrons in the $$(\pi^{*}_{2p})$$ antibonding orbitals are unpaired (Hund's rule). Therefore, $$O_2$$ is paramagnetic. This is a significant success of MO theory, as Lewis structures predict $$O_2$$ to be diamagnetic.
6. Fluorine Molecule ($$F_2$$):
- Electronic configuration of F: $1s^2 2s^2 2p^5$.
- MO Configuration (order for $$O_2$$ and $$F_2$$): $$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^{*}_{2p})^4 $$
Electrons in BMOs ($$\sigma_{1s}, \sigma_{2s}, \sigma_{2p}, \pi_{2p}$$): 2 + 2 + 2 + 4 = 10
Electrons in ABMOs ($$\sigma^{*}_{1s}, \sigma^{*}_{2s}, \pi^{*}_{2p}$$): 2 + 2 + 4 = 8
Bond Order = $$(10 - 8) / 2 = 2 / 2 = 1$$.
Nature: $$F_2$$ has a bond order of 1, corresponding to a single bond (F-F). All electrons are paired, making $$F_2$$ diamagnetic.
7. Neon Molecule ($$Ne_2$$):
- Electronic configuration of Ne: $1s^2 2s^2 2p^6$.
- MO Configuration: All atomic orbitals combine to form filled bonding and antibonding MOs. $$ (\sigma_{1s})^2 (\sigma^{*}_{1s})^2 (\sigma_{2s})^2 (\sigma^{*}_{2s})^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi^{*}_{2p})^4 (\sigma^{*}_{2p})^2 $$
Electrons in BMOs: 2+2+2+4+2 = 12
Electrons in ABMOs: 2+2+4+2 = 10
Bond Order = $$(12 - 10) / 2 = 2 / 2 = 1$$. Wait, Neon is a noble gas and does not form stable diatomic molecules.
Correction: The MO diagram for Ne₂ would have all orbitals filled up to the $2p$ level. The highest occupied molecular orbitals would be the bonding $\pi_{2p}$ and $\sigma_{2p}$ orbitals which are filled, and the lowest unoccupied molecular orbitals would be the antibonding $\pi^{*}_{2p}$ and $\sigma^{*}_{2p}$ orbitals. The bond order is indeed calculated to be 0.
Nature: The bond order of $$Ne_2$$ is 0. This indicates that a stable $$Ne_2$$ molecule does not form under normal conditions, which is consistent with Neon being a noble gas with a stable electron configuration.
MO theory provides a quantitative and qualitative understanding of the stability and properties of diatomic molecules by considering the delocalization of electrons in molecular orbitals.