Why are the pi bonding molecular orbitals ( $π_{2 p_{x}}$ $pi_(2p_x)$ , $π_{2 p_{y}}$ $pi_(2p_y)$ ) lower in energy than sigma bonding molecular orbital ( $σ_{2 p_{z}}$ $sigma_(2p_z)$ ) in the MO diagram of the $N_{2}$ $"N"_2$ molecule?

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Why are the pi bonding molecular orbitals ( $π_{2 p_{x}}$ $pi_(2p_x)$ , $π_{2 p_{y}}$ $pi_(2p_y)$ ) lower in energy than sigma bonding molecular orbital ( $σ_{2 p_{z}}$ $sigma_(2p_z)$ ) in the MO diagram of the $N_{2}$ $"N"_2$ molecule?

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Answer 1 · 2016-02-11T07:38:21

Because from right to left, the orbital mixing interaction between two compatible orbitals of similar energies decreases the energy of the lower-energy orbital (relative to what it would be without mixing) and increases the energy of the higher-energy orbital (relative to what it would be without mixing).

This effect is still barely present on $N_{2}$ $N_2$ , so its $σ$ $sigma$ bonding MO is still higher in energy than its $π$ $pi$ bonding MOs.

In $N_{2}$ $"N"_2$ , the $σ_{g (2 s)}$ $sigma_g(2s)$ and $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ molecular orbitals are compatible (they are both symmetric with respect to infinite rotation and inversion), AND they are close enough in energy (generally within $\pm 12 eV$ $pm "12 eV"$ ), so they mix.

Not surprisingly, this effect is called orbital mixing, and the result of it enhances the bonding with additional electron stabilization. This is often shown for the second-period elements.

Inorganic Chemistry, Miessler et al., Ch. 5.2.3

From ${Li}_{2}$ $"Li"_2$ to $N_{2}$ $"N"_2$ , we would actually see a trend of the $σ_{g (2 s)}$ $sigma_g(2s)$ decreasing in energy (faster than) the $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ decreases in energy; the effect of orbital mixing decreases as the $σ_{g (2 s)}$ $sigma_g(2s)$ and $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ get farther and farther apart in energy.

Therefore, the effects of the mixing become less and less significant from left to right.

Nitrogen is the last element in the second period for which the diminishing effects of the orbital mixing are still significant enough for the energies of the $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ to still be higher than the $π_{u (2 p_{x})}$ $pi_(u(2p_x))$ and $π_{u (2 p_{y})}$ $pi_(u(2p_y))$ in energy.

From $O_{2}$ $"O"_2$ through ${Ne}_{2}$ $"Ne"_2$ , the $σ_{g (2 s)}$ $sigma_g(2s)$ and $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ are too far apart in energy to interact, so the orbital mixing effects are no longer as significant.

At that point, the $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ has crossed the $π_{u (2 p_{x})}$ $pi_(u(2p_x))$ and $π_{u (2 p_{y})}$ $pi_(u(2p_y))$ orbitals in energy, so the orbital ordering "switches" and these diatomics have a $σ_{g (2 p_{z})}$ $sigma_g(2p_z)$ MO lower in energy than the $π_{u (2 p_{x})}$ $pi_(u(2p_x))$ and $π_{u (2 p_{y})}$ $pi_(u(2p_y))$ MOs.

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Why are the pi bonding molecular orbitals ( $π_{2 p_{x}}$ $pi_(2p_x)$ , $π_{2 p_{y}}$ $pi_(2p_y)$ ) lower in energy than sigma bonding molecular orbital ( $σ_{2 p_{z}}$ $sigma_(2p_z)$ ) in the MO diagram of the $N_{2}$ $"N"_2$ molecule?

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Why are the pi bonding molecular orbitals (π2pxpi_(2p_x), π2pypi_(2p_y)) lower in energy than sigma bonding molecular orbital (σ2pzsigma_(2p_z)) in the MO diagram of the N2"N"_2 molecule?

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Why are the pi bonding molecular orbitals ( $π_{2 p_{x}}$ $pi_(2p_x)$ , $π_{2 p_{y}}$ $pi_(2p_y)$ ) lower in energy than sigma bonding molecular orbital ( $σ_{2 p_{z}}$ $sigma_(2p_z)$ ) in the MO diagram of the $N_{2}$ $"N"_2$ molecule?