2 Improved Formatting, bolding mathematics with markdown has no effect, used mhchem for chemistry elements edited Mar 21 '18 at 3:53 Martin - マーチン♦ 34.9k99 gold badges118118 silver badges245245 bronze badges I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $$\sigma^{}_{C-H}$$ MO's$$\sigma_{\ce{C-H}}$$ MOs and a $$\sigma^{}_{C-C}$$$$\sigma_{\ce{C-C}}$$ MO, lower in energy. My best guess would be the position of the $$\sigma^{}_{C-H}$$$$\sigma_{\ce{C-H}}$$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $$\sigma^{}_{C-H}$$$$\sigma_{\ce{C-H}}$$ of one C-H$$\ce{C-H}$$ bond and the $$\sigma^{*}_{C-H}$$$$\sigma^{*}_{\ce{C-H}}$$ of C-H$$\ce{C-H}$$ bond anti to it - you don't really need to be worried about the $$\sigma^{}_{C-H}$$$$\sigma_{\ce{C-H}}$$ of that anti bond because it cannot overlap with the other $$\sigma^{}_{C-H}$$$$\sigma^{}_{\ce{C-H}}$$. That brings us to the second part of your query. A $$\sigma^{}_{C-H}$$$$\sigma_{\ce{C-H}}$$ can donate electron density to an empty $$\sigma^{*}_{C-H}$$$$\sigma^{*}_{\ce{C-H}}$$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference. I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $$\sigma^{}_{C-H}$$ MO's and a $$\sigma^{}_{C-C}$$ MO, lower in energy. My best guess would be the position of the $$\sigma^{}_{C-H}$$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $$\sigma^{}_{C-H}$$ of one C-H bond and the $$\sigma^{*}_{C-H}$$ of C-H bond anti to it - you don't really need to worried about the $$\sigma^{}_{C-H}$$ of that anti bond because it cannot overlap with the other $$\sigma^{}_{C-H}$$. That brings us to the second part of your query. A $$\sigma^{}_{C-H}$$ can donate electron density to an empty $$\sigma^{*}_{C-H}$$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference. I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $$\sigma_{\ce{C-H}}$$ MOs and a $$\sigma_{\ce{C-C}}$$ MO, lower in energy. My best guess would be the position of the $$\sigma_{\ce{C-H}}$$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $$\sigma_{\ce{C-H}}$$ of one $$\ce{C-H}$$ bond and the $$\sigma^{*}_{\ce{C-H}}$$ of $$\ce{C-H}$$ bond anti to it - you don't really need to be worried about the $$\sigma_{\ce{C-H}}$$ of that anti bond because it cannot overlap with the other $$\sigma^{}_{\ce{C-H}}$$. That brings us to the second part of your query. A $$\sigma_{\ce{C-H}}$$ can donate electron density to an empty $$\sigma^{*}_{\ce{C-H}}$$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference. 1 answered Mar 18 '18 at 7:04 Sagnik 18188 bronze badges I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $$\sigma^{}_{C-H}$$ MO's and a $$\sigma^{}_{C-C}$$ MO, lower in energy. My best guess would be the position of the $$\sigma^{}_{C-H}$$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $$\sigma^{}_{C-H}$$ of one C-H bond and the $$\sigma^{*}_{C-H}$$ of C-H bond anti to it - you don't really need to worried about the $$\sigma^{}_{C-H}$$ of that anti bond because it cannot overlap with the other $$\sigma^{}_{C-H}$$. That brings us to the second part of your query. A $$\sigma^{}_{C-H}$$ can donate electron density to an empty $$\sigma^{*}_{C-H}$$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference.