2 Improved Formatting, bolding mathematics with markdown has no effect, used mhchem for chemistry elements
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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.

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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.