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Assume the x-axis to be the internuclear axis. Hence, both $1s-1s$ and $2p_x-2p_x$ will form $\sigma$ bonds.

Also consider the fact that $p$ orbitals have more directional characteristics.

Given the above facts, shouldn't the $2p_x-2p_x$ overlapping be more effective than $1s-1s$ overlapping? Why is $1s-1s$ orbital overlapping stronger than $2p_x-2p_x$ orbital overlapping?

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  • $\begingroup$ Who says this is true in the first place? $\endgroup$ – Ivan Neretin Apr 4 at 11:21
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    $\begingroup$ @IvanNeretin an exercise in my book, Resonance Inorganic Chemistry for IIT-JEE page #127 $\endgroup$ – user76377 Apr 4 at 11:24
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Your question is hinged on a false premise. On a deeper level, the premise makes no sense at all and can't even be classified as true or false. If a book says otherwise, then maybe it contains some very specific piece of context that would turn the situation upside down and make everything fall in place. Or maybe the book just says nonsense; such things happen, after all.

To begin with, orbital overlap depends on the internuclear distance more than slightly. Now, $1s$ orbitals are a good deal smaller than $2p$ (not like "you vs Dwayne Johnson", but more like "you vs an elephant"). Naturally, at a distance where $2p_x$ orbitals are in their best overlap, the $1s$ orbitals barely feel each other. Likewise, if the distance is good for $1s$, it is way too short for $2p_x$ and makes them feel pretty uncomfortable. So you see that considering one of these overlaps to be universally stronger than the other is plain untrue.

There is more to it, however. See, the $1s$ and $2p$ orbitals of the same atom are pretty far apart in energy. They never mix. In other words, they can never be important at the same time.

  • If both your atoms are $\rm H$, then their $1s$ orbitals overlap and make a good strong covalent bond, while the $2p$ orbitals are empty and hence effectively don't exist.
  • On the other hand, if both your atoms are (say) $\rm O$, then their $2p_x$ participate in bonding (together with other $2p$'s and $2s$), while the $1s$ orbitals are pretty much kept apart. What's more important, both $1s$'s are fully occupied, hence whatever little interaction they have will end up producing a bonding and an antibonding orbital which are also both occupied, thus negating the effect.

At this point you might be wondering what if one of your atoms is $\rm H$ and another $\rm O$, but that's a story for another day.

So it goes.

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