what's the importance of "s" character in hybrid orbitals?

I am a high school student and I am learning about hybridization, my teacher told me that it is due to the "s" character in the hybrid orbital that they can arrange themselves so that the molecule faces minimum repulsion {lone pair-lone pair, lone pair-bond pair and bond pair-bond pair} but I am a little confused as "s" orbital is uniformly distributed throughout the space in a spherical form so then even if the hybridization is say "sp2" involving mixing of "pz","px" and "s" orbital then why those three orbitals can arrange in x-z plane only? I know they have "pz" and "px" character but still due to s character can't it do so? or is it a rule that hybrid orbitals can show its s character by only arranging in planes of p or d orbitals involved? as I am learning it at my level it seems that its only qualitative talk but does it have any mathematical explanation also proving that it can't arrange in x-y or y-z plane? or I have to just memorize it?




Your questions can be divided into the following two parts:


1. Since "$\ce {s}$" orbital is uniformly distributed throughout the space in a spherical form, how come your teacher told you that in the $\ce{sp^2}$ hybridization, it is due to the "$\ce {s}$" character in the hybrid orbitals that they can arrange themselves so that the molecule faces minimum repulsion {lone pair-lone pair, lone pair-bond pair and bond pair-bond pair}?
2. Why do the three $\ce{sp^2}$ hybridized orbitals involving mixing of "$\ce {p\_z}$","$\ce {p\_x}$", and "$\ce {s}$" orbitals arrange in the $xz$-plane only? Since the "$\ce {s}$" orbital with a spherical form points in all directions, why can't the $\ce{sp^2}$ hybridized orbitals point in other directions, e.g., in the $xy$-plane or the $yz$-plane? Are there any mathematical explanations for this?


What you are asking is basically that, since "$\ce {s}$" orbital is spherical, how can it dictate orientation? And if it can dictate orientation, why can't it cause the orientations to be out of the ones of the components?


Let's start by watching a video on interpreting $\ce{sp^2}$ hybridization based on Quantum Chemistry taken from the YouTube: [Quantum Chemistry $\ce{sp^2}$ hybridization](https://www.youtube.com/watch?v=YY9vZUdNQt8&t=170s). To present a self-contained answer, we need to do a little mathematical homework to arrive at the expressions for the three $\ce{sp^2}$ hybridized orbitals shown in the video.


To that end, let's express the $\ce{sp^2}$ hybridized orbitals as linear combinations of the three original atomic orbitals as:
$$\psi\_{sp^2\_1}=a\_1\psi\_{2s}+b\_1\psi\_{2p\_z}+c\_1\psi\_{2p\_x}$$
$$\psi\_{sp^2\_2}=a\_2\psi\_{2s}+b\_2\psi\_{2p\_z}+c\_2\psi\_{2p\_x}$$
$$\psi\_{sp^2\_3}=a\_3\psi\_{2s}+b\_3\psi\_{2p\_z}+c\_3\psi\_{2p\_x}$$
where the three atomic orbitals are the three basis functions. The coefficients in the above-given equations can be obtained by using the normalization and orthogonalization of each atomic orbital and hybridized orbital, and the property that the "$\ce {s}$" orbital contributes $\frac{1}{3}$ and "$\ce {p}$" orbitals contribute $\frac{2}{3}$ to the integrated probability density. This is trivial and can be done on back of an envelope so I will just leave it to interested students.


Note that the solutions are not unique and depend on the coordinate system chosen. However, transformation between different coordinate systems does not change the electron densities of the quantum mechanical system. For convenience and the purpose of illustration, we can place the three atomic orbitals in the way as shown in the figure given below (left panel). As you pointed out, the "$\ce {s}$" orbital is spherical. In addition, the "$\ce {p\_z}$" and "$\ce {p\_x}$" orbitals point along the $z$ and $x$ Cartesian axes, respectively. Then you can understand that the "$\ce {p\_y}$" orbital (not shown) should point along the $y$ Cartesian axis, which by the right-hand rule should point out of the screen. By aligning the atomic orbitals that way, we will obtain the solutions as shown in the video.


Putting the obtained coefficients back to the above-given equations, the three hybridized orbitals are expressed as linear combinations of the three original atomic orbitals in the following form:
$$\psi\_{sp^2\_1}=\sqrt{\frac{2}{6}}\psi\_{2s}+\sqrt{\frac{4}{6}}\psi\_{2p\_z}+\sqrt{\frac{0}{6}}\psi\_{2p\_x}$$
$$\psi\_{sp^2\_2}=\sqrt{\frac{2}{6}}\psi\_{2s}-\sqrt{\frac{1}{6}}\psi\_{2p\_z}+\sqrt{\frac{3}{6}}\psi\_{2p\_x}$$
$$\psi\_{sp^2\_3}=\sqrt{\frac{2}{6}}\psi\_{2s}-\sqrt{\frac{1}{6}}\psi\_{2p\_z}-\sqrt{\frac{3}{6}}\psi\_{2p\_x}$$
As can be seen from the figure shown below (right panel), the three hybridized orbitals point to right, upper left, and lower left, respectively, with their corresponding labels shown. So you can see the key point here: **the atomic orbitals are base vectors pointing along certain directions as defined by the coordinate system, and hybridized orbitals are also vectors with orientations determined by the coefficients and the base vectors in the linear combinations**.


For the first part of your question, it would be better to say that it is due to the "participation" of the "$\ce {s}$" orbital in the hybridization, the hybridized orbitals can arrange themselves in the way they are, i.e., with certain directions and 120 degree apart. The "$\ce {s}$" orbital is spherical so it does not have any preferred direction. You can say that it does not dictate the orientation of the hybridized orbitals, but its participation in the hybridization causes the mixing of the three original basis functions and then the "cards" need to re-shuffled, so to speak, to give new orbitals with preferred directions.


For the second part of your question, as can be seen from the mathematical expressions of the three hybridized orbitals given above, there is no basis function coming from the "$\ce {p\_y}$" orbital. This is the reason that the three hybridized orbitals do not point in the $xy$-plane or the $yz$-plane that involves the $y$ axis. For the same reason, the two $\ce{sp}$ hybridized orbitals only remain on the axis of the "$\ce {p}$" orbital, depending on whether it is "$\ce {p\_x}$", "$\ce {p\_y}$", or "$\ce {p\_z}$" orbital. In contrast, the four $\ce{sp^3}$ hybridized orbitals will point in four different directions forming tetrahedral geometry. The difference between these different hybridizations can be understood as 1D, 2D, and 3D.


Here is a little bit more information regarding the 120 degree. Since you are a high school student, you can use geometry and algebra to prove that's the case by using the coefficients shown above. Basically, $\alpha = arctan(\frac{\sqrt{\frac{1}{6}}}{\sqrt{\frac{3}{6}}})=arctan(\sqrt{\frac{1}{3}})=30$ degree. Likewise, $\beta$ is also $30$ degree. Then it can be readily shown that the three hybridized orbitals are 120 degree apart. One more thing, not for now but when you enter college if you are still interested, you can practice to prove that each of these three hybridized orbitals is indeed normalized and they are orthogonal to each other.


[![enter image description here](https://i.stack.imgur.com/EICwX.png)](https://i.stack.imgur.com/EICwX.png)


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