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What happens to the lobe of the p-orbital during sp² hybridization?

During the formation of $\ce{C2H4}$ molecule:

We know that both the carbon atoms will have a total of 3 p-orbitals and they will be involved in hybridization.

Let us number the carbon atoms as $\ce{C}1$ and $\ce{C}2$.

Now the p-orbital of $\ce{C}1$ along $x$-axis will form a $\unicode[Times]{x3C3}$ bond with the p-orbital of $\ce{C}2$ and a $\unicode[Times]{x3C3}$ bond with the s-orbital of hydrogen.

And similarly the p-orbital of $\ce{C}2$ along x-axis will form a $\unicode[Times]{x3C3}$ bond with hydrogen.

So both the lobes of the p-orbital of $\ce{C}1$ and $\ce{C}2$ along $x$-axis are involved in sigma bond formation.

Lets say the p-orbital of $\ce{C}1$ and $\ce{C}2$ along the $y$-axis forms a $\unicode[Times]{x3C0}$ bond.

So here also both the lobes of p-orbital of $\ce{C}1$ and $\ce{C}2$ are involved in hybridization.

Let say that the p-orbital of $\ce{C}1$ which is along the $z$-axis forms a $\unicode[Times]{x3C3}$ bond with the s-orbital of hydrogen.

And similarly the p-orbital of $\ce{C}2$ along $z$-axis forms a $\unicode[Times]{x3C3}$ bond with the s-orbital of the other hydrogen.

Now my confusion is that here two p-orbitals along the $x$ and $y$-axes have four lobes and all of them are involved in bond formation.

But the p-orbital which is along $z$-axis has only one of its lobes involved in bonding. What happens to the other lobe of this p-orbital?

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You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.

While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only one lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital will not.

This link provides a nice visualisation of the different orbitals involved in ethene bonding.

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