# What Causes Atomic Orbitals to Hybridise?

I'm finding it difficult to visualise the process of hybridisation. Taking the example of formation of $\ce{CH4}$, can it viewed as the collapsing of the p orbitals of the Carbon atoms as they get "heavier" on excitation even before the introduction of Hydrogen atoms or is it the collision with Hydrogen atoms that causes the fusion?

• Hybridisation is purely a mathematical construct, the p orbitals do not actually collapse or anything like that. (To be honest the entire concept of an orbital is also a mathematical construct.) – orthocresol Jun 26 '15 at 11:23

Carbon has valence configuration: $2s^2 2p_x ^1 2p_y^1$. So, VB theory predicts that it can form two bonds. This is one of the main defect of VB theory. Then promotion comes to the rescue! One electron from $2s$ gets promoted to unoccupied $2p_z$ orbital, so as to make the configuration: $2s^12p_x^12p_y^12p_z^1$ . So, $\ce{C}$ can form four bonds. But still there is a problem: experiments show that the four bonds in methane are exactly equivalent which is not possible by this configuration since $s$ & $p$ orbitals are not same. To deal with the problem, concept of hybridization was developed. Hybridization gives rise to four equivalent orbitals but having different orientations (four corners of the tetrahedron.) The non-normalized wavefunctions of four orbitals are as follows: $$\psi_1 = \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_2 =\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\\\\\ \psi_3=\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_4 = \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}$$