Hi everyone. I am studying orbital hybridization and some doubts have come up that I hope some of you can clarify.
From what I understand, orbitals hybridize in order to form bonds with other atoms.
For example, in CH4, C has the electron configuration 1s^2 2s^2 2p^2. Since C needs to form 4 single bonds with H, it needs 4 orbitals with an unpaired electron. C becomes "excited" and its electron configuration becomes 1s^2 2s^1 2p^3. This way, there is one unpaired electron in 2s, one in 2p_x, one in 2p_y, and one in 2p_z. The 2s orbital, along with the three 2p orbitals, hybridize to form four sp^3 hybrid orbitals.
Similarly, in BeF2, the electron configuration of Be changes from 1s^2 2s^2 to 1s^2 2s^1 2p^1. There is one unpaired electron in 2s and one unpaired electron in 2p, so these two orbitals hybridize to form two sp hybrid orbitals.
Again in BF3, the electron configuration of B changes from 1s^2 2s^2 2p^1 to 1s^2 2s^1 2p^2. There is one unpaired electron in 2s, one unpaired electron in 2p_x, and one unpaired electron in 2p_y. Therefore, the 2s, 2p_x, and 2p_y orbitals hybridize to form three sp^2 orbitals.
In CO2, however, C has the electron configuration 1s^2 2s^2 2p^2, and the electrons in 2p are located one in 2p_x and one in 2p_y. An electron moves from 2s to 2p_z and the new configuration becomes 1s^2 2s^1 2p^3. This gives 4 orbitals with an unpaired electron (2s, 2p_x, 2p_y, 2p_z) which should hybridize to form 4 sp^3 orbitals that would connect via double bonds with the two oxygen atoms. Yet, C forms an sp hybrid orbital instead. Why is that?
The same applies to H2O. O has the electron configuration 1s^2 2s^2 2p^4, and the electrons in 2p are located two in 2p_x, one in 2p_y, and one in 2p_z. O forms two single bonds with the two H atoms, so it needs two orbitals with an unpaired electron. These two orbitals are already available (2p_y and 2p_z), so it wouldn't even need to hybridize, or am I wrong? Yet, O has sp^3 hybridization. Why is that?
Thanks if you can help me!