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QUESTION

# Hybridization of "HCl"?

"HCl" has no orbital hybridization. Chlorine's 3s is too low in energy to interact with hydrogen's 1s, but chlorine's 3p_z can interact with hydrogen's 1s atomic orbital just fine.

A good general rule is that being less than about 12 eV apart in energy is required for orbitals to be close enough in energy.

The 3s and 3p orbitals of "Cl" are apparently too far apart in energy to interact for orbital hybridization.

Based on these atomic orbital energies (Inorganic Chemistry, Miessler et al., Table 5.2):

"E"_(1s)^"H" = -"13.61 eV"

"E"_(3s)^"Cl" = -"25.23 eV"

"E"_(3p)^"Cl" = -"13.67 eV"

... here is the MO diagram for "HCl":

• 1b_1 and 1b_2 are nonbonding because the 3p_x and 3p_y atomic orbitals of chlorine weren't compatible with hydrogen's 1s, and the 1a_1 is nonbonding because the 3s of chlorine is too low in energy to interact with hydrogen's 1s.
• 2a_1 and 3a_1 are the sigma_z and sigma_z^"*" bonding and antibonding MOs, respectively. They form because the 1s of hydrogen is compatible and close enough in energy to the 3p_z of chlorine.

It would contradict the Lewis structure to say that there is orbital hybridization in "HCl". Here's why:

1. Nonbonding orbitals arise when two orbitals are NOT compatible and/or NOT close enough in energy.
2. Although the 3s and 3p atomic orbitals are less than "12 eV" apart, if they were to be able to interact, then given the closeness in energy to the 1s atomic orbital of "H", that 1s of "H" would be able to interact with the 3s of "Cl" too.
3. The 3s and 1s are compatible (just know that they are compatible; the why is outside the scope of your level of education).
4. So, if they were close enough in energy, they would interact, and then there would not be one nonbonding orbital from the 3s to hold one of the three lone pairs of electrons on "Cl", when we know that lone pair should be accounted for by the Lewis dot structure.

Therefore, there is no orbital hybridization in "HCl".