What is the hybridization in #"HCl"#?
1 Answer
A good general rule is that being less than about
The
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
#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
- Nonbonding orbitals arise when two orbitals are NOT compatible and/or NOT close enough in energy.
- 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. - The
#3s# and#1s# are compatible (just know that they are compatible; the why is outside the scope of your level of education). - 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