THEORY FOR CH2 IS SAME AS CH3 SO I HAVE EXPLAINED IT IN SHORT IF U STILL HAVE DOUBT COMMENT BELOW.
• Generate group orbitals for CH2 group.
• Compare linear vs. bent orientation of the two
H atoms relative to C.
• Linear configuration
• Angled/bent configuration
• Expect secondary mixing between C and E
producing C’ and E’.
• σ(CH2) w/ σ symmetry,
• π(CH2) (a degenerate pair of orbitals) w/ π
symmetry.
• σout(CH2) w/ σ symmetry, pointed away from the
hydrogens.
• As with CH3 = MH3... there is an MH2.
How do shifts in positioning of H atoms impact the energetics of the group orbitals?
• Water M = O, brings 6 valence σ(CH3) w/ each H atom contributing 1 ē. (total of 8 valence ē) . Fill
starting with lowest energy group orbital, σ(CH2).
• Water prefers bent geometry because of energy gains from lowering the energy of C/C’, which is
occupied. Oxygen lone pairs of ē are not equivalent and are best thought of as being in C’ and D
MO’s.
• Generate group orbitals for CH2 group.
• Compare linear vs. bent orientation of the two
H atoms relative to C.
• Linear configuration
• Angled/bent configuration
• Expect secondary mixing between C and E
producing C’ and E’.
• σ(CH2) w/ σ symmetry,
• π(CH2) (a degenerate pair of orbitals) w/ π
symmetry.
• σout(CH2) w/ σ symmetry, pointed away from the
hydrogens.
• As with CH3 = MH3... there is an MH2.
How do shifts in positioning of H atoms impact the energetics of the group orbitals?
• Water M = O, brings 6 valence σ(CH3) w/ each H atom contributing 1 ē. (total of 8 valence ē) . Fill
starting with lowest energy group orbital, σ(CH2).
• Water prefers bent geometry because of energy gains from lowering the energy of C/C’, which is
occupied. Oxygen lone pairs of ē are not equivalent and are best thought of as being in C’ and D
MO’s.
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