hybridization of acetic acid

acetic acid | orbital hybridization

First draw the structure diagram for acetic acid, and for every non-hydrogen atom, count the σ-bonds. Note that double and triple bonds consist of one σ-bond together with one or two π-bonds: Identify those atoms with lone pairs: Find the steric number by adding the lone pair count to the number of σ-bonds: Consult the following chart to determine the hybridization from the steric number: steric number | hybridization 2 | sp 3 | sp^2 4 | sp^3 5 | dsp^3 6 | d^2sp^3 7 | d^3sp^3 Assign the provisional hybridization based on the table: Next identify any sp^3 atoms with lone pair electrons which can participate in a conjugated π-bond system. These atoms can lower their energy by placing a lone pair in a unhybridized p orbital to maximize overlap with the neighboring π-bonds. Note that halogens and elements from the third period and below do not engage in bond conjugation, except in the case of aromaticity: Adjust the provisional hybridizations to arrive at the result: Answer: | |

formula | CH_3CO_2H Hill formula | C_2H_4O_2 name | acetic acid alternate names | acetic acid, glacial | ethanoic acid | ethylic acid | glacial acetic acid | methanecarboxylic acid | vinegar acid mass fractions | C (carbon) 40% | H (hydrogen) 6.71% | O (oxygen) 53.3%