hybridization of miconazole

miconazole | orbital hybridization

First draw the structure diagram for miconazole, 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 | C_18H_14Cl_4N_2O name | miconazole IUPAC name | 1-[2-(2, 4-dichlorophenyl)-2-[(2, 4-dichlorophenyl)methoxy]ethyl]imidazole alternate names | daktarin iV | minostate | monistat mass fractions | C (carbon) 52% | Cl (chlorine) 34.1% | H (hydrogen) 3.39% | N (nitrogen) 6.73% | O (oxygen) 3.84%