Determine the hybridization of the atoms labeled A, B and C in the molecule and the bond angles
3. The Attempt at a Solution
A. Normally, I would say it is sp3 for oxygen since there are 4 groups attached to it including the lone pairs. However because it is attached to an sp2 carbon; the lone pair could de-localize and the charge spread. Thus I would think the O's hybridization is actually sp2. Also, I would say the bond angles are less than 109.5 (I don't know if there is an actual or absolute value) because a tetrahedral arrangement has this bond angle but the lone pairs push the bond angles closer together than a tetrahedral bond angle?
B. Since it is bonded to 3 groups including it's lone pair, and has a double bond, I would say sp2? Bond angle less than 120 degrees since it is bent but greater than 109.5 because it's not tetrahedral?
C. Sp3 since the N is attached to 4 groups including it's lone electrons and is not attached to an sp2 carbon which could de-localize any charges. Bond angles would be 109.5 because it is in a tetrahedral arrangement?
Thanks in advance for the help.