I was told that that in phenoxide ion, benzene ring is very strongly activated. The order my teacher told me was phenol < amine < 2 degree amine < 3 degree amine < phenoxide (for activation of benzene)
Example: Reimer–Tiemann reaction happens only in sodium phenoxide
So, what is so special in phenoxide that makes it so much activating? Also, why does oxygen contribute more in resonance than nitrogen, when it is more electronegative than the latter?
I assume by your notes that your teacher was mentioning electrophilic aromatic substitution reactions. These are two-step reactions. The product-determining step (whether it is ortho-, meta- or para-) in the substitution mechanism is the first step, which is also the slow (high activation energy), and hence the rate determining step (see [energy Diagram] below). The second step is relatively fast, because positively charged intermediate ion loose a proton to get back the resonance stabilized aromatic ring again. Accordingly, the rate of the reaction is roughly depend on the influence of each of the given substituents to stabilize the benzenonium intermediate (positively charged adduct). You can best see it by looking at the capability of each substituent to delocalize the positive charge within the benzenonium intermediate.
Each substituent given except for the phenolate ion has a lone pair. Thus, each can share it with the benzenonium nucleus to reduce the positive charge, reducing the energy of intermediate. However, phenolate ion ($\ce{ph-O^-}$) has full negative charge, which can neutralize benzenonium intermediate completely, to give least enery among all substituents considered here.
