15 $\ce {SO3}$ molecule has three double bonded oxygen to the central sulfur atom. Sulfur has $\ce {sp^2}$ hybridization and it has 6 outer electrons which make the bonds with the oxygen. So shouldn't the bond order be 2?
For one thing, $\ce {SO3^ {2-}}$ has two extra electrons, so it's an ion. $\ce {SO3}$ is a neutral molecule. There are a different number of electrons, so they're just different.
Structure of $\ce{SO3}$ (sulfur trioxide): In the molecule, if each oxygen atom shares two electrons with sulfur atom then how does the sulfur atom remain stable? It already has 6 valence electrons...
I've drawn a more correct mechanism for the reaction of dilute $\ce {SO3}$ with water in the liquid phase: $$\ce {SO3 (aq) + 3H2O (l) -> SO4^2- (aq) + 2H3O+ (aq)}$$ $\ce {SO3}$ is a strong electrophile, enough to react quickly with water, which is a relatively weak nucleophile. A water molecule is added to the structure, facilitated by the dislocation of a $\ce {S=O}$ bonding electron pair in ...
Is it $\ce {SO3}$ or $\ce {SO3H+}$? According to what my professor taught me, both $\ce {SO3}$ and $\ce {SO3H+}$ may be the electrophile involved - and mechanisms with both species can be successfully written. I'm looking for some source (s) that confirms the existence of both electrophiles in sulfonation reaction of benzene.
In this case, if I want to decrease the amount of $\ce {SO3}$, I need to shift the equilibrium to the left, meaning that I need to decrease the concentration of $\ce {SO3}$. Based on my understanding of Le Chatelier's principle, changing the temperature will have the opposite effect on the equilibrium because the reaction is exothermic.