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.
Nature: Methanesulfonic acid (MSA) and SO3 formation from the addition channel of atmospheric dimethyl sulfide oxidation
Methanesulfonic acid (MSA) and SO3 formation from the addition channel of atmospheric dimethyl sulfide oxidation
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 ...
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?
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...
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.
Here, in $\ce {SO3}$, we have no lone pairs on the sulfur atom whereas, in $\ce {SO2}$, the central atom has a pair of lone electrons. According to Bent's rule, the lone pairs tend to occupy the orbital with more s-character which in turn decreases the s-character in the $\ce {S-O}$ bonds and increases the p-character.