The very first picture reflects both sulfurs v an increased octet.
The problem is the in the top half of the below picture, us are supposed to finish up v sulfuric acid, but we actually finish up v sulfuric acid plus 2 lone bag (if I attracted the system correctly). Go I? (I actually separated the end the top component of the snapshot into two components - first, hydrogen attacking oxygen; second, oxygen attack sulfur - for this reason the arrow pushing wouldn"t acquire confusing).
The 2nd picture adheres to updated chemistry fact. Follow to Wikipedia, sulfur trioxide has one double bond and two dative solitary bonds. Sulfuric acid is likewise a fee separated species with four single bonds, not two solitary bonds and two double bonds. Ns seem to end up with the proper product here without any type of hand-waving.
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edited might 8 "17 at 4:40
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I"ve attracted a an ext correct device for the reaction that dilute $\ceSO3$ through water in the fluid phase:
$$\ceSO3 (aq) + 3H2O (l) -> SO4^2- (aq) + 2H3O+ (aq)$$
$\ceSO3$ is a strong electrophile, sufficient to react easily with water, i beg your pardon is a reasonably weak nucleophile. A water molecule is added to the structure, promoted by the dislocation the a $\ceS=O$ bonding electron pair in $\ceSO3$ come the oxygen atom, forming the intermediate structure shown. Extr water molecules then quickly abstract hydrogen atoms from the intermediate (actually, the very first abstraction is far simpler than the second, therefore it can have been more proper to display each proton abstraction independently in sequence). The last product of the reaction is sulfate anions and also hydronium cations ($\ceH2SO4$ is a strong acid in water, so the doesn"t really exist together a molecule. The 2nd proton isn"t almost as acidic as the first, yet in dilute equipment both are ionized). The formal an international reaction is shown listed below the mechanism. Notification that in ~ every suggest the formal charge on the main sulfur atom is specifically the same, zero.
Now, with regards to our small discussion around the actual framework of $\ceSO3$. Timeless valence bond theory, amplified with the ide of hypervalency and also VSEPR theory, characteristics the framework shown, with three twin bonds. The Wikipedia article for $\ceSO3$ mentions another structure, based on modern-day valence bond theory computational calculations, through one double and two single bonds. This framework is quite amazing as it does no invoke hypervalency. Indeed, hypervalency is assumed to be an old concept, and not really vital in more modern theoretical treatments of chemical bonding.
However, if one opts to screen the structure according to modern-day valence theory, they would do well to make it explicit. Virtually all structure formulae this day are attracted in the paper definition of classic valence bond theory. Details assumptions native the timeless theory perform not carry over come the modern-day theory. For example, the writer of the post "Modern valence link theory", Chem. Soc. Rev., 1997,26, 87-100 (free access mirror) uncover that the bonding in $\ceN2O$, defined classically as a mixture that the canonical develops $\ce^-N= N^+=O$ and also $\ceN≡N^+-O^-$, is ideal described together $\ceN≡N=O$. Even with five bonds, the main nitrogen atom has actually a calculation occupancy close to 7, i.e. That is no hypervalent. Because the structural building rules of timeless and modern-day valence bond concept aren"t the same, the structures created from both execute not mesh and cannot be compared directly. The easiest thing to do, then, is to attract $\ceSO3$ according to classic valence bond theory, v three dual bonds.