What is the nature of the bond in sodium dodecylbenzene sulfonate?

  • Thread starter Thread starter chemstudent123
  • Start date Start date
  • Tags Tags
    Bond Explain
AI Thread Summary
Sodium dodecylbenzene sulfonate features both ionic and covalent bonds. The sodium ion (Na+) forms an ionic bond with one of the oxygen atoms in the sulfonate group, which does not require a visible "stick" to indicate the bond. This bond is characterized by the polarity of the interaction rather than a physical attachment. The discussion clarifies that in a non-polar solvent, the Na+ ion remains bonded to the molecule through the sulfonate group. Understanding these bond types is crucial for accurately reporting on the compound's structure.
chemstudent123
Messages
13
Reaction score
0
I have to write a report on the bonds of the following element sodium doecylbenzene sulfonate. The problem is in the pic (see pic) the sodium doesn't have a bond to anything is this an ionic bond? Plz help I'm extremely confused:bugeye:
Ballnstick(fin wif double bonds).jpg
 
Physics news on Phys.org
Do you mean the compound sodium decylbenzene sulfonate or sodium dodecylbenzene sulfonate? Either way, you'll have one ionic bound like you said, but there are other covalent bonds that you can explain.
 
ok thanks so the sodium doesn't have an actaul physical bond to the rest of the structure (no stick attaching to the rest of the structure) because its an ionic bond right
 
chemstudent123 said:
ok thanks so the sodium doesn't have an actaul physical bond to the rest of the structure (no stick attaching to the rest of the structure) because its an ionic bond right
An ionic bond is no less of an actual physical bond than a covalent bond. In fact, the only difference between what are called covalent and ionic bonds is in the polarity of the bond.

In the absence of a polar solvent, the Na+ ion does not float about, detached from the rest of the molecule - it is bonded at one of the three oxygen (red) atoms of the sulfonate group. In fact, if you look carefully, the picture shows a white stick on one of the oxygen atoms, so that's the "stick" that attaches it to the rest of the structure. Of course, there is no reason (other than for the sake of representation) to specifically choose one of the O-atoms over the other two.
 
oh so that is an ionic bond, there is no need for a "stick" to be attached to the sodium
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top