Because the lone pairs of the oxygen are delocalized and participate in resonance they are BOTH sp^2 not one being sp^2 and the other sp^3. Resonance doesn’t change the configuration of the molecule but only shows the movement of electrons within a molecule. But thank you for the video it really helped a lot 🤗.
I see that Kent's answer can possibly be a solution, but if we take in consideration of the expression pi bond = (6n+2-#)/2, where n is the number of atoms (excluding any hydrogens) and # is the amount of available electrons then it will amount to (6x3+(2-20))/2=0. This means there should not be an extra bond. So, O-Cl-O with single bond would be satisfy. This is why some of you get confused. To check that this is correct, you can balance the formal charge O= -1, -1 and Cl= +1 and get a -1. Hopefully, this helps.
If you were to draw an electron level diagram for the chlorine atom here to indicate hybridization, what would it look like. I'm just stumped because to my knowledge (gr12) i know you can only form pi bonds with p orbitals, and if the 3 p orbitals are hybridized into sp3 orbitals, how can there be pi bonds in the double bond?
kentchemistry.com I don't understand one thing, when you gave the double bond to the left oxygen, the right oxygen still had a formal charge of -1, so does that mean that the '"right" Lewis structure for the Chlorite ion is the resonance hybrid version?
I was debating this structure in a study session and I agree with your final product. One of my buddies argued that there should be a double bond for each oxygen so that FC = 0, leaving CL with FC= -1. Although the FC in each structure is even, I thought that because O is more electronegative, it would be more probable to find ClO2- in the form you explained (with more lone pairs). Is my reason the correct reason for said structure?
No he is absolutely right.. In order to get the most correct Lewis structure formal charge must equal the charge of the ion and thus it is an exception to the octet rule. Give your "Chemistry instructor" the finger for me
You are right about the expanded octet rule on period three, but isn't the hybridization of Cl Sp2d? Because it is in the third period? I may be wrong, but i thought that is how that worked. If not can you explain why?
Athough his answer makes sense according to the formal charge rule, however, since O is more electronegative, it won't be willing enough to share an extra pair of electrons with Cl by forming a double bond. Anyways, people who are familiar with Its tetrahedral symmetry (sp3 hybridization), you'll notice that this structure won't be possible if there was a double bond.
For the formal charges of polyatomics, shouldn't they equal the charge of the polyatomic ion itself? Like why wouldn't the first one work? Because both Oxygen's have charges of -1 and the Chlorine has a +1 charge, so the net is -1, which is the charge of the actual ion, ClO2- Can someone explain please?
Period 3 elements(Starting from Na, Mg...Cl,Ar) can have more than 8 elections. Similar to that Be,B and H can have less than 8 electrons. These are the exceptions for Lewis dot structure, so he is doing it right.
@adrenalinerushr certain periodic table elements are capable of "bullying" elements into taking more than an octet. O and F are the most common. This is correct.
In reality a resonance hybrid of those two resonance forms of the ion exist, where the charge is indeed spread over the two oxygens and the bondlengths are between that of a single and double bond length. check resonance(chemistry) on wikipedia
i thought this structure was correct too, but my chemistry book says its like O-Cl-O, no double bonds... and when you calculate the bound electrons, which are 4 [8*3 ( 3 not-hydrogen-atoms) - 20 (total electrons) = 4 bound electrons], everything made sense
Lewis is most famous for his rule of 2 (or now better known as the octet rule). Your approach does not only violate that, but it also does not take into account the difference in electronegativities. In addition to that, an atomic centre that has a pi bond can never be just sp3 hybridised (if there even was something like hybridisation in the first place).
+Martin Christoph Schwarzer there are exceptions to the octet rule. This is because of the d orbital, which allows for some elements, like sulfur in SF6, to have more than 8 electrons in its outer shell.
+Gem Elle The involvement of d-orbitals in bonding is marginal if not completely wrong. For example see this answer: chemistry.stackexchange.com/a/5242/4945 Hybridisation itself is a flawed concept, tragically oversimplifying the true nature of chemical bonding.
he is wrong guys, do not follow his lead. I asked my chemistry instructor, he said we make double bond when we have to. the structures which preserve 8-electron octets are more valuable for determining which compound has a "coordinate" covalent bond, because such a bond is defined as one formed from electrons "belonging" to only one of the atoms. so O--CL--O is right. just single bond for both oxygen
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Thank you for taking the time to work out these structures and explaining them!
You are a better teacher than my chem teacher.
i understand that the formals werent good at first but how did you know that you had ot pull up two electrons from oxygen in order to balance it out?
Because the lone pairs of the oxygen are delocalized and participate in resonance they are BOTH sp^2 not one being sp^2 and the other sp^3. Resonance doesn’t change the configuration of the molecule but only shows the movement of electrons within a molecule. But thank you for the video it really helped a lot 🤗.
I see that Kent's answer can possibly be a solution, but if we take in consideration of the expression pi bond = (6n+2-#)/2, where n is the number of atoms (excluding any hydrogens) and # is the amount of available electrons then it will amount to (6x3+(2-20))/2=0. This means there should not be an extra bond. So, O-Cl-O with single bond would be satisfy. This is why some of you get confused. To check that this is correct, you can balance the formal charge O= -1, -1 and Cl= +1 and get a -1. Hopefully, this helps.
good video
If you were to draw an electron level diagram for the chlorine atom here to indicate hybridization, what would it look like. I'm just stumped because to my knowledge (gr12) i know you can only form pi bonds with p orbitals, and if the 3 p orbitals are hybridized into sp3 orbitals, how can there be pi bonds in the double bond?
kentchemistry.com I don't understand one thing, when you gave the double bond to the left oxygen, the right oxygen still had a formal charge of -1, so does that mean that the '"right" Lewis structure for the Chlorite ion is the resonance hybrid version?
I was debating this structure in a study session and I agree with your final product. One of my buddies argued that there should be a double bond for each oxygen so that FC = 0, leaving CL with FC= -1. Although the FC in each structure is even, I thought that because O is more electronegative, it would be more probable to find ClO2- in the form you explained (with more lone pairs). Is my reason the correct reason for said structure?
No he is absolutely right.. In order to get the most correct Lewis structure formal charge must equal the charge of the ion and thus it is an exception to the octet rule. Give your "Chemistry instructor" the finger for me
You are right about the expanded octet rule on period three, but isn't the hybridization of Cl Sp2d? Because it is in the third period? I may be wrong, but i thought that is how that worked. If not can you explain why?
Athough his answer makes sense according to the formal charge rule, however, since O is more electronegative, it won't be willing enough to share an extra pair of electrons with Cl by forming a double bond. Anyways, people who are familiar with Its tetrahedral symmetry (sp3 hybridization), you'll notice that this structure won't be possible if there was a double bond.
Thanks for your video!
chlorine is in the 3rd period which gives it access to d-level orbitals so it can break the octet rule.
For calculating formal charges don't you add the amount of bonded electrons as opposed to subtracting them?
For the formal charges of polyatomics, shouldn't they equal the charge of the polyatomic ion itself?
Like why wouldn't the first one work? Because both Oxygen's have charges of -1 and the Chlorine has a +1 charge, so the net is -1, which is the charge of the actual ion, ClO2-
Can someone explain please?
Hybridization for cl is actually sp2d because of the pi bond. I entered your answer in my online hw and it said its sp2d
Period 3 elements(Starting from Na, Mg...Cl,Ar) can have more than 8 elections. Similar to that Be,B and H can have less than 8 electrons. These are the exceptions for Lewis dot structure, so he is doing it right.
is formal charge the same as oxidation number?
thank you so much! I finally understand
Not sp3 hybridization. It is sp2d because when you expand the octet of Cl- it would expand to the d orbital
@adrenalinerushr certain periodic table elements are capable of "bullying" elements into taking more than an octet. O and F are the most common. This is correct.
How do you know that Cl goes in the center?
because there are two O
Should this be bent?
Why doesn't the chlorine have an octave?
The chlorine have 10 electrons....U broke the octet rule...sth wrong?
In reality a resonance hybrid of those two resonance forms of the ion exist, where the charge is indeed spread over the two oxygens and the bondlengths are between that of a single and double bond length.
check resonance(chemistry) on wikipedia
My book refers to 3 possible structures for this to achieve resonance and I have no idea what the third one would look like.... Anyone?
i thought this structure was correct too, but my chemistry book says its like O-Cl-O, no double bonds...
and when you calculate the bound electrons, which are 4 [8*3 ( 3 not-hydrogen-atoms) - 20 (total electrons) = 4 bound electrons], everything made sense
Watch at 0.75 playback speed if it is too fast
How can Oxygen have 10 e??
Lewis is most famous for his rule of 2 (or now better known as the octet rule). Your approach does not only violate that, but it also does not take into account the difference in electronegativities. In addition to that, an atomic centre that has a pi bond can never be just sp3 hybridised (if there even was something like hybridisation in the first place).
+Martin Christoph Schwarzer there are exceptions to the octet rule. This is because of the d orbital, which allows for some elements, like sulfur in SF6, to have more than 8 electrons in its outer shell.
+Gem Elle The involvement of d-orbitals in bonding is marginal if not completely wrong. For example see this answer: chemistry.stackexchange.com/a/5242/4945
Hybridisation itself is a flawed concept, tragically oversimplifying the true nature of chemical bonding.
I believe you meant to write "Period 3" elements
@GreenDayPunkChick Cl is the least electronegative
thankss!!
Didn't think it was possible to burn someone so hard with one word
@BenStinard No
he is wrong guys, do not follow his lead. I asked my chemistry instructor, he said we make double bond when we have to. the structures which preserve 8-electron octets are more valuable for determining which compound has a "coordinate" covalent bond, because such a bond is defined as one formed from electrons "belonging" to only one of the atoms. so O--CL--O is right. just single bond for both oxygen
subtitle the video, please :(
Use oxygen as a central atom and see
Did not help
ur wrong
you are wrong. so so wrong!