Just a heads up, this formula works for molecules that have the following: 1. Molecules with only 1 central atom 2. Molecules that do not contain hydrogen.
Too anyone who is complaining about how the formula doesn't work with hydrogen, change 8n to 2n in the formula. If the molecule has hydrogen and another element just do (Ve + 8n + 2m)/2 where n is the number of atoms sorrounding the center atom(excluding hydrogen) and m is the number of hydrogen atoms sorrounding the atom.
Holy shit dude. I was struggling so much with VSEPR just because I was having a hard time finding the no. of Lone pairs. Your formula works like magic dude ! It's so easy now to find the no. of Lone pairs ! Thank you
Professor Organic Chemistry Tutor, thank you for a short explanation on How to Calculate the number of Lone Pairs using a Formula in AP/General Chemistry. Although I took General Chemistry many years ago , this is the first time I can remember seeing a formula used to calculate lone pairs in Modern Chemistry. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
I think the formula changes according to the orbital structure so for NH3, for example, the formula in the video would be ((5+3(1))-8(3))/2 = -8 but that wouldn't make any sense so what I do is replace the 8(n) with 2(n) due to the 2 electrons in the s orbital of hydrogen. so the formula is ((5+3(1))-2(3))/2 = 1 and that would mean 1 electron lone pair which is correct. I know this doesn't sound very scientific but I tried it and it works every time even for CH4 where there are 0 lone pairs and HF where there are 3 lone pairs of fluorine.
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Great video. I was getting lost on the valencies and then later figured out that you used the outer electrons in the octate of the surrounding atoms. Coz the valency of oxygen is 2 and you used 6 for example.
I think Xe has more electrons than it should have. In your example it was has, 2 (LP)+2+2+2+2 = 12 electrons !!!! Please if there is something wrong from my side let me know.. Thanks a lot this video was helpful.
it should be Ve-2(n)/2. It is given 2 as a constant since the terminal atoms attached to the centrAL atom Hydrogen cannot have more than 2 electrons on its outermost shell!
@@jennylynosorio556 Ohhhhhh so this I can use for like H2O? I was like OMG this is not working for CH4 and NH3 and H2O!! ahhhh thank you so much for clarifying
@@hillz00 Yeah... The constant you put on "n" depends on the max number of electrons the element can have in its valence shell. Hydrogen cannot have an octet structure so the formula Ve-8n/2 will not work Hydrogen can only attain a duplet structure... You'll use Ve-2n/2 rather... Hope this helps?
Hi, not too sure but I think it works with all compounds except those with hydrogen in them. Well that’s what’s happened to me for the most part but I know it doesn’t work on ones with hydrogen for sure
Try: Lone Pairs=(Number of Valence electrons - Nearest multiple of 8 to the number of valence electrons) divided by 2. For Example: - NH3. ((3+5)- 8(The nearest multiple of 8 to Ve)) divided by 2 =8-8=0 - S02 (see answer at 2:10) ((6+2(6))- 16(The nearest multiple of 8 to Ve)) divided by 2=(18-16) divided by 2 = 2/1 =1 Works for every one i have tried so far
One more limitation beside hydrogen presence: it only works for molecules with one central atom (meaning all the other atoms are bonded to the central one)
No, it's just... Not like this. Every atom that bonds to another has only 1 possible configuration, unless specific circumstances, and there's physics and chemical reasons for that. However, this may not be too obvious for you, so there you are. If you got Br3-, there's only one configuration possible. The bonds are determinated by the valence atoms. So, here you got 2 atoms of Br with 7 valnece electrons and anothrr ion with 6 valence electrons. Keep in mind that every atom has to reach, usually, the 8 electrons configuration. Covalent bonds, as the ones that bromine creates, make sure that an atom shares one, two or three electrons with another. In this case, the central atom shares his 2 electrons with the ones that are on the side, meanwhile the ones on the side shares their electrons to the central atom. So we got 8 electrons per atom, with a total of 4 electrons shared. So yeah, when an atom shares an electron to another, it forms one single bond. If you got two electrons shared, you got a double bond. Same with the triple bond. The central molecule is determinated by the elctronegativity. The less electronegative molecule (with less atoms), goes in to the center. In this case, the central atom is the bromine atom with 1 electron in less. I hope you understood. Have a great day
That formula works for molecules that don't contain hydrogen atoms since hydrogen can only hold 2 electrons and not 8. For molecules that contain all outer hydrogen atoms with the exception of the central atom, the formula needs to change from (Ve - 8n)/2 to (Ve - 2n)/2. N = 5, H = 1 Ve = N + 3H = 5 + 3(1) = 8 n = 3 for the 3 outer hydrogen atoms (Ve - 2n)/2 = (8 - 2*3)/2 = (8 - 6)/2 = 2/2 = 1 This make sense since Nitrogen only has 1 valence electron
Yea cool formulae. But can I get the formulae that works for organic compounds to determine the lone pair. It does not work for organic compound especially when the centre atom is not one.
Thanks for saying this because I was trying to use this calculation for H2O and wound up with -4; and I’m here like what is wrong why doesn’t it work🥴😩😩😩. I will have to keep this in mind
he talks about the covalent bond in this video only. H2O is hydrogen bond, some other compounds start with H should be acids so that is why this formula doesn't work for all types of molecules.
For BeCl2, Be = 2, Cl = 7. Ve = Be + 2Cl = 2 + 2(7) = 16 n = 2 for the 2 outer chlorine atoms (Ve - 8n)/2 = (16 - 8*2)/2 = (16-16)/2 = 0/2 = 0 This make sense since BeCl2 doesn't have any lone pairs on the central Be atom.
Just a heads up, this formula works for molecules that have the following:
1. Molecules with only 1 central atom
2. Molecules that do not contain hydrogen.
Too anyone who is complaining about how the formula doesn't work with hydrogen, change 8n to 2n in the formula. If the molecule has hydrogen and another element just do (Ve + 8n + 2m)/2 where n is the number of atoms sorrounding the center atom(excluding hydrogen) and m is the number of hydrogen atoms sorrounding the atom.
This actually worked for NH3 man ... thank you so much. Why doesn't this comment have more likes 😅
upvoting this comment
do u mean Ve - 8n - 2m?
He even explained it at 3:57, do people not pay attention to his videos or something?
would this be formula be applicable to atoms that have two or more central atoms?
Holy shit dude. I was struggling so much with VSEPR just because I was having a hard time finding the no. of Lone pairs. Your formula works like magic dude ! It's so easy now to find the no. of Lone pairs !
Thank you
You are the best tutor on RUclips
Dude you have carried me through senior maths and science thank you 💗🙏
This formula works bro I've calculated for XeOF4, XeO2F2, XeF3+,and so many other molecules
Hi I'm having some problems with calculating with SCl4 can you please help me with it?
Oh my god!!!! This helped a lot.....thank you
Professor Organic Chemistry Tutor, thank you for a short explanation on How to Calculate the number of Lone Pairs using a Formula in AP/General Chemistry. Although I took General Chemistry many years ago , this is the first time I can remember seeing a formula used to calculate lone pairs in Modern Chemistry. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
Thank you ... organic chemistry tutor.Your videos are always helpful...I recommend organic chemistry tutor's videos to every chemistry beginner.
I think the formula changes according to the orbital structure so for NH3, for example, the formula in the video would be ((5+3(1))-8(3))/2 = -8 but that wouldn't make any sense so what I do is replace the 8(n) with 2(n) due to the 2 electrons in the s orbital of hydrogen. so the formula is ((5+3(1))-2(3))/2 = 1 and that would mean 1 electron lone pair which is correct. I know this doesn't sound very scientific but I tried it and it works every time even for CH4 where there are 0 lone pairs and HF where there are 3 lone pairs of fluorine.
No man it doesn't work with HCN unfortunately. It says 3 if I work with your strategy, bu the actual is zero.
.
UgliestBagel m
coz the sample problems and that formula only works for a molecule with 2 different elements! i supposed!
@@jennylynosorio556 try NH3 IT HAS TWO DIFFERENT ELEMENTS
I have no word to express that how happy I am right now thanks
Your videos are super clear and helpful. Anytime I need help with something I see if you have a video about it. Keep up the good work! Thank you!
This is sooo simple..woww!!! I understood soooo well.. thankyou sir😃😃😃
Love from India ❤
It is very helpful sir ji thanks ❤️❤️❤️
Thank you so much for the explanation! Would this be still valid to ones like NO3?
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Thank you very much . Your videos are helpful 💓
Great video. I was getting lost on the valencies and then later figured out that you used the outer electrons in the octate of the surrounding atoms. Coz the valency of oxygen is 2 and you used 6 for example.
I think Xe has more electrons than it should have. In your example it was has, 2 (LP)+2+2+2+2 = 12 electrons !!!!
Please if there is something wrong from my side let me know..
Thanks a lot this video was helpful.
This is why I have a 25 in Chemistry
what you mean ?
@@debabrataroy345 he had a 25 percent in chemistry lol thats what he means. Some people are slower learners cant blame them
you must be a god for making this video thx to your steel balls ❤❤❤
great work!!!!!!!!!!!!!!!!! i have learnt so much.
🙏 I thought I would Never get this!
thanks a lot for the vid, any chacne of uploading an easy way to calculate the bond pair? I need it most bro
do you teach private online sessions math and chemistry ?
I need more examples idk how this formula would work for CH4 NH3 HF etc
it should be Ve-2(n)/2. It is given 2 as a constant since the terminal atoms attached to the centrAL atom Hydrogen cannot have more than 2 electrons on its outermost shell!
@@jennylynosorio556 Ohhhhhh so this I can use for like H2O? I was like OMG this is not working for CH4 and NH3 and H2O!! ahhhh thank you so much for clarifying
You cannot use this formula if there are Hydrogens in your molecule
@@hillz00
Yeah...
The constant you put on "n" depends on the max number of electrons the element can have in its valence shell.
Hydrogen cannot have an octet structure so the formula Ve-8n/2 will not work
Hydrogen can only attain a duplet structure...
You'll use Ve-2n/2 rather...
Hope this helps?
Yeah it did ... I was so confused about H2O
Try to use this formula on NH3
I was trying on it too but got 1.5?
It not worked for nh3
CH3 alsoooo
Just use the equation to get the number of nonbonding electrons:
Total valence - 2(number of hydrogen atoms in the molecule). 👌
I hate chemistry with a passion.
Uss
Why it's damn interesting
😂 😂 😂 😂 that nice perfume you use is because of chemistry.
@@elijahtenywa3063then I am okay without perfume just remove chemistry from my life.
Same man, same😤
Thank you , great explanation:)
Does this formula applies to all the structures or there is certain limitation for using this formula?
Hi, not too sure but I think it works with all compounds except those with hydrogen in them. Well that’s what’s happened to me for the most part but I know it doesn’t work on ones with hydrogen for sure
@@w_ahidaI know that I am late but you should change the 8n to 2n for compounds containg hydrogen
What about BrF3 ??
How would you calculate it for PH3?
Thank you very much 💕
Hi could you help me understand how you can know if it has a double bond or not like you know for oxygen
Try: Lone Pairs=(Number of Valence electrons - Nearest multiple of 8 to the number of valence electrons) divided by 2.
For Example:
- NH3.
((3+5)- 8(The nearest multiple of 8 to Ve)) divided by 2 =8-8=0
- S02 (see answer at 2:10)
((6+2(6))- 16(The nearest multiple of 8 to Ve)) divided by 2=(18-16) divided by 2 = 2/1 =1
Works for every one i have tried so far
Mate nh3 has 1 lone pair
5:27
Why we didnt place(n) by 4 why you filled it by 3 only where is the atom of N?
I think it's cuse nitrogen is the central atom . We have to take the number of those attached to the central atom.
@@sabihamukrim5119 YEP U RIGHT BRO, [IDK WHY I AM MSGGING TO YEAR OLD COMMENT]
amazing formula for expanded octet
One more limitation beside hydrogen presence: it only works for molecules with one central atom (meaning all the other atoms are bonded to the central one)
U're very great .......
I wish I live somewhere round your HOUSE.
Erm ☠️
Regarding Br3-, is it possible to form double bonds with the two bromine atoms? Or would that be too many bonds?
@@Childish_Eli what exactly is wrong with you?
No, it's just... Not like this. Every atom that bonds to another has only 1 possible configuration, unless specific circumstances, and there's physics and chemical reasons for that. However, this may not be too obvious for you, so there you are. If you got Br3-, there's only one configuration possible. The bonds are determinated by the valence atoms. So, here you got 2 atoms of Br with 7 valnece electrons and anothrr ion with 6 valence electrons. Keep in mind that every atom has to reach, usually, the 8 electrons configuration. Covalent bonds, as the ones that bromine creates, make sure that an atom shares one, two or three electrons with another. In this case, the central atom shares his 2 electrons with the ones that are on the side, meanwhile the ones on the side shares their electrons to the central atom. So we got 8 electrons per atom, with a total of 4 electrons shared. So yeah, when an atom shares an electron to another, it forms one single bond. If you got two electrons shared, you got a double bond. Same with the triple bond. The central molecule is determinated by the elctronegativity. The less electronegative molecule (with less atoms), goes in to the center. In this case, the central atom is the bromine atom with 1 electron in less. I hope you understood. Have a great day
i don't understand when you said that you want to put one lone pair but you put two can you please explain??
I am not sure but i think i understand those 1 lone pair mean you add 2 electrons on the sulfur atom
Then what's going to be for H2O?
why would anyone dislike this
And what would be number of lone pairs for NH3. Am getting -8
That formula works for molecules that don't contain hydrogen atoms since hydrogen can only hold 2 electrons and not 8. For molecules that contain all outer hydrogen atoms with the exception of the central atom, the formula needs to change from (Ve - 8n)/2 to (Ve - 2n)/2.
N = 5, H = 1
Ve = N + 3H = 5 + 3(1) = 8
n = 3 for the 3 outer hydrogen atoms
(Ve - 2n)/2 = (8 - 2*3)/2 = (8 - 6)/2 = 2/2 = 1
This make sense since Nitrogen only has 1 valence electron
@@TheOrganicChemistryTutor thank you 🙏
Why is sulphur the center atom?
For ions??
(Oh3)+
Wait but this doesn’t work for hydrogen
Hydrogen is not a molecule ok
@@ashmitmishra5948 ohk if not hydrogen, it doesn't work for H20...
@@PranavRKumar it will not work for molecules with hydrogen because hydrogen does not form octet
For hydrogen = valence electro - 2(n) /2....
exactly
great work by a great teacher
Is there a formula to know how many bonds?
Yea cool formulae.
But can I get the formulae that works for organic compounds to determine the lone pair. It does not work for organic compound especially when the centre atom is not one.
can use for C2O ??
Oh my this is so good
Sulphur has six valence electrons so then why does it form bonds with its five electrons only in SO2?
i hope u r my lecture. 😄
Does this formula apply to ions
Yes, if you adjust the number of valence electrons based on the charge of the ion.
THANK YOU
Thank you so much love I wish I had money to give you. I will make a statue in your honor and remember you to my dying breath.
What about bonds with more than one central atom? It doesn't works for them.
What we can do in case of methane
You are legend
thank you!!!
Thank uu ❤️❤️😭
What if there are a few central atoms?
What about the bond pair?
It does not work with compounds containing H (hydrogen )
Thanks for saying this because I was trying to use this calculation for H2O and wound up with -4; and I’m here like what is wrong why doesn’t it work🥴😩😩😩. I will have to keep this in mind
i get a negative 3 for HCN
H = 1 ve-
C = 4 ve-
N = 5 ve-
1 + 4 + 5 = 10 ve-
LP = (ve - 8n)/2
= (10 - 8(2))/2
= 10 - 8 (multiply by 2 and divide by 2 cancel)
= 2 LP
@@Sara-bg8yt what a Idiot calculation
@@Sara-bg8yt dude when I divided by 2 the 10 should've become 5...simple math brother
Do all have lone pairs?
No
But how do you find lone pairs for something like N2H2?
It won't work for that molecule since it has more than 1 central atom.
thanks 🙏🙏🙏🙏🙏
What if its NH2CONH2(Urea)?
The formula works well for molecules with 1 central atom. For molecules with multiple central atoms, it's a different story.
Thanks 😊
What does that n mean ?
No of atoms attached
Why doesn't this work for the H2O molecule? I got -4. which I'm assuming is 4 lone pairs but H2O molecule only has 2 lone pairs
he talks about the covalent bond in this video only. H2O is hydrogen bond, some other compounds start with H should be acids so that is why this formula doesn't work for all types of molecules.
How to do with H3O
this formula doesn't work for all molecules, you should indicate what types of a compound suit for.
Thank you
Pf5 solution nikal kr btana step vise
i have learnt the structure of SO2
Doesn't work for hydrogens or duets
It works even for organic compounds
Somebody help me do the lone pair of CO
Ur formula doesn't work on many molecules
Is that true?
@@Tag_Gwthere are exceptions when Hydrogen is present, which the chemistry tutor mentioned
If hydrogen is there then instead of 8n use 2n
Hydrogen can only hold 2 electrons in its valance shell! Not 8, so don't divide by 8 when Hydrogen is present.
🥺❤ thanks
What about XEOF4
I got 1lonepair for that.
why doesnt school fkn teach me this
Sir N2o , doesn't match the formula
It doesn't work on ccl4
It doesn't work for BeCl2
For BeCl2, Be = 2, Cl = 7.
Ve = Be + 2Cl = 2 + 2(7) = 16
n = 2 for the 2 outer chlorine atoms
(Ve - 8n)/2 = (16 - 8*2)/2 = (16-16)/2 = 0/2 = 0
This make sense since BeCl2 doesn't have any lone pairs on the central Be atom.
The formula couldn’t work for H2O and HCN , why
It does not work for molecules having hydrogen in it .
He himself tells this in this video👇
ruclips.net/video/9B5FGPDwX_E/видео.html
For XeF6
please incease your volume
Doesn't work on hydronium ion
It doesn't work on all compounds. For example, H2O the number of lone pairs using this formula is negative 4 !!!
in lone pair formula use 2 instead of 8 you will get the answer.
Uh yeah this formula definitely does not work. A for effort though.
it doesnt work for h2o
See my response to the comment below.
it's not working on NH3
I does not wrk fr H2..species..
Hey, you made a little mistake with the SO2 example. Each oxygen atom has a double bond with the sulfur atom, otherwise it doesn't make sense
No it's right
@@Denji-xc3cy Good to know
Lewis dot structure is the easiest topic in chemical bonding lol
Just it's for only elements doesn't attached with hydrogen and molecules with only one central atom.
This formula doesn't work for Every compound
Plz say the proof
Nice