Btw sir, it is a super video. I'm revising the syllabus very easily with the help of your videos. Thankyou so much for helping. It's very easy to understand now as it was a very confusing topic. Thanx a ton sir.
Super sir.. i tried to solve these type of questions from reference books even i never got the half information... your explation is awesome... thank you so much sir...
Nice video . But there , I guess +/- signs depend upon whether the system is symmetric or antisymmetric w.r.t plane of symmetry passing through the nuclear axis. Actually I read somewhere. Kindly acknowledge.
Sm days before i watched ur video n aftr that i reglrly used to rly vry useful ....plz sir make a video on lanthanide n actnides based on the qstn of csir net n plz keep uploading more videos ...
also the two O atoms has the 2p^4 configuration so after forming the molecular orbitals they are redistributed and also there is a difference there the unpaired electron is in antibonding molecular orbital (pi-star 2px and pistar 2py) while the sigma orbitals are formed by 2pz orbitals of each of it in hybradised statewhich as a pair of 2 electrons while the pi-bond is formed by the 2py and 2px orbitals as bonding molecular orbitals having a pair of electrons while antibonding molecular orbitals in are singly occupied
Thank you so much ..Ur videos are really helpful.. whenever I want to understand a new concept...I search for Ur videos first..and the best part of Ur videos are that it includes previous year questions..which are must for a better concept understanding...
kashmina akhtar ..Since there are two possibilities 1. Either both the electrons are of same spin then 2s+1=3 which is not in option or 2. Both the electron of opposite spin then 2s+1=1 which is given in the option
For hydrogen problem option 1st and 4th both are possible.. 1st is ....when spin doesn't change we get 2s+1=2×0+1=1 4th is .... when spin also changes .. We get 2s+1= 2×1+1=3 Tell me which is more probable.. I think no spin change is more probable...
I think when there is the formation of the molecular orbitals then there is splitting in them where the molecular orbitals in sigma bonding is having lower energy than the molecular orbitals involved in pi-bond formation so there they are not degenerate and there is redistribution of electrons in the molecular orbitals
Thank you for explaining this topic. However some suggestions for future please do not repeat the sentences continuously, some times you explain very simple calculation like gxg=g again and again, even you explain 2x1+1=3. We are here to understand molecular term symbols it means we already know something. This is only taken as my feedback and don't mind
Dear sir.... In the question asked in csir Dec 2016. 4 marks.... I have a problem... Since the two different electrons are present in two different orbitals... So why the multiplicity isn't counted to be 3 (2.1+1) because spin must be 1/2+1/2=1... Both the electrons have same spin... Plz clear this doubt..
If you watch carefully, I've repeated that its possible value, Now we are getting lambda value 1 so it could be Pi only, and with pi as option we only have option 1. yes you are also right that multiplicity could be 3 also but as it is not mentioned in options so we will not take it.
All 'Bout Chemistry... Sir if i want to go for multiplicity i.e for these two electrons.... Multiplicity will be only three... That's only present in option 2nd.... That's why raised this problem...
@@AllBoutChemistryDear sir, in csir-december 2016 question....if we put both the electrons in px orbital, then the L value becomes 2 and the multiplicity will be 1, so option 3 can be the correct answer. Please answer, little anxious to know this
@@sachinlingampally4600, but you cannot put both of them in Px, coz electronic config is already given to you in the question. You've to fill accordion to that.
@@AllBoutChemistry Thanks for your quick reply sir, However it was asked for the possible molecular term symbol in the question cant we treat this question as similar case, which you discussed in the previous vedieo for calculating the possible atomic term symbol for p4
so the resultant angular momentum is 0 because what ever be the angular momentum of pi-antibonding molecular orbital along x direction will be negative of angular momentum of pi-antibonding molecular orbital along y- direction . similar to case in Px and Py which have the magnetic quantum no as +1 and -1 . and the angular momentum is due to the magnetic field only
also there is the one that sir you had shown there for the P- subshell filling in case of atomic state to be filled in first and then third orbital is not possible at all as that will violate the hund rule which says the electrons are always filled in such a way to give highest spin multiplicity in order to have highest exchange energy and also having highest value of orbital angular momentum possible that is highest value of L= 1+0 = 1 not L=1-1=0 so that is not possible for ground state term at all. at the time 15.35 of the video
when possible term symbol is asked the you have to consider all possible values of spin multiplicity with given number of electrons, with 2 electrons you can get either 1 or 3 but in option 1 was available so
Deepshikha, it's a very well known misconception when you refer different books for this. Now to be very particular with this just remember that I've taken sigma bonds along z Axis, so Pz will correspond to sigma bond formation and it'll be denoted as sigma Pz. Whereas other two that is Px and Py will form pi bonds coz they're perpendicular to z axes and their energy is same so its Written Px=Py You can consider sigma bond in X Axis, then the things will become the way you are telling. But remember, in most of the standard books you will find sigma bonds along z Axis.
nicely explained sir, but I have an query I have read somewhere that minus sign with symbol sigma is used when multiplicity is triplet but if it is singlet ( when both electron are present in same pi orbital) in that case (+) sign will be used with Sigma symbol plz clarify the confusion .
In case of allowed and forbidden transition, you said that the "3rd rule" is having same subscript of +/-. I answered the question of CSIR JUN 2018 by following this rule but it is not matched with csir answer key.......
KRISHANU BERA, I think you're making some minor mistake in solving that question. Actually in June 2018 Question you don't have to worry about the sign. It could be done by two rules itself. th-cam.com/video/C_A_EqdOlPQ/w-d-xo.html I've solved that problem here In this link and it will be around 14:30, so just check how it will be solved.
In Dec 2016 problem spin multiplicity is 3...bt u gave ans as 1...how it is so? And I m not satisfied with that explanation...can't we say the ans with spin multiplicity itself?...pls clarify it sir...
When we have to find out the possible term symbol, there we have to focus on lamda value and we can neglect spin multiplicity. but only when possible term symbol is asked
after filling the electrons in electronic configuration as suggested in the video, you will find that both 2 electrons are there in 1(sigma)^2 which means there is no unpaired electrons in it, so multiplicity will be n+1 where n is number of unpaired electrons, you will get 1 sigma plus g
But for excited state the electronic config is 1(sigma)^1(g) 1(sigma)*^1(u) which makes the Spin multiplicity to be 3(2*1+1). The question was asked for first excited state not the ground state.And also it should be 'u' not 'g' [ u * g = u ] . Also for allowed transition the ANS should be 1 sigma plus u which is not there in the option.Please clarify !
Arghya Sarkar, in the above comment I was talking about ground state of Hydrogen molecule. Yes you're absolutely correct about first excited state of H2 molecule, and you yourself answered everything, but you got confused, in the above comment of urs you yourself mentioned spin multiplicity 3,then how come answer will be 1 sigma u plus. It is 3 sigma u plus and option D is correct option
Thanks a lot, sir for the prompt reply and also for the clarification. I got confused because I thought that first excited state should have an allowed transition which I gather is not necessarily true. So the answer is 3 sigma u plus but a transition from ground state to the first excited state is forbidden. Right?
Sir pls make more vedios on csir syllabus .... it will help to poor people like me and others .....bcz we dont have money to join other private institutions or online ......its request sir please
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Bro,most wonderful and most easiest method which you have understood us which i was never seen
Sir.the way u explain..is amazing..first time am thinking that I may clear csir exam with the help of your video lectures.thank you..so much sir
kiran prabha, thank you for nice words, it means alot to me 😊
Btw sir, it is a super video. I'm revising the syllabus very easily with the help of your videos. Thankyou so much for helping. It's very easy to understand now as it was a very confusing topic. Thanx a ton sir.
Good command on english
Solved that problem exited. Hydrogen is 3 €u+
Thank you so much for this video!! So much better than my teacher at university :)
Best video on molecular term symbol starts from basic and cover all the aspects for exam.thnxs
thank you for kind words, it means a lot to me
Thanks a lot, this explanation is better than every other bit explained online
Super sir.. i tried to solve these type of questions from reference books even i never got the half information... your explation is awesome... thank you so much sir...
Nice video . But there , I guess +/- signs depend upon whether the system is symmetric or antisymmetric w.r.t plane of symmetry passing through the nuclear axis. Actually I read somewhere. Kindly acknowledge.
You are so much better than my professor.
best explanation.....ur teaching is so vivid and clear that even the tough concept seems to be easy .. hats off ...sir
Ur classes are very clearly understand of topics. Thank u sir
Sm days before i watched ur video n aftr that i reglrly used to rly vry useful ....plz sir make a video on lanthanide n actnides based on the qstn of csir net n plz keep uploading more videos ...
Thank You for being a regular viewer,
I'll try to make a video on Lanthanide and Actinide in upcoming days.
But sir according to rule multiplicity should be 3.....
Right bt given is 1
also the two O atoms has the 2p^4 configuration so after forming the molecular orbitals they are redistributed and also there is a difference there the unpaired electron is in antibonding molecular orbital (pi-star 2px and pistar 2py) while the sigma orbitals are formed by 2pz orbitals of each of it in hybradised statewhich as a pair of 2 electrons while the pi-bond is formed by the 2py and 2px orbitals as bonding molecular orbitals having a pair of electrons while antibonding molecular orbitals in are singly occupied
Oooo9
Very simple and easy way to solve these problems. Wonderful tutorial sir.
Once I saw ur video..now for any topic I used to search .. understands well...
In first excited state of hydrogen I was getting multiplicity 2..plz explain
Yeah..
I too
Nice sir... The video and the way of ur teaching just superb... Mind blowing...
Thank you 😊
Thank you so much ..Ur videos are really helpful.. whenever I want to understand a new concept...I search for Ur videos first..and the best part of Ur videos are that it includes previous year questions..which are must for a better concept understanding...
Really very easy explaination.i understand very well and thank you very much for this video.
You're welcome 😊
Awesome teaching😃😃👏👌👌
thankyouuuuuuuuu very much!❤❤❤❤❤❤
You are a very excellent professor, all the best sir
Sir at 19:26 sec 2s+1=3 but in case of pi 2s+1=1 then how 1st option is correct
Yes
how?
Because 3pi not in option
kashmina akhtar ..Since there are two possibilities
1. Either both the electrons are of same spin then 2s+1=3 which is not in option or
2. Both the electron of opposite spin then 2s+1=1 which is given in the option
It is asked possible term
For hydrogen problem option 1st and 4th both are possible..
1st is ....when spin doesn't change we get 2s+1=2×0+1=1
4th is .... when spin also changes ..
We get 2s+1= 2×1+1=3
Tell me which is more probable..
I think no spin change is more probable...
wonderful
Sir, at time 22:40 , December 2016 problem, kindly explain how multiplicity came 1 to give answer 1pi. Thank you.
they have mentioned in the excited state....so spin will be reversed in sigma2pz* orbital give zero spin value
Outstanding sir.
gopalkrishna niyogi, thanks brother
Thank you very much for this explanation. It was very helpful.
Superb superb....
I think when there is the formation of the molecular orbitals then there is splitting in them where the molecular orbitals in sigma bonding is having lower energy than the molecular orbitals involved in pi-bond formation so there they are not degenerate and there is redistribution of electrons in the molecular orbitals
Thank you for explaining this topic. However some suggestions for future please do not repeat the sentences continuously, some times you explain very simple calculation like gxg=g again and again, even you explain 2x1+1=3. We are here to understand molecular term symbols it means we already know something. This is only taken as my feedback and don't mind
Thank you so much for the feedback, will take care of these in future videos 😊
Keep up the good work sir👌👌
Thank you 😇
Very helpful sir, nice explanation
Thank you 😊
Simply superb sir
thank you 😊
So good sir
Dear sir.... In the question asked in csir Dec 2016. 4 marks.... I have a problem... Since the two different electrons are present in two different orbitals... So why the multiplicity isn't counted to be 3 (2.1+1) because spin must be 1/2+1/2=1... Both the electrons have same spin... Plz clear this doubt..
If you watch carefully, I've repeated that its possible value, Now we are getting lambda value 1 so it could be Pi only, and with pi as option we only have option 1.
yes you are also right that multiplicity could be 3 also but as it is not mentioned in options so we will not take it.
All 'Bout Chemistry... Sir if i want to go for multiplicity i.e for these two electrons.... Multiplicity will be only three... That's only present in option 2nd.... That's why raised this problem...
@@AllBoutChemistryDear sir, in csir-december 2016 question....if we put both the electrons in px orbital, then the L value becomes 2 and the multiplicity will be 1, so option 3 can be the correct answer. Please answer, little anxious to know this
@@sachinlingampally4600, but you cannot put both of them in Px, coz electronic config is already given to you in the question. You've to fill accordion to that.
@@AllBoutChemistry Thanks for your quick reply sir, However it was asked for the possible molecular term symbol in the question
cant we treat this question as similar case, which you discussed in the previous vedieo for calculating the possible atomic term symbol for p4
so the resultant angular momentum is 0 because what ever be the angular momentum of pi-antibonding molecular orbital along x direction will be negative of angular momentum of pi-antibonding molecular orbital along y- direction . similar to case in Px and Py which have the magnetic quantum no as +1 and -1 . and the angular momentum is due to the magnetic field only
also there is the one that sir you had shown there for the P- subshell filling in case of atomic state to be filled in first and then third orbital is not possible at all as that will violate the hund rule which says the electrons are always filled in such a way to give highest spin multiplicity in order to have highest exchange energy and also having highest value of orbital angular momentum possible that is highest value of L= 1+0 = 1 not L=1-1=0 so that is not possible for ground state term at all. at the time 15.35 of the video
Super explanation sir
Thank You :)
Superb explanation
thank u sir for uploading this vedio this was really helpful for me that u sir thank u so much
you're welcome :)
Thank you very much sir.... please do regular videos......
due to research work I'm quite irregular bt I'm trying to be regular... :)
This video helps me a lot....Thanks Sir...
@22:30 , how did u get multiplicity equals to 1?
I get 3 (2×1+1=3) for 2 unpaired electron🤔
when possible term symbol is asked the you have to consider all possible values of spin multiplicity with given number of electrons, with 2 electrons you can get either 1 or 3 but in option 1 was available so
All 'Bout Chemistry, thanks sir. I understood
good explanation ..........thank u very much sir
Best explanation
thank you
Outstanding
Ans is 4 as for sigma orbital L=0 which corresponds to € , 2s+1=3, g x u= u, since no pi therefore +. Hope this is the correct answer
How the spin multiplicity is 3?
Do we have to consider half spins from each of the sigma orbitals??
super teaching sir thank you so much
You're welcome 😇
Sir in example of O2 molecule, u write that π px orbital= π pz orbital. But I know that π py = π pz..pls explain
Deepshikha, it's a very well known misconception when you refer different books for this.
Now to be very particular with this just remember that I've taken sigma bonds along z Axis, so Pz will correspond to sigma bond formation and it'll be denoted as sigma Pz. Whereas other two that is Px and Py will form pi bonds coz they're perpendicular to z axes and their energy is same so its Written Px=Py
You can consider sigma bond in X Axis, then the things will become the way you are telling.
But remember, in most of the standard books you will find sigma bonds along z Axis.
Ok ..thanks sir
You're welcome
nicely explained sir, but I have an query I have read somewhere that minus sign with symbol sigma is used when multiplicity is triplet but if it is singlet ( when both electron are present in same pi orbital) in that case (+) sign will be used with Sigma symbol plz clarify the confusion .
Very well explained 👍👍
nice...sir
good morning sir ,this video is surperb sir
The term of 1st excited state of hydrogen molecule is 1singa u which is not including in 4 options
very nice explained.. and june 2018 also had a que on transition take place or not
Very nice lecture sir thanks 🙏🙏
Your all lectures are very clear, thank you for sharing your knowledge
you're welcome :)
keep watching our videos and share with your frnds...
@@AllBoutChemistry sir please reply...22:40 answer cannot be π^1 ...plzz give d correct answer
@@shafnat3262 check the answer key
Really very helpfullll
Question in 22:36. If we calculate spin , we get 3. But in the option 3 is for sigma. Sir calculated only l value 🤔
Grt sir
For the question first excited of h2 molecule, if the answer is option 4, then change in spin multiplicity will not be zero. Sir please explain
Sir... Why they use these type of symbols for molecular term symbols???... What is link between these symbols and molecular term symbols???
Thank you for making video
You're welcome 😇
At 26:59 the ans is (4) right?? Pls reply.
yess..correct only
@@nehruramaiyan2435 bt that transition is spin forbidden,right?
Sir, ∆l =0 is laporte forbidden eg. d to d is forbidden but it is allowed in case spectroscopic term symbol... How it is possible????
why we don't use plus or minus subscript except sigma orbitals ?
Sir ∆L=0,+-1...is it right or ∆L=+-1
∆L = 0, ±1 is right!
Sorry sir i'm not understood why in oxygen spis is a 1/2 not a 1...i think 1 is because sum 1/2 spin + 1/2 spin in orbital pi
Thank you sir its helping
You're welcome 😇
In case of allowed and forbidden transition, you said that the "3rd rule" is having same subscript of +/-. I answered the question of CSIR JUN 2018 by following this rule but it is not matched with csir answer key.......
KRISHANU BERA, I think you're making some minor mistake in solving that question. Actually in June 2018 Question you don't have to worry about the sign. It could be done by two rules itself.
th-cam.com/video/C_A_EqdOlPQ/w-d-xo.html
I've solved that problem here In this link and it will be around 14:30, so just check how it will be solved.
Could you please explain what is the term symbol of He2 and what the charge is + or -?
Thanks sir,
sir, if the condition is like that for homonuclear diatomic g to g or u to u. what will b the transition for that?
Thank you
sir plz make a video on the topic of lanthanide and actinide..
Alright 👍
Sir delta vale example K que karaiye na
Very nice classes sir, explain molecular term symbols of hetero nuclear di atomic molecule
Sir the diatomic term symbol for c2 molecule will be sigma plus or minus??
Thank you so much sir
you're welcome... :)
Tysm sir, nice explanation
Hi sir I have doubt in
how to write electronic terms of diatomic molecules you can see the IRA N.Levine page 407 sir.
Sir please make a video on huckel molecular orbital theory for net
Alright 👍
Sir, I have a doubt.
In the question DEC-2016; 4 Marks, which comes around the minute, 19.41, why cant we consider only 3 sigma ?
There itself I explained in detail that you can't get L=0 there, you'll get L=1, that gives pi as term.
Thanks a lot,Sir.
I will watch it again.
In Dec 2016 problem spin multiplicity is 3...bt u gave ans as 1...how it is so? And I m not satisfied with that explanation...can't we say the ans with spin multiplicity itself?...pls clarify it sir...
16:29 sir why didn't we follow hunds rule L shouldn't be 1+0=1?
Sir confusion with dec 2016.. if we consider spin multiplicity it is found to be 3, but your explanation towards answer has multiplicity 1.
Pls clear it.
When we have to find out the possible term symbol, there we have to focus on lamda value and we can neglect spin multiplicity. but only when possible term symbol is asked
@@AllBoutChemistry ohk.. sir.. thank you...
you're welcome :)
Thanks sir😊
Thanks
Sir how can we get the term for singlet O2 1£g+ if rule 3 is concerned..... (+) sign......????
4 option fr june 2016
Sir I have a confusion on the question of 26:58, for sigma1s1*, how the spin multiplicity is 3 in the answer?
Oh found my mistake, 1s1 + 1s1* so two unpaired electron
And H2 me kitne unpaired electrons lenge???
In ground state zero unpaired, in excited 2 unpaired.
How do we calculate multiplicity of hydrogen molecule..actually it does not match the options..I got the answer 3 sigma plus g
after filling the electrons in electronic configuration as suggested in the video, you will find that both 2 electrons are there in 1(sigma)^2 which means there is no unpaired electrons in it, so multiplicity will be n+1 where n is number of unpaired electrons,
you will get 1 sigma plus g
But for excited state the electronic config is 1(sigma)^1(g) 1(sigma)*^1(u) which makes the Spin multiplicity to be 3(2*1+1). The question was asked for first excited state not the ground state.And also it should be 'u' not 'g' [ u * g = u ] . Also for allowed transition the ANS should be 1 sigma plus u which is not there in the option.Please clarify !
Arghya Sarkar, in the above comment I was talking about ground state of Hydrogen molecule.
Yes you're absolutely correct about first excited state of H2 molecule, and you yourself answered everything, but you got confused, in the above comment of urs you yourself mentioned spin multiplicity 3,then how come answer will be 1 sigma u plus.
It is 3 sigma u plus and option D is correct option
Thanks a lot, sir for the prompt reply and also for the clarification. I got confused because I thought that first excited state should have an allowed transition which I gather is not necessarily true. So the answer is 3 sigma u plus but a transition from ground state to the first excited state is forbidden. Right?
Yes that will be spin forbidden.
sir ,,plz explain for non equivalent pi electron. Questions Dec 2013
very nice
Sir pls make more vedios on csir syllabus .... it will help to poor people like me and others .....bcz we dont have money to join other private institutions or online ......its request sir please
Answer- First excited state of Hydrogen molecule is option number 4
Sir plz make videos on f block elements..
Alright... 👍
In 2016 question ans is 4. thank you sir