The atomic number of titanium(Ti) is 22 and not 23 as used in the video. So in the electronic configuration it should be 3d2 and not 3d3. Thus, the Z(eff) = 8.85. Otherwise, pretty good video, enhanced my understanding quite a lot. Thanks!
Yeah I noticed that too, as I was solving it before him keeping the video paused to practice. The value I got for S was 11.95 and Z* was 9.05 with the corrected values. Please let me know if I was correct or wrong?
s=18*1.00 + 1*0.35=18.35 Zeff=22-18.35=3.65 The atomic number not being 22 is not the only error, the electrons in the 3s and 3p were supposed to be multiplied by 1 and not 0.35.Literally all the electrons before 3d are supposed be multiplied by 1 and added to the one other electron in the 3d orbital,which is multiplied by 0.35. To err is human. Thank you Sir.
I feel this, lost some marks on a test because of that. Oh well, that's life. For anyone else, I used this resource to learn this: en.wikipedia.org/wiki/Slater's_rules
This is not the sort of 'group' Slaters Rules are referring to. An important step of Slaters Rules he did not mention in the video is to group the electron configuration into the following order: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f ) (5s, 5p) (5d) (and so on). When Slater's Rules refers to electrons in the same 'group' as the electron of interest, it is referring to the groups written out above in parentheses, not the group related to the n value. If we look at the groups in relation to Slaters Rule, we can see that 3s and 3p are NOT in the same 'group' as 3d, and should therefore be multiplied by 1.00 to get the S value @@felixkambwili7754
Professor Organic Chemistry Tutor, thank you for showing How to use Slater's Rule to Estimate the Effective Nuclear Charge in AP/General Chemistry. From the video, I found the examples/practice problems confusing/problematic, however I will review this material from start to finish. Professor Organic Chemistry Tutor, thanks to the great viewers for finding and correcting the error(s) in this video.
This is a great help. well explained but I'm just confused, isn't Ti atomic number is 22?. I think the formula/solution is referring to Vanadium. But its okay I corrected it on my notes. Thank you
When calculating for the d sub shell you do not consider s and p to have the same constant of 0.35 if they have the same quantum number, n. I was marked down for that.
You've missed a very important step/concept in Slaters Rules!!! After writing out the electron configuration, it should be rearranged and grouped into the following order: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f ) (5s, 5p) (5d) (and so on) When Slater's Rules refers to electrons in the same 'group' as the electron of interest, it is referring to the groups written out above in parentheses, NOT the group related to the principal quantum number (n). For example, take the Titanium calculation. If we look at the groups in relation to Slaters Rule, we can see that 3s and 3p are NOT in the same 'group' as 3d, and should therefore be multiplied by 1.00. Please like this comment if you agree!
but what i learned from other sources said that for d and f orbitals any electrons in the same principle quantum number but lower angular quantum number, such as 3s,3p with 3d. the electrons in 3s and 3p will have a effect of 1.00 to the electron in 3d orbital, not the 0.35
Titanium's atomic number is 22. Shouldn't its electron configuration be 1s2 2s2 2p6 3s2 3p6 4s2 3d2? Having a 3d3 makes it atomic number 23 which is Vanadium
Watch this video with a grain of salt as I found there were calculation errors while calculating 3d shielding constant for titatium (he multiplied the 3s and 3p levels by 0.85 and according to my prof. and a few other commenters its 1.00)
While removing the Zeff of Ti why did u give the shielding effect of 2s2 and 2p6 orbitals 1? Aren’t they supposed to be (n-1) so their screening effect should be 0.85? Am I wrong?
@@Shloka-np6ov i study 1st grade in algeria and i find the answer of our problem which is 0.85 we use it only with S and P in (n-1) the others D, F and G... in (n-1) we use 1.00
@@ahmedmaadoune1245 I would have but I don’t have any social media account.. I am currently preparing for an important exam so I am staying away from any kind of distraction.. I am really sorry though 🙁
Z eff is caused by the electrons which are shielding the electron under study from nucleus in this case 1s,2s,2p,3s,3p orbital e- shielding the 3d e- but 4s orbit is present outside of 3d so it didn't involve in shielding 3d,don't take the energy lvls to serious instead you can consider shielding e- and the inner orbits.hope it helps🤗🤗.
@@shujriz2286 You are confusing with orbital energy lvl with shell/orbit lvl.n+ l rule is for orbital(subshell more precisely)energy not for orbit. e- are filled according to the n+l energy rule but their is no need for them get arranged in a progressive shell according to n+l rule i.e.,3d e- are located in 3rd shell which is closer to nucleus than 4s e- located in 4th shell,so 4s e- aren't shielding the the 3d e- therefore 4s e- aren't reducing the attractive force(Z eff) b/w 3d e- and the nucleus,this is the reason y we exclude the 4s e- from calculation.
Hi Nick I don't know if you still need an explanation but when we are calculating Shielding constant for a d or f orbital all lower orbitals will contribute a 1 to S not 0.85
Why are you using the value of (n-2) for the 2s and 2p electrons in your example for a 3d electron in Titanium? Shouldn't these electrons have the values of 0,85 since we are looking for a (3)d electron. Aren't (2)s and (2)d (n-1) und thus = 0,85?
Doubt you still needs this answered, but since someone else might read this and also be confused, the rules change if you are working with an electron in a d- or f-orbital. Slaters rules states as follows: 1:Electrons with the same principal quantum number contributes 0.35 to slaters screening constant, BUT if the electron studied is in a d- or f-orbital, then the electrons in the s- and p-orbital count 1.00 each. (This was not done in the video) 2: Electrons with one less principal quantum number (n-1) contributes 0.85 to slaters screening constant, BUT if the electron studied is in a d- or f-orbital, then the electrons in the s- and p-orbital count 1.00 each. (Note: This varies on some of my sources, my old prof told me this as written but other sources claims that it's set to 1.00 for ALL n-1 orbitals and not just s- and p-orbitals). 3: Everything lower is set to 1.00 What this means is that for titanium (which actually has an atomic charge of 22 and not 23 as in the video) the effective charge of an 3d orbital is as follows: [Ti]: 1s2, 2s2, 2p6, 3s2, 3p6, 3d2, 4s2 The 4s have a higer quantum number and can therefore be ignored, since 3d is being studied, one electron is ignored, so we get one 3d electron with a value of 0.35. Since we are studying an d-orbital, the 3s and 3p electrons are set to 1.00, same goes for the 2s and 2p electrons. Every electron that are (n-2) or lower are automatically set to 1.00 as in the video. so Zeff = Z - ((2 x 1.00) + (8 x 1.00) + (8 x 1.00) + ( 1 x 0.35)) = 22 - 18.35 = 3.65 Hope this helps (Source: Descriptive Inorganic Chemistry, 5th ed, G.Canham, T.Overton)
Can you please fix your mistake with the effective nuclear charge of Titanium? Not only is the atomic number incorrect, but so is your formula... this has most likely been confusing students and hurting their test scores for years now. At this point, it has become irresponsible. I am a big fan of your channel and I agree you explain things very well but this needs to be fixed.
The calculation for one of the 3d electrons is wrong. All the electrons on the lower levels (1s2...3p6) should be multiplied by 1, none of them by 0.85, because it's a different rule for d and f groups.
Final Exams and Video Playlists: www.video-tutor.net/
The atomic number of titanium(Ti) is 22 and not 23 as used in the video. So in the electronic configuration it should be 3d2 and not 3d3. Thus, the Z(eff) = 8.85. Otherwise, pretty good video, enhanced my understanding quite a lot. Thanks!
Thank you, I noticed that too
Right but the screening constant would be 18.85 and hence Zeff would be 3.15
Correct me if I'm wrong
@@I_willdoit no bro, screening constant is 13.15 and effective nuclear charge is 22-13.15=8.85
Yeah I noticed that too, as I was solving it before him keeping the video paused to practice. The value I got for S was 11.95 and Z* was 9.05 with the corrected values.
Please let me know if I was correct or wrong?
Yes I agree with @I'm unstoppable todayyy mine answer is also coming 18.85 for screening constant and hence 3.15 for effective nuclear charge 👍👍
Everytime I google something calculation on chem, you are always here! And that's awesome !
Sir u made a mistake Ti atomic number 22
Ti is 22
Yes
Yes
Ya it's 22
ya
s=18*1.00 + 1*0.35=18.35
Zeff=22-18.35=3.65
The atomic number not being 22 is not the only error, the electrons in the 3s and 3p were supposed to be multiplied by 1 and not 0.35.Literally all the electrons before 3d are supposed be multiplied by 1 and added to the one other electron in the 3d orbital,which is multiplied by 0.35.
To err is human.
Thank you Sir.
I feel this, lost some marks on a test because of that. Oh well, that's life. For anyone else, I used this resource to learn this: en.wikipedia.org/wiki/Slater's_rules
Yes your solution is absolutely right. 👍👍👍👍
3s3p and 3d are in the same group with principle quantum number of n=3, they must be multiplied by 0.35 and all other groups by 1.00 to obtain 8.85
u r right...
This is not the sort of 'group' Slaters Rules are referring to. An important step of Slaters Rules he did not mention in the video is to group the electron configuration into the following order: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f ) (5s, 5p) (5d) (and so on). When Slater's Rules refers to electrons in the same 'group' as the electron of interest, it is referring to the groups written out above in parentheses, not the group related to the n value. If we look at the groups in relation to Slaters Rule, we can see that 3s and 3p are NOT in the same 'group' as 3d, and should therefore be multiplied by 1.00 to get the S value @@felixkambwili7754
Definitely....it's a wonderful explanation.it had made me seem Slater's rule a bit sensible but the electronic configuration of titanium is 3d2
Our inorganic chem professor actually recommended your videos to us. Thanks for the educational content! :D
Professor Organic Chemistry Tutor, thank you for showing How to use Slater's Rule to Estimate the Effective Nuclear Charge in AP/General Chemistry. From the video, I found the examples/practice problems confusing/problematic, however I will review this material from start to finish. Professor Organic Chemistry Tutor, thanks to the great viewers for finding and correcting the error(s) in this video.
This is a great help. well explained but I'm just confused, isn't Ti atomic number is 22?. I think the formula/solution is referring to Vanadium. But its okay I corrected it on my notes. Thank you
Ya I got that confused too...
Yeah titanium atomic no is 22
It will be 3d^2
Atomic no. For titanium is 22 sir, but i do really appreciate you have made me understand the zeff concept very well.
Finally understood Slaters rules after watching this...
Brilliant 👍
I still have no idea what’s going on
edit- three years later, i finally understood what’s going on. doing my masters in orgo chem now
rip
Mee too
Go study then
I love you so much the number of classes you've helped me pass is insane
Which class are u in?
@@jhakkasnews2011 Nice question actually
When calculating for the d sub shell you do not consider s and p to have the same constant of 0.35 if they have the same quantum number, n. I was marked down for that.
Yeah same, realized that today :)
right
Your teaching methods are incredible 🙏
Thank you bro, saved me a lot of headaches
Thanks for your nice explanation. The atomic number of titanium is 22 and not 23. Kindly rectify this since many may be misled. Thanks
thank you so much, this is really helpful especially during exams. your way of explaining is just perfect!
For 3d e of Ti, effective nuclear charge would be 8.85, which is close to the actual one ( Z of Ti is 22)
good correction!
You've missed a very important step/concept in Slaters Rules!!! After writing out the electron configuration, it should be rearranged and grouped into the following order:
(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f ) (5s, 5p) (5d) (and so on)
When Slater's Rules refers to electrons in the same 'group' as the electron of interest, it is referring to the groups written out above in parentheses, NOT the group related to the principal quantum number (n). For example, take the Titanium calculation. If we look at the groups in relation to Slaters Rule, we can see that 3s and 3p are NOT in the same 'group' as 3d, and should therefore be multiplied by 1.00. Please like this comment if you agree!
Thank you your lessons are very understandable
Glad you like them!
but what i learned from other sources said that for d and f orbitals any electrons in the same principle quantum number but lower angular quantum number, such as 3s,3p with 3d. the electrons in 3s and 3p will have a effect of 1.00 to the electron in 3d orbital, not the 0.35
Titanium's atomic number is 22. Shouldn't its electron configuration be 1s2 2s2 2p6 3s2 3p6 4s2 3d2? Having a 3d3 makes it atomic number 23 which is Vanadium
thats why he got the estimate a bit far from the actual value... but its all ok
brilliant video btw, helps me a lot on my kinetics exam
Yayyy Slater's Rule finally makes sense. Thank you!!!
This 12 min video made be understand the rule better than the 45 hrs in class
u saved my day sir thank you very much!! love from india :)
Thank you so much for clearing the doubt 🎉
Beryllium 2S2 why we are taking only 1 for the purpose of calculating screening constant?
Watch this video with a grain of salt as I found there were calculation errors while calculating 3d shielding constant for titatium (he multiplied the 3s and 3p levels by 0.85 and according to my prof. and a few other commenters its 1.00)
While removing the Zeff of Ti why did u give the shielding effect of 2s2 and 2p6 orbitals 1? Aren’t they supposed to be (n-1) so their screening effect should be 0.85? Am I wrong?
i have the same question
@@ahmedmaadoune1245 which exam are u preparing for?
@@Shloka-np6ov i study 1st grade in algeria and i find the answer of our problem which is 0.85 we use it only with S and P in (n-1) the others D, F and G... in (n-1) we use 1.00
@@ahmedmaadoune1245 ooo thanks for leaving the comment!!
@@ahmedmaadoune1245 I would have but I don’t have any social media account.. I am currently preparing for an important exam so I am staying away from any kind of distraction.. I am really sorry though 🙁
Thanks for sharing the knowledge sir🎉🎉
You are awesome and so are your videos!!
Keep up the great work!
Thank you my good sir. You just saved my grade.
My also
It was helpful. Thank you.
Brilliant explanation. Thank you so much 💓
Atomic number of Ti is 22
S= (18*1) + (1*0.35) =18.35
Zeff= Z - S
= 22-18.35
Zeff = 3.65
(Since for d orbital electron n=0.35
n-1/2/3= 1)
Thank you so much
Zeff of K(potassium) is 2.2 in my course book but in the video its 18.7
Which one is correct?. I'm confused
Please continue to make videos for Inorganic chem and phys chem... I am taking inorg right now and you helped me get through all of ochem T_T
@8:14 4s at higher energy level than 3d?nplus l for 4s is 4 and for 3d is 5..
And titanium wont be having 3d2 instead od 3d3
i needed to see someone to agree with me
There's a mistake Actually he took vanadium and not titanium
Z eff is caused by the electrons which are shielding the electron under study from nucleus in this case 1s,2s,2p,3s,3p orbital e- shielding the 3d e- but 4s orbit is present outside of 3d so it didn't involve in shielding 3d,don't take the energy lvls to serious instead you can consider shielding e- and the inner orbits.hope it helps🤗🤗.
@@rohith6245 4s is outside of 3d?
N plus rule has gone for a vacation?
@@shujriz2286 You are confusing with orbital energy lvl with shell/orbit lvl.n+ l rule is for orbital(subshell more precisely)energy not for orbit. e- are filled according to the n+l energy rule but their is no need for them get arranged in a progressive shell according to n+l rule i.e.,3d e- are located in 3rd shell which is closer to nucleus than 4s e- located in 4th shell,so 4s e- aren't shielding the the 3d e- therefore 4s e- aren't reducing the attractive force(Z eff) b/w 3d e- and the nucleus,this is the reason y we exclude the 4s e- from calculation.
The electronic configuration for titanium must be 3d² not 3d³ or am I wrong?
9:40 why is it there 8x1 for 2s2 2p6... why it isn't 0.85x8
Thank you very much professor
truth be told you be doing way more than my lecturer
For the last example (specfically for Li), are we not supposed to multiply .3 by 2 because we have two electrons in subshell 1 (i.e., 1s^2)?
Nice explanation. Keep it up. 👍👍👍
Tks for the video to clarify my doubt
You can show rules in your video..it will help understanding better👍🏻
As usual, a lifesaver!
It's awesome ur still making vids
for number 3 arent the 2s2 and 2s6 electrons one energy level aways shouldnt that be .85?
that's what i thought?
Hi Nick I don't know if you still need an explanation but when we are calculating Shielding constant for a d or f orbital all lower orbitals will contribute a 1 to S not 0.85
@@sabeehahmed5148 Thank U So Much My Doubt Cleared
Can you please check again the electronic configuration of titanium, i think it was mistaken
I was so confused about this rule. Now I am very clear. Thank you so much
I still don't understand Slater's law after watching this. Where did you pull the numbers you used to multiply by the Slater rules numbers from?
THANK YOU SO MUCH...
This really helped in clearing my concepts.....thank you @theorganicchemistrytutor
If you calculated it for 3d3 configuration then it should be approximately 4.3 which is of vanadium your calculation is wrong
Who the f*ck are you
Veryy well explained 👍👍👍👍👍👍👍👍👍
Hello there! Is the atomic number of Titanium equal to 23? Isn't it 22? and 23 is for Vanadium? Please enlighten me..
the best! thank you!
3d orbitals are not in the same group as 3s 3p (n = .35 for 3d and n=1 for everything to the left) Inorganic Chemistry Gary L. Miessler
In the titanium problem why did you take 1 for 2s and 2p rather than taking 0.85 ?
It was very helpful, thank you !
Very useful,keep it up
Bro I literally just now figured out how to do this because of you! Great tutorial! Thank you!
thank u so much sir
U are great teacher ❤️❤️❤️❤️❤️
Thanks alot Sir ❤
i think you just mixed up Titanium and Vandiums electronic configuration, still great examples and video though!
Please sir make video on nmr, mass, ER spectroscopy
Thank u sir
Why are you using the value of (n-2) for the 2s and 2p electrons in your example for a 3d electron in Titanium? Shouldn't these electrons have the values of 0,85 since we are looking for a (3)d electron. Aren't (2)s and (2)d (n-1) und thus = 0,85?
Doubt you still needs this answered, but since someone else might read this and also be confused, the rules change if you are working with an electron in a d- or f-orbital. Slaters rules states as follows:
1:Electrons with the same principal quantum number contributes 0.35 to slaters screening constant, BUT if the electron studied is in a d- or f-orbital, then the electrons in the s- and p-orbital count 1.00 each. (This was not done in the video)
2: Electrons with one less principal quantum number (n-1) contributes 0.85 to slaters screening constant, BUT if the electron studied is in a d- or f-orbital, then the electrons in the s- and p-orbital count 1.00 each. (Note: This varies on some of my sources, my old prof told me this as written but other sources claims that it's set to 1.00 for ALL n-1 orbitals and not just s- and p-orbitals).
3: Everything lower is set to 1.00
What this means is that for titanium (which actually has an atomic charge of 22 and not 23 as in the video) the effective charge of an 3d orbital is as follows:
[Ti]: 1s2, 2s2, 2p6, 3s2, 3p6, 3d2, 4s2
The 4s have a higer quantum number and can therefore be ignored, since 3d is being studied, one electron is ignored, so we get one 3d electron with a value of 0.35. Since we are studying an d-orbital, the 3s and 3p electrons are set to 1.00, same goes for the 2s and 2p electrons. Every electron that are (n-2) or lower are automatically set to 1.00 as in the video.
so Zeff = Z - ((2 x 1.00) + (8 x 1.00) + (8 x 1.00) + ( 1 x 0.35)) = 22 - 18.35 = 3.65
Hope this helps
(Source: Descriptive Inorganic Chemistry, 5th ed, G.Canham, T.Overton)
thank you!
@@alexyoung2886 Thank you so much!
@@alexyoung2886 Thank you so much for helping me understand. Your wisdom will forever be etched in my notes in ur honor😂
Easy,thank you
Very useful class💞💞
The electron configuration is wrong. If you are calculating for Titanium, the electron configuration should end with 3d^2 not 3d^3
I was confused
Yeah,😅
i lowkey kinda wanna know the person behind this voice over... ngl im a bit distracted bruh ur voice is soothing
For titanium i dont understand on the 2S and 2P use 1 instead of 0.85
How will l know if he is in the same level energy please answer me bc in this point l have difficul?
Titanium atomic number is 22 not 23 .It may be topografic mistake.
Math & Science 2024 sounds just like you!
I'm confused in number 3. I thought it should be .85 instead of 1 since it's in s and p orbital
I understood thanks a lot
Love you so much👍👍👍
Can you please fix your mistake with the effective nuclear charge of Titanium? Not only is the atomic number incorrect, but so is your formula... this has most likely been confusing students and hurting their test scores for years now. At this point, it has become irresponsible. I am a big fan of your channel and I agree you explain things very well but this needs to be fixed.
The calculation for one of the 3d electrons is wrong. All the electrons on the lower levels (1s2...3p6) should be multiplied by 1, none of them by 0.85, because it's a different rule for d and f groups.
For Titanium question it should have been 3d² ?
S = 18*1 + 1*0.35 = 18.35
Zeff = 22 - 18.35 = 3.65
This is accurate, all electrons before those in 3d (except the 4s ones of course) were all supposed to be multiplied by 1
18*1 + 1*0.35=18.35
Is this happens only for d or f shell.?
Really really lovely lecture. Made slater rules really easy. Thanks alot!!!!!
Thank you so much sir love from india 😎😎😎🤘🤘🎊🎉🎉🎉🎉🎊
Electronic configuration of titanium is 3d2 4s2
yeah it is lol
Please make physical chemistry videos 😭
i think 4s level ,, there was no need to show it before 3d and it was wrong to show it before 3d. or am i missing something?
Thank you
helpful, thanks
Quite helpful
How is the value for S same for all the elements in last example ?
What is the program that you use to make the video tutorials?
The best video on UA-cam about slaters rule 🙏👍
Thanks
But you made a mistake in Ti =22
Sir,2nd shell of atom the value will 0.85