Hey dude, I am rly bad at chem and atm we are doing topic 2 and I do not get the electromagnetic spectrum. Can you explain how the spectrometer can even see the isotopes present in an element.
So because one more electron and one more proton are added to the outer orbital across a group, the force of attraction increases and pulls the outer orbital closer to the nucleus so the atom gets smaller but rather since going down a group although also increases the valence electrons on the previous element in its groups’ outer orbital as well as the effective nuclear charge gets bigger, because another orbital is added the distance between the new outer orbital and nucleus just so happens to be farther out than the previous elements emptier outer orbital?
Since this is obviously the case, is the rate at which the atomic radii gets smaller across a period greater or less than the rate at which the atomic radii gets bigger down a group?
For ex. Na vs K. K fills all of the possible electrons it could have to be filled on its 2nd energy levels and across the period of Na elements atomic radii get smaller bc of greater effective nuclear charge, so you’d expect K to be as big as Ar, ok Nice makes sense! But noooo K wants to have one more proton and, more importantly, one more electron than Ar making the orbital stick out farther than even Na’s 2p orbital sticks out! Why was adding one more electron and one more proton after filling the 2nd energy levels the determining factor in being able to violate effective nuclear charge theory which shows that the more electrons and protons, the greater the effective nuclear charge, therefore the smaller the atomic radii?
Ok well I know that it’s because K has a 3p orbital that it now uses because (for whatever reason) only 8 electrons can fit within a certain orbital (with some exceptions) but so basically the 8 electrons on K’s 2nd energy orbitals repel the electron on K’s 3rd energy level more than the nucleus (and added proton) can attract it, so much more so that the 3p orbital of K sticks out farther than Na’s emptier 2p orbital.
Across a period effective nuclear charge dominates but down a period the number of occupied energy levels dominates. So atoms get smaller across a period and larger down a group.
@@MSJChem ok first off sorry for typos instead use Li and Na in my ex and Ne as the noble gas. But i guess I answered my own question about how only 8 e-‘s can fit on an orbital and so adding another one to Na from Ne means it has to put it in the 3p orbital space, thus making radius bigger. But I guess my new question is why can only 8 e-‘s fit on an energy level
Extremely helpful and clear. Thanks.👍🙏
perfect, explanation is really understandable
Thank you for your very clear and concise explanation
pray for me to get a 7 in IB chem😭😭😭
tqsm... really helpful!!!!
Thank you ❤️
for the alg
Can you please do videos where you answer past papers?
Thank u
Hey dude, I am rly bad at chem and atm we are doing topic 2 and I do not get the electromagnetic spectrum. Can you explain how the spectrometer can even see the isotopes present in an element.
3 years later, did you find your answer lmao
4 years later, did you find your answer lmao
5 years later, did you find your answer lmao
@@cookin_curry06 we still waiting bro- R U M25????
6 years later, did you find your answer lmao
Thank you sir
Does the IB require knowledge on the effective nuclear charge?
Not how to calculate it, no.
So because one more electron and one more proton are added to the outer orbital across a group, the force of attraction increases and pulls the outer orbital closer to the nucleus so the atom gets smaller but rather since going down a group although also increases the valence electrons on the previous element in its groups’ outer orbital as well as the effective nuclear charge gets bigger, because another orbital is added the distance between the new outer orbital and nucleus just so happens to be farther out than the previous elements emptier outer orbital?
Since this is obviously the case, is the rate at which the atomic radii gets smaller across a period greater or less than the rate at which the atomic radii gets bigger down a group?
For ex. Na vs K. K fills all of the possible electrons it could have to be filled on its 2nd energy levels and across the period of Na elements atomic radii get smaller bc of greater effective nuclear charge, so you’d expect K to be as big as Ar, ok Nice makes sense! But noooo K wants to have one more proton and, more importantly, one more electron than Ar making the orbital stick out farther than even Na’s 2p orbital sticks out! Why was adding one more electron and one more proton after filling the 2nd energy levels the determining factor in being able to violate effective nuclear charge theory which shows that the more electrons and protons, the greater the effective nuclear charge, therefore the smaller the atomic radii?
Ok well I know that it’s because K has a 3p orbital that it now uses because (for whatever reason) only 8 electrons can fit within a certain orbital (with some exceptions) but so basically the 8 electrons on K’s 2nd energy orbitals repel the electron on K’s 3rd energy level more than the nucleus (and added proton) can attract it, so much more so that the 3p orbital of K sticks out farther than Na’s emptier 2p orbital.
Across a period effective nuclear charge dominates but down a period the number of occupied energy levels dominates. So atoms get smaller across a period and larger down a group.
@@MSJChem ok first off sorry for typos instead use Li and Na in my ex and Ne as the noble gas. But i guess I answered my own question about how only 8 e-‘s can fit on an orbital and so adding another one to Na from Ne means it has to put it in the 3p orbital space, thus making radius bigger. But I guess my new question is why can only 8 e-‘s fit on an energy level
I thought that effective nuclear charge only applies to Successive ionisation energies
Effective nuclear charge can be applied to any trend in the periodic table such as atomic radius, etc.
ty!!!
yes