effective nuclear-charge

[inaccensa]

I know someone else had asked the same Q, but I didn't find the replies helpful.

My q is the effective nuclear charge should decrease as we go down the periodic table, since the distance from the nucleus increases and acc to columbs law, the as r increases, F decreases

EK states that it increases from top to bottom.
I have been doing this by the formula Zeff = #protons- core electrons, assuming that the electrons in the same shell don't shield

I hope MCAT tells us whether the e- in the same shell shield or not. What do u think is a safe assumption

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[wanderer]

My q is the effective nuclear charge should decrease as we go down the periodic table, since the distance from the nucleus increases and acc to columbs law, the as r increases, F decreases

It's effective nuclear charge, not effective nuclear force.

 

[CaliforniaKid]

 

[inaccensa]

It's effective nuclear charge, not effective nuclear force.

Well, let me say this again, Since the electrostatic force is decreasing as the radius is increasing, does that affect the effective nuclear charge? We are adding shells to the elements

 

[CaliforniaKid]

Well, let me say this again, Since the electrostatic force is decreasing as the radius is increasing, does that affect the effective nuclear charge? We are adding shells to the elements

As Atomic Radii increases, Z effective also increases.


Zeff = Z − S (Z is atomic number and directly proportional to Zeff).

keep it simple.

 

[inaccensa]

As Atomic Radii increases, Z effective also increases.


Zeff = Z − S (Z is atomic number and directly proportional to Zeff).

keep it simple.

When atomic radius decreases, Zeff increases. It increases from Left to Right. I know that the Zeff with increase with increasing no of protons, but what about the increasing distance? Does that affect the Zeff.

EK says that Zeff increases and posts on SDN says that it remains same, i'm confused.

 

[dapmp91]

When atomic radius decreases, Zeff increases. It increases from Left to Right. I know that the Zeff with increase with increasing no of protons, but what about the increasing distance? Does that affect the Zeff.

EK says that Zeff increases and posts on SDN says that it remains same, i'm confused.

ok, I'm gonna try to tackle this very easy concept that is being asked like its a quantum mechanic question. When the atomic radius decreases, Zeff increases. This is because as you go from left to right on a periodic table, the # of protons increases, as the # of protons increases, it increases the pull on the valence (outermost electrons) thus effectively decreasing the radius (also note the electrons are increasing also, but the force is negligible, the proton force overcomes this). However, as you go down a period, the shell's increases on an atom. As the atom gains more and more shells, the valence electrons feel less of a Zeff because now there are more electrons in between the proton (which is the nucleus) and the valence (outermost) electrons. When radius is increased the valence electrons feel a low Zeff. Note Zeff, is the effective nuclear charge; basically its the charge the proton has that is in the nucleus. hope this helps.

 

[wanderer]

Zeff increases as you go down because the higher the shell the electron resides in, there is a greater probability for the electron to be closer to the nucleus than predicted on the basis of the shell's radius. If it's closer to the nucleus, then the inner electrons won't shield the valence electron from the nucleus as much. Try to apply Gauss's Law to this concept.

As Atomic Radii increases, Z effective also increases.


Zeff = Z − S (Z is atomic number and directly proportional to Zeff).

keep it simple.

As you go down the periodic table, S fails to "keep pace" with Z, so Zeff goes up.

 

[RogueUnicorn]

this is all wrong.

Zeff INCREASES the higher you are in the periodic table and INCREASES as you go to the right. Period. There is less distance and less electron shielding. All the periodic trends can be correlated to this.

 

[RogueUnicorn]

Zeff increases as you go down because the higher the shell the electron resides in, there is a greater probability for the electron to be closer to the nucleus than predicted on the basis of the shell's radius. If it's closer to the nucleus, then the inner electrons won't shield the valence electron from the nucleus as much. Try to apply Gauss's Law to this concept.


As you go down the periodic table, S fails to "keep pace" with Z, so Zeff goes up.

this is wrong and is not a proper understanding of quantum. sorry.

 

[Charles_Carmichael]

Zeff increases as you go down because the higher the shell the electron resides in, there is a greater probability for the electron to be closer to the nucleus than predicted on the basis of the shell's radius. If it's closer to the nucleus, then the inner electrons won't shield the valence electron from the nucleus as much. Try to apply Gauss's Law to this concept.


As you go down the periodic table, S fails to "keep pace" with Z, so Zeff goes up.

As you go up the periodic table, the electron is likely to be closer to the nucleus. Smaller principal quantum number means that the electron is closer to the nucleus. That's why the elements closer to the bottom of the table (ie. cesium) are more reactive: since their valence electrons are much further away from the nucleus and are easier to ionize.

Edit: As a sidenote, bleargh, your signature is one of the funniest things I've seen!

 

[wanderer]

As you go up the periodic table, the electron is likely to be closer to the nucleus. Smaller principal quantum number means that the electron is closer to the nucleus.

This does not contradict my earlier post. I merely said that the larger the shell the more likely a valence electron is going to be have the same radius as an electron in an inner shell (due to a larger distribution), thus leading to less shielding.

this is all wrong.

Zeff INCREASES the higher you are in the periodic table and INCREASES as you go to the right. Period. There is less distance and less electron shielding. All the periodic trends can be correlated to this.

Check your chemistry textbook; Zeff increases with increasing atomic number (in a given group).
The periodic trends are correlated to F= k *Zeff* e / r². The change in r is much more significant than the change in Zeff, which is why metals with higher atomic numbers are more reactive within the same group.

 

[RogueUnicorn]

this does not contradict my earlier post. I merely said that the larger the shell the more likely a valence electron is going to be have the same radius as an electron in an inner shell (due to a larger distribution).

Check your chemistry textbook; zeff increases with increasing atomic number (in a given group).
the periodic trends are correlated to f= k *zeff* e / r². The change in r is much more significant than the change in zeff, which is why metals with higher atomic numbers are more reactive within the same group.

no

 

[RogueUnicorn]

\
Edit: As a sidenote, bleargh, your signature is one of the funniest things I've seen!

why thank you.. i tried to come up with something clever instead of dark side.. but i failed..

 

[wanderer]

no

Great argument you have there. 🙄

 

[Elmnt]

Straight from my gen chem book.

"Going down a column, the effective nuclear charge experienced by valence electrons changes far less than it does across a row."

Which we all agree on, however...

"The effective nuclear charge INCREASES SLIGHTLY as we go down a family because larger electron cores are less able to screen the outer electrons from the nuclear charge."

Wanderer100 is correct

 

[RogueUnicorn]

no. if the effective nuclear charge increases down a family, Br would be more electronegative than F, S more than O, etc.

 

[Charles_Carmichael]

I don't think Zeff really increases when going down a family and if it does, it would be very slight increase. For example, Na has 11 protons and 11 electrons in a neutral atom. But there's only one valence electron and the other ten electrons are core electrons that contribute to shielding; each of those core electrons would cancel out a proton. So, Zeff for Na is 11-10 = 1. Then, for example, K, which has 19 protons and 19 electrons in a neutral atom. Once again, there's only one valence electron and the rest are core electrons contributing to shielding. So, Zeff for K is 19-18 = 1.

I guess if you used extremely sensitive machines to very accurately measure Zeff, there probably will be very small differences, to the point of them being pretty insignificant. But in general, the Zeff is pretty much the same for all atoms within a family. At least, this is how I learnt it. Hope this helps.

 

[Elmnt]

no. if the effective nuclear charge increases down a family, Br would be more electronegative than F, S more than O, etc.

So my book tells lies. 🙄

Ill post an explanation in a moment

 

[Elmnt]

I don't think Zeff really increases when going down a family and if it does, it would be very slight increase. For example, Na has 11 protons and 11 electrons in a neutral atom. But there's only one valence electron and the other ten electrons are core electrons that contribute to shielding; each of those core electrons would cancel out a proton. So, Zeff for Na is 11-10 = 1. Then, for example, K, which has 19 protons and 19 electrons in a neutral atom. Once again, there's only one valence electron and the rest are core electrons contributing to shielding. So, Zeff for K is 19-18 = 1.

I guess if you used extremely sensitive machines to very accurately measure Zeff, there probably will be very small differences, to the point of them being pretty insignificant. But in general, the Zeff is pretty much the same for all atoms within a family. At least, this is how I learnt it. Hope this helps.

I agree. As far as what you need to know for an MCAT, I think it is best to just assume that the Zeff is the same down a family which answers the OP question.

From what it seems in my text, the change in Zeff down the group is very minuscule but it does in fact increase down a group. It is due to the probability of core electrons shielding valence electrons. They are not as effective in doing so when n increases.

 

[CaliforniaKid]

As Atomic Radii increases, Z effective also increases.


Zeff = Z − S (Z is atomic number and directly proportional to Zeff).

keep it simple.

I Stand corrected here. After looking through my P Chem notes, here is what i have written down.

Z eff = Z (atomic #) - (# of core electrons) - (some shielding from valence)

What this means:

As you go from LEFT to RIGHT Z is increasing faster than the shielding and so Z effective increases.

As you go down Valence electrons are in shells with increasing n and valence electrons are further from the nucleus.
so Z effective decreases.

In summary
left to right = Z eff increases
going down = Z eff decreases

The above is correct and we can apply it to predict atomic radii.

Since from left to right Z eff increases, this means that electrons are held tighter and radii decrease.

since going down z eff decreases, electrons are held less tightly and lose, so the radii are larger.


Hope this helps.

 

[RogueUnicorn]

I agree. As far as what you need to know for an MCAT, I think it is best to just assume that the Zeff is the same down a family which answers the OP question.

From what it seems in my text, the change in Zeff down the group is very minuscule but it does in fact increase down a group. It is due to the probability of core electrons shielding valence electrons. They are not as effective in doing so when n increases.

how does this make sense? the core electrons are confined to the same orbitals regardless of however many n electrons you add on top of that.

 

[Elmnt]

how does this make sense? the core electrons are confined to the same orbitals regardless of however many n electrons you add on top of that.

Electrons don't just sit still and put up a fence around protons. They are moving non stop, sometimes into different energy levels. Thus sometimes they are able to shield valence electrons and sometimes the valence electrons are not shielded. When the atomic radius increases rapidly... like it does down a group... the ability of core electrons to shield the protons charge is diminished. Thus the Zeff is increased.

NOW, something that may help this make sense.

As the principal quantum number increases the orbitals become larger and the electron spends more time away from the nucleus. An increase in n also means that the electron has a higher energy and therefore less tightly bound to the nucleus.

I made that line bold to make it clear that electrons aren't bound to energy levels, they have the ability to make their way close to the nucleus.

All of this is based on radial probability density functions.


As far as the MCAT goes I found this link that gives similar advice as to assuming the Zeff is constant down a group.
http://www.eduturca.com/mcat-exam/periodic-trends-effective-nuclear-charge-z-or-zeff.html

 

[RogueUnicorn]

Electrons don't just sit still and put up a fence around protons. They are moving non stop, sometimes into different energy levels. Thus sometimes they are able to shield valence electrons and sometimes the valence electrons are not shielded. When the atomic radius increases rapidly... like it does down a group... the ability of core electrons to shield the protons charge is diminished. Thus the Zeff is increased.

NOW, something that may help this make sense.

As the principal quantum number increases the orbitals become larger and the electron spends more time away from the nucleus. An increase in n also means that the electron has a higher energy and therefore less tightly bound to the nucleus.

I made that line bold to make it clear that electrons aren't bound to energy levels, they have the ability to make their way close to the nucleus.

All of this is based on radial probability density functions.

these are for electrons ADDED. the electrons that were previously there maintain their same quantum number (e.g. n=2). did you even take quantum? electrons may not be "fenced in" but resting atom electrons do not jump shells, subshells, orbitals, etc.

 

[Elmnt]

these are for electrons ADDED. the electrons that were previously there maintain their same quantum number (e.g. n=2). did you even take quantum?

No I did not take a class on quantum besides what was covered in gen chem. But from the information gathered, I have a better understanding of this than you.

Let me state it again, as it is in the textbook.

"The effective nuclear charge increases slightly as we go down a family because larger electron cores are less able to screen the outer electrons from the nuclear charge."

If you do not agree with this statement, take it up Brown, LeMay Jr. Bursten, and Murphy. All with their Ph.D's in chemistry and likely understand this better than you and I combined.


Also see the edit above with the link

 

[wanderer]

no. if the effective nuclear charge increases down a family, Br would be more electronegative than F, S more than O, etc.

No. Electronegativity has to do with the Coulomb force on the outer electrons. The increase in Zeff going down a group does not compensate for the decrease of 1/r².
http://www.wou.edu/las/physci/ch412/Periodic%20trends/periodic_trends.htm
You see in that graph that a 2s orbital has two maxima, one of which is close in distance to the 1s maximum. The higher the principal quantum number the more maxima and the more likely an electron is going to be closer to the nucleus and therefore the nucleus will not be shielded by those electrons as efficiently.

 

[Elmnt]

No. Electronegativity has to do with the Coulomb force on the outer electrons. The increase in Zeff going down a group does not compensate for the decrease of 1/r².
http://www.wou.edu/las/physci/ch412/Periodic trends/periodic_trends.htm
You see in that graph that a 2s orbital has two maxima, one of which is close in distance to the 1s maximum. The higher the principal quantum number the more maxima and the more likely an electron is going to be closer to the nucleus and therefore the nucleus will not be shielded by those electrons as efficiently.

Thanks for clearing that up.

 

[RogueUnicorn]

As far as the MCAT goes I found this link that gives similar advice as to assuming the Zeff is constant down a group.
http://www.eduturca.com/mcat-exam/periodic-trends-effective-nuclear-charge-z-or-zeff.html

with regards to the MCAT Kaplan teaches the trend to go down the group i believe.

 

[RogueUnicorn]

No. Electronegativity has to do with the Coulomb force on the outer electrons. The increase in Zeff going down a group does not compensate for the decrease of 1/r².
http://www.wou.edu/las/physci/ch412/Periodic%20trends/periodic_trends.htm
You see in that graph that a 2s orbital has two maxima, one of which is close in distance to the 1s maximum. The higher the principal quantum number the more maxima and the more likely an electron is going to be closer to the nucleus and therefore the nucleus will not be shielded by those electrons as efficiently.

hmm now that i see the wave function graphs i think i'm wrong.
i will consult my p-chem bible and return when i find it..

 

[inaccensa]

😕