Why is there sudden increase in atomic radii when moving from neon to sodium?

Discussion in 'Chemistry' started by ash64449, Jun 1, 2013.

1. ash64449Registered Senior Member

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Hello friends,here is a question from the periodic table!!

I know that across the period atomic radii decreases and down the group atomic radii increases.

Across the period,Shielding Effect is very less from sub-orbitals and as a result it doesn't compensate the increasing nuclear charge as atomic number increases and as a result atomic radii decreases across the period.

Down the group, Shielding Effect is more effective because normally there are completely filled inner electrons particularly in s-block elements as a result nuclear charge experienced by the valence electrons in very low. So added electrons move to new energy level. So atomic radii increases down the group.

But think of the Neon which has atomic radii of 38 pm. But when we increase the atomic number by 1,it becomes Sodium and a sudden increase in atomic radii to 180 pm.

How can this trend be explained?

I know that as we go to sodium, added electrons move to new energy level.But i cannot understand how that can explain the sudden increase in atomic radii. Are there any additional effects accompanying? OR in other words,i want to know why there is sudden increase in Shielding Effect.

3. exchemistValued Senior Member

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Ash, it's because you are starting a new shell with new principal quantum number, n. Each principal shell starts some way out beyond the previous one. It's a whole new series of solutions of the wave equation, i.e. a different series of "harmonics" of the "vibration" of the electrons in the atom. You take a step up in energy, which is then modified in various ways as you go across the succeeding period (increasing nuclear charge, shielding vs.penetration of the various sub-shells and so so on.) Think of the line spectrum of hydrogen: Lyman, Balmer, Paschen, Pfund, Brackett etc.

5. ash64449Registered Senior Member

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So the Short answer is while electrons entering the new principal quantum number,It is a whole series of new solutions which is accompanying,making a new start in chemical and physical properties.(i mean in each period,the variation in physical and chemical properties takes place in same pattern).

How does the line spectrum of hydrogen help me to understand this concept?

7. wellwisherBannedBanned

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Another way to think of this is sodium is a very reactive element and wants to lose an electron. This reactivity implies that extra electron is not being held very tightly and therefore is farther away from the nucleus. Neon is not reactive so the electron is tighter and closer.

The reason has to do with the electromagnetic force. The electro-static aspect of the EM force, has a balance between protons and electrons. But the magnetic force aspect of the EM force is somewhat repulsive sending the electron off where it will leave easier.

If we have two wires with current going in opposite directions, these will attract, even though both have negative charge in motion. The magnetic aspect is very strong causing all the electrons to attract; two wires attract. With sodium we add a third wire winding, which repels one of the two wires and attracts the other wire. The best place to optimize or minimize energy is to move that third wire away.

The magnetic is very important and is why oxide or O-2 is stable even with too many electrons.

8. exchemistValued Senior Member

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Yes, the start of each new period represents electrons starting to fill a new principal shell, which has a higher energy than the one that has just been filled, due to being at a greater average distance from the nucleus.

Re the hydrogen emission spectrum, this may or not be helpful, depending on how much you know about it. Each series represents electrons making transitions to different principal shells:

Ultraviolet: Lyman series, representing transitions to n=1

Visible: Balmer, transitions to n=2

Infra red: Paschen: transitions to n=3

Far IR: Brackett: transitions to n=4

This illustrates that the greatest energy is given off by electrons returning to n=1, less by those returning to n=2 and so on, and thus shows that the shells have higher energy as n increases. In principle, higher energy has to mean that electrons in these shells have either or both of more kinetic energy and/or more potential energy. More potential energy implies a greater distance from the nucleus and this is what we find in practice. The spectral series converge to a limit, beyond which there is a continuum (i.e no more quantisation of energy levels). This corresponds to ionisation of the electron from the atom - it ceases to be bound at all and can wander off to any distance it likes from the nucleus.

9. ash64449Registered Senior Member

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Yes.higher energy level indicate higher potential energy of the electrons and that indicate higher distance from the nucleus.

In My Textbook,it is given that Electrons enter into new energy level despite increase in nuclear charge because:

inner energy levels are filled with electrons,which serve to shield the outer electrons from the pull of the nucleus.

My actual question was why was there some spontaneous increase in Shielding effect and as a result lead that electron to recede in new principle quantum number?

Now i think i have found a solution to this but i don't know whether it is 100% correct. So i need your help to confirm whether it is right or not.

I think that after the noble gas configuration,when we go one step by adding one proton,the electron that is added cannot accommodate the inner shell as a result,it is forced to move a little bit higher(because Of Shielding Effect). But now only certain fixed orbitals are allowed for electrons and as a result it moves to higher energy level. So as a result there is sudden increase in energy level despite Shielding effect was small. What do you think?

10. exchemistValued Senior Member

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Ah, now I see your problem. No, it's the Pauli Exclusion Principle.You can't have 2 fermions in the same state. So once you have 2 electrons with opposite spin in each orbital, they are full and the next electron MUST go into an empty one, even if the next available orbital is of significantly higher energy. This is how the Aufbau Principle works.

11. araucaBannedBanned

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Let see if this might help ? The proton in the nucleus charge decrease with distance . The nuclei is a fixed size as you add electron clouds the distance if next electron to the nucleus increases and the strength between nucleus and electon decreases.

12. ash64449Registered Senior Member

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Ok. Sorry,I didn't study Pauli Exclusion Principle. Thank you for the information!!

Now let me summarize: In Neon's Electronic configuration,all electrons are fully filled ,each orbitals already accommodated all the two electrons.

So when we increase the atomic number,New electron cannot accommodate inner shell,So it is forced to go to new shell and new shell is little bit far away from the inner shell and that is the reason for increase in atomic radii.

Correct?

13. ash64449Registered Senior Member

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I would like to ask questions to everyone.

Does Stable Electronic Configuration can do any changes in atomic radii?

I am not talking about noble gas configuration. I mean i saw different changes taking place in d-orbitals. I think it is to provide stable arrangement(or symmetric arrangement)

14. exchemistValued Senior Member

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Yes, that's it exactly.

15. exchemistValued Senior Member

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Ash, I don't understand your question, I'm afraid. Can you rephrase it?

16. exchemistValued Senior Member

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Ash, I don't understand your question, I'm afraid. Can you rephrase it?

17. ash64449Registered Senior Member

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OK.

For Example, take the case of Chromium. It's sub shell electronic configuration is $[Ar]3d^5 4s^1$
But if we follow the pattern of arrangement in terms of periodic table, we should actually get $[Ar]3d^4 4s^2$

But it is arranged in the way of the first one. Because the arrangement of electrons like $3d^5$ is stable electronic configuration. Elements always want stability.so they are arranged like this.

My question is does arranging electrons in $[Ar] 3d^5 4s^1$ have different atomic radii than arranging electrons in $3d^4 4s^2$ for the same element?

18. ash64449Registered Senior Member

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Making an another thread for this question seems not right. Since it is related to understanding the periodic table,I would like to ask here to everyone:

Does ionization enthalpy mean energy required to remove the outermost electron or the valence electron? I am asking this because:
In d-block elements, valence electrons are present in penultimate d-orbital.But outer most electron present in s-orbital. So which electron we mean?

According to me,i think it must be valence electrons because they are the one that participate in chemical reactions. Valency of d-block(don't know) according to me depends on inner d-orbital. But i have heard they exhibit variable oxidation state. But i don't know how to calculate that variable oxidation state by looking periodic table.

19. exchemistValued Senior Member

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That's interesting. I had to look this up. I see the ground state of V is [Ar] 3d4 s2, whereas, increasing atomic number by one, Cr has indeed a ground state configuration of [Ar] 3d5 4s1, as you say. The 3d and 4s orbitals are very close in energy at this point in the Periodic Table. The reason why Cr prefers the configuration with 2 unpaired electrons is presumably because the orbitals are so close in energy that the mutual repulsion created by pairing 2 electrons in 4s is enough to make the unpaired configuration the more stable of the two.

By the way, you might be interested to read the attached article on the subject of the "myth" of the special stability of half-filled and filled subshells.

As to the atomic radius of the marginally less stable [Ar] 3d4 4s2 configuration, I must say I don't know the answer to this. I presume not much different, since the energy of this configuration will be very close to that of the actual ground state configuration.

20. exchemistValued Senior Member

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I think the term "valence electron" probably needs to be used with care. All it really means is an electron that can take part in chemical bonding. This can apply to more than one, of course. But there is only one truly outermost electron, the removal of which corresponds to the 1st ionization energy. As to which subshell this electron comes from, I'm not sure how to answer this, because the configuration of the resulting ion will be determined in a new way, due to the excess nuclear charge that is now present, which will modify the energy levels of all the orbitals, so some rearrangement may take place. It's a good question: I don't think I'd ever thought about this before!

N.B. In the transition elements, both s and d electrons can take part in bonding. For example, the highest oxidation states of Ti, V and Cr involve all the 3d and 4s electrons.

21. ash64449Registered Senior Member

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Exchemist,i have noticed this one strange trend:

Cr=140 pm.
Mn=140 pm!!! (both elements are present in same period. Mn is the succeeding element of Cr)

Can you explain this trend?

22. ash64449Registered Senior Member

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Sorry, I know that there are sub shells within sub shells. but i will study them later when my teacher teaches them. Can you write the Sub shell configurations of 3d5 4s1?

Sorry,i am unable to open the link. Care to post it again?

What change would you expect if the energy of that type of configuration is not close to actual ground state configuration?

23. ash64449Registered Senior Member

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So it is not possible to say which electron will be removed when 1st ionization enthalpy takes place?

I don't understand.. Doesn't rearrangement take place after electron is removed as a result why can't we determine?

I mean ionization enthalpy to remove inner electrons is greater than outermost electron.(in case of transition elements,i think inner electrons require less energy to remove than outer electrons) Correct?