Question

Can anyone pls...help me in understanding lanthanide contraction....???

Answer 1

@priyar pls...help me

Answer 2

@mayankdevnani pls...help

Answer 3

from lanthanum to hafnium there is a huge decrease in atomic radius. this is due to lanthanide contraction. this happens because of the poor shielding effect of f - orbitals..
do u know what is sheilding effect..?

Answer 4

Bt i found dat der has been written greater decrease than expected in d Wikipedia... Is dat true?

Answer 5

ya..thats what i meant when i said.."huge" decrease..this anomaly is due to the poor dhielding effect of f orbital..

Answer 6

Okk...bt if u check d radius of lanthanides n dat of normal period u will find that there is a gr8 diff in case of period elements rather than lanthanides...

Answer 7

what is ur doubt? i didn't get u..

Answer 8

U told dat der is larger decrease dan if we assumed der wr no lanthanides n if u check d periodic table n compare their radii u wud observe dat in a period there is more decrease dan in lanthanides....

Answer 9

^is this english?

Answer 10

i think priyar meant that the increase in radii is less than expected because of poor shielding by f electrons and as a consequence, the valence electrons experience a higher effective nuclear charge. All these concepts are explained in detail online, here's an example http://chemwiki.ucdavis.edu/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/aLanthanides%3A_Properties_and_Reactions/Lanthanide_Contraction

Answer 11

we study lanthanide contraction separately to understand the different placing of certain elements....for ex: Because of the lanthanoid contraction, heavier hafnium, which immediately follows the lanthanoids, possesses a radius nearly identical to the lighter zirconium.

Answer 12

But this is exactly what happens in general
I mean to say that across a period also there is a decrease in size of elements due to increasing effective nuclear charge.....n here also as we move towards right in lanthanide series ....there is a decrease in size.....

Answer 13

The key point is that the increase in radii IS LESS THAN EXPECTED. There is a constant increase in size, but this increase constant falls as we reach the lanthanides..



Ill give you an analogy. Say you always get A's, and so you expect A's, but after a test you get a C because you didn't study enough.

(here the mark is the change in radius, which fell short of expectations, and studying is the shielding by electrons)

Answer 14

But there is no increase in the radii of elements as we move across period...

Answer 15

as you move down a group

Answer 16

these are just two different ways of looking at the same thing. if you go across a period, then there is a larger decrease in atomic radius than what would be expected.

Answer 17

Lanthanides r in a group or in period

Answer 18

the lanthanides are in period 5 (horizontal direction) but are in several groups. my previous argument "The key point is that the increase in radii IS LESS THAN EXPECTED.." was directed at the comparison from the period above to the lanthanide period (i.e. down a group), as in Y compared to La.

Answer 19

Sorry if changing the point of view made it more confusing.

Please state any of your doubts clearly so we can help you more accurately.

Answer 20

@aaronq sir would you please let me know lanthanides are in a period or group.....
U said in 5th period....n then u said in group...

Answer 21

they are in both, independent of each other. you can think of them like coordinates in a cartesian plane (or battleship if that floats your boat (no pun intended)).




This isn't really all that relevant to the initial question.

Answer 22

Bt Si is not a lanthanide...

Answer 23

It was an example

Answer 24

Sir can u plss explain this question by giving an example of lanthanides and the radii comparison with the normal period elements ... ?

Answer 25

i'm not even sure what you're asking, "an example of lanthanides and radii comparison of normal elements" ?

look at this page, it has some good clear graphs to illustrate what I think you're asking
http://chemwiki.ucdavis.edu/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/4_f-Block_Elements/The_Lanthanides/aLanthanides%3A_Properties_and_Reactions/Lanthanide_Contraction

Answer 26

http://crystalmaker.com/support/tutorials/crystalmaker/atomic-radii/index.html

Answer 27

Sir wud u pls...go through this page it gives radii of all the elements

Answer 28

N as u said that there is a huge decrease in the radii of lanthanides when we compare them to normal period elements ....here you wud be able to see that it's not true since the decrease is more for nornal period elements

Answer 29

Normal*

Answer 30

Sir did u understand what my doubt is ???

Answer 31

I see what your doubt is, i think.

I took the data from there and graphed it, I labelled where the large decrease in radius actually is, what is termed "lanthanide contraction". It occurs between Ba and La.


You can see the changes from Sr to Y and Ca to Sc are pretty much constant, and thus as the radius is much less than expected (as defined by previous trends).

Answer 32

Here this graph shows that there is a huge difference in sizes between Ba and La which shows that there is a great decrease in lanthanides ...Is it ?

Answer 33

Bt sir i want among lanthanides ...u compared it with one lanthanide element and other non-lanthanide

Answer 34

Okk sir i will tell my doubt clearly..now ...
From the page i send u as reference u can see that difference in the radii of La n Ce is jst 0.10 n that between Li n Be is 0.55 which is greater than that in lanthanides....

Answer 35

that great decrease you just mentioned is called the lanthanide contraction, it's relative to other elements not between the lanthanides themselves.

amongst lanthanides you can see the pattern is not clear, its all scattered with varying differences, which \(likely\) has to do with electrons entering orbitals of different geometries and electron repulsion effects.

Right but you're not making a fair comparison. If you compare Li to Be, you have to compare Cs to Ba, which show approximately the same decrease in radius.

This is because the electrons are in s orbitals.

Answer 36

That's what i m asking with respect to which elements do we compare the lanthanides. . .
Okk..we can't compare the f block vid s block then whom to compare for lanthanides..

Answer 37

you compare them according to the previous trends, just look at the data graphically, it's pretty obvious that group 1 elements are the starting point for any comparison.

Answer 38

Group 1 or period 1...

Answer 39

V can compare them with any period or group as per our choice...
Is it d case with them?

Answer 40

group 1.

you can make any comparison you want, but that doesn't mean it's a valid comparison, nor that it really means anything. Look at the graph, label where the orbitals start and finish. you'll see why this is a valid comparison

Answer 41

Okk...so in ur diagram u r comparing the radii of lanthanides with Sr and Y
N Ca and Sc...

Answer 42

Yeah, i'm essentially comparing the "jump" from the s orbitals to the d orbitals in the case Sr-Y and Ca-Sc, then we would expect the same "jump" from Ba to La, but this "jump" is much larger than we expected.

Answer 43

That is absolutely correct with me. ..that means we need to make the comparison jst like that ....whenever n v don't have to make comparison with s block of 1st period...

Answer 44

Thank you so much sir....i really needed a help in this part ...seriously thanks a lot......

Answer 45

no problem! i do hope it's cleared up for ya

Answer 46

Yep sure...it's a great deal of time dat i wasted n even u too ...jst for the sake of letting me understand