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Dear Sirs
 
The question is arrange the given in increasing ionic radii.
Na+,   Li+,  Mg2+,    Be2+ (The given sequence is actually also the ansewer).
 
While comparing Na+ and Mg2+ I know that Na+ is bigger as both are isoelectronic so Na+ (less atomic no.) will be greater.Same is the case with Li+ and Be2+ .  But the problem arises when i compare Li+ and Mg2+.    Although the correct answer says Li+  >  Mg2+ but shouldn’t Mg2+ be biggger as it has one EXTRA shell with it..
Please clarify
 
Thanks
 

Ajitesh Garg , 9 Years ago
Grade 11
anser 3 Answers
Harishwar

Last Activity: 9 Years ago

Hi Student
For isoelectronic species the cation with high positive charge has less atomic radii because of increase in the Z effective from the nucleus

Shiv Shukla

Last Activity: 9 Years ago

thanks sir for the answer. But  i already know how to compare isoelectronic species . What i want to know is that how to compare Mg2+(10e) and Li+(2) in terms of ionic radii. they are not isoelectronic .
Please clarify sir

Neeti

Last Activity: 9 Years ago

see, these are exceptions kind of. Li is in the second period and we know that second period elements show anomalous behaviour. you are right in saying that Mg has an extra shell but you should also consider the electronic configuration, position of the element etc. we know that second period elements are smaller than first period elements and 2 electrons are removed from Mg which actually makes it extra small, and smaller than Li.

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