Question

Arrange the following ions in decreasing oder of thier site N3-;O2-;Na+F-Mg2+Al3+

Answer 1

Hello
-ve sign means more of electrons and +ve sign means lack of electrons.
Here N3- with atomic no. 7 has 7+3= 10 electrons
O2- witha tomic no. 8 has 8+2 = 10 electrons
F- with atomic no. 9 has 9+1= 10 electrons
Na+ with atomcatomic no. 11 has 9-1= 10 electrons
Mg2+with atomcatomic no. 12 has 12-2= 10 electrons
Al3+with atomcatomic no. 13 has 13-3= 10 electrons
Since all these ions have 10 electrons in their shell therefore these areisoelectronic speicies
The more + the charge, the smaller the ionic radius. Remember that - means adding electrons. These electrons go in the outermost shells. Also, when an atom loses electrons, it clings ever more tightly to the ones it has left, further reducing the ionic radius. therefore the order of ionic radii will be →

AL3+ Mg2+ Na+ F- O2- N3- (increasing order)

3 years ago

For finding which of these(C ^4- ,N^3-, O^2-,F-) has more bigger size ,first we have to find( Z/E) .

   Where Z is atomic number and E is the present number of electrons in the atom.

  for C^-4 (atomic no=6 (Z) , present no of electrons is =6+4(E) .

   Therefore Z/E of C^4- = 6/10 = 0.6 

  Now same for every ISO electron 

   lesser is the Z/E of an atom more bigger 

  Is its size .

  E.g :Z/E of O^2- = 0.8 

    Z/E of N^3- =0.7 

   Z/E of C^4- =0.6

     Since 0.6

    Therefore C^4->N^3->O^2- (size)

 Hope this may help you.

 

 

Answer 2

~*HEY FRIEND*~
~* THE ANSWER OF YOUR QUESTION IS..*~

✍✍✍✍✍✍✍✍________________________________________
all the above ions are
ISOELECTRONIC

and size order in isoelectronic is directly proportional to number of negative ion and inversly proportional to the no. of positive ion

so , in the above question size order will be

N-3 > O-2 > F- > Na+ > Mg+2 > Al+3

HOPE IT WILL HELP YOU