Question

Question 3.36 The size of isoelectronic species — F–, Ne and Na+ is affected by

(a) Nuclear charge (Z )

(b) Valence principal quantum number (n)

(c) Electron-electron interaction in the outer orbitals

(d) None of the factors because their size is the same.

Class XI Classification of Elements and Periodicity in Properties Page 95

Answer 1

concept :-
we know,
$size \: \: of \: species \: \: \: \alpha\: \frac{1}{nuclear\: charge}$

The size of isoelectronic species; F^-, Ne and Na^+ is effected by nuclear charge (z).
if nuclear charge ( atomic number ) of species increases then the size of the these species decreases.

here, F^-, atomic number = 9
Ne, atomic number =10
Na^+, atomic number=11
hence, order of atomic number { e.g nuclear charge} is F^- < Ne < Na^+

so, order of size of these species is Na^+ < Ne< F^-

hence, option (a) is answer

Answer 2

Answer:

Explanation:

Well, the correct order should be: Ne > F- > Na+  

If you notice carefully all of these are isoelectronic species with a total of 10 electrons. Now, Ne is a noble gas therefore it’s radius is Van Der waal radius which is greater than ionic radius, so Ne has the greatest radius. Out of F- and Na+, since F gains one electron, the hold of F nucleus on outer shell electrons decreases per electron because the number of electrons in outer shell or the valence shell has increased, hence it expands whereas in case of Na, it loses 1 electron so the hold of nucleus on each outer shell electron increases (due to less outer shell electrons now) so it contracts which gives us this order.