JEE MAIN - Chemistry (2015 (Offline) - No. 21)
The ionic radii (in Å) of N3–, O2– and F– are respectively:
1.36, 1.71 and 1.40
1.71, 1.40 and 1.36
1.71, 1.36 and 1.40
1.36, 1.40 and 1.71
Selitys
Here all of them are isoelectric and for isoelectric species size of anion increases as negative charge increases.
So, correct order is :
N3– > O2– > F–
$$ \therefore $$ Radius of N3– = 1.71 Å
and Radius of O2– = 1.40 Å
and Radius of F– = 1.36 Å
So, correct order is :
N3– > O2– > F–
$$ \therefore $$ Radius of N3– = 1.71 Å
and Radius of O2– = 1.40 Å
and Radius of F– = 1.36 Å