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JEE Main 20154 April
Exam
JEE Main
Year
2015
Date
4 April
Subject
Chemistry
Chapter
Classification of Elements and Periodicity in Properties

JEE Main 2015 · 4 April · Chemistry · Classification of Elements and Periodicity in Properties

The ionic radii ( in Å ) of N ^ 3 - , O ^ 2 - and F ^ - are...

The ionic radii (inÅ)( \text{in} Å ) of N3−,O2−{N}^{3 -}, {O}^{2 -} and F−{F}^{-} are respectively:

Options

  1. A1.71,1.36and1.401 . 71, 1 . 36\text{and}1 . 40
  2. B1.36,1.40and1.711 . 36, 1 . 40\text{and}1 . 71
  3. C1.36,1.71and1.401 . 36, 1 . 71\text{and}1 . 40
  4. D1.71,1.40and1.361 . 71, 1 . 40\text{and}1 . 36

Correct answer

D. 1.71,1.40and1.361 . 71, 1 . 40\text{and}1 . 36

Step-by-step solution

As Ze↑\dfrac{\text{Z}}{\text{e}} ↑ ionic radius decreases for isoelectronic species.

Z=numberofprotonse=numberofelectronsZ = numberofprotonse = numberofelectrons

N3−(Ze)=710{\text{N}}^{3 -} \left(\dfrac{\text{Z}}{\text{e}}\right) = \dfrac{7}{1 0}

O2−(Ze)=810{\text{O}}^{2 -} \left(\dfrac{\text{Z}}{\text{e}}\right) = \dfrac{8}{1 0}

F−(Ze)=910{\text{F}}^{-} \left(\dfrac{\text{Z}}{\text{e}}\right) = \dfrac{9}{1 0}

Trend of ionic radius: N3−>O2−>F−{\text{Trend of ionic radius: N}}^{3 -} > {\text{O}}^{2 -} > {\text{F}}^{-}

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