Periodic Classification and Periodic Properties 2 Question 6
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6. The ionic radii (in $\AA$ ) of $\mathrm{N}^{3-}, \mathrm{O}^{2-}$ and $\mathrm{F}^{-}$respectively are
======= ####6. The ionic radii (in $\AA$ ) of $\mathrm{N}^{3-}, \mathrm{O}^{2-}$ and $\mathrm{F}^{-}$respectively are
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $1.36,1.40$ and 1.71
(b) $1.36,1.71$ and 1.40
(c) $1.71,1.40$ and 1.36
(d) $1.71,1.36$ and 1.40
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Answer:
Correct Answer: 6. (c)
Solution:
- Number of electrons in $\mathrm{N}^{3-},=7+3=10$
Number of electrons in $\mathrm{O}^{2-}=8+2=10$
Number of electrons in $\mathrm{F}^{-}=9+1=10$
Since, all the three species have each 10 electrons, hence they are isoelectronic species.
Periodic Classification and Periodic Properties 43
It is considered that, in case of isoelectronic species as the negative charge increases, ionic radii increases and therefore the value of ionic radii are
$$ \begin{aligned} & \mathrm{N}^{3-}=1.71 \ & \mathrm{O}^{2-}=1.40 \end{aligned} \quad \mathrm{~F}^{-}=1.36 \text { (lowest among the three) } $$
Time Saving Technique There is no need to mug up the radius values for different ions. This particular question can be solved through following time saving.
Trick The charges on the ions indicate the size as $\mathrm{N}^{3-}>\mathrm{O}^{2-}>\mathrm{F}^{-}$. Thus, you have to look for the option in which the above trend is followed. Option(c) is the only one in which this trend is followed. Hence, it is the correct answer.
