The ionic radii of `N^(3-), O^(2-)` and `F^(-)` are respectively given by:

Options

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

Answer

C

Explanation

Number of electrons in `N^(3-), =7+3=10`
Number of electrons in `O^(2-)=8+2=10`
Number of electrons is `F^(-)=9+1=10`
Since, all the three species have each 10 electrons, hence they are isoelectronic species.
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
`N^(3)=1.71` (highest among the three)
`O^(2-)=1.40`
`F^(-)=1.36` ( 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 `N^(3-) gt O^(2-) gt 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.