The ionic radii of \(\mathrm{F^-}\) and \(\mathrm{O^{2-}}\) respectively are \(1.33\ \mathrm{\AA}\) and \(1.4\ \mathrm{\AA}\), while the covalent radius of N is \(0.74\ \mathrm{\AA}\).
The correct statement for the ionic radius of \(\mathrm{N^{3-}}\) from the following is :
The correct statement for the ionic radius of \(\mathrm{N^{3-}}\) from the following is :
Show Hint
- It is smaller than \(\mathrm{F^-}\) and N
- It is bigger than \(\mathrm{O^{2-}}\) and \(\mathrm{F^-}\)
- It is bigger than \(\mathrm{F^-}\) and N, but smaller than of \(\mathrm{O^{2-}}\)
- It is smaller than \(\mathrm{O^{2-}}\) and \(\mathrm{F^-}\), but bigger than of N
The Correct Option is B
Solution and Explanation
For isoelectronic species, radius decreases as nuclear charge increases: \(\mathrm{N^{3-} \gt O^{2-} \gt F^-}\).
Also, an anion is larger than its neutral atom, so \(\mathrm{N^{3-}}\) is bigger than N (0.74 \(\mathrm{\AA}\)); its actual ionic radius is about 1.71 \(\mathrm{\AA}\).
So \(\mathrm{N^{3-}}\) is bigger than both \(\mathrm{O^{2-}}\) and \(\mathrm{F^-}\).
Answer: (2) It is bigger than \(\mathrm{O^{2-}}\) and \(\mathrm{F^-}\)
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