Which is the correct order of size of {O}, {O}²-}, {F}^{-} and {F} ?

Correct Answer is : O2−>F−>O>F{O}²-} > {F}^{-} > {O} > {F}O2−>F−>O>F

Correct Answer is : O2−>F−>O>F{O}²-} > {F}^{-} > {O} > {F}O2−>F−>O>F

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UPSEE 2019 Solved Paper

Question : 91 of 150

Marks: +1, -0

\mathrm{O}

\mathrm{O}^{2-}, \mathrm{F}^{-}

\mathrm{F}

Solution:

Radii of anion is more than neutral element. Here,

\mathrm{O}^{-2}

has

10 e^{-}

and

\mathrm{F}

has

10 e^{-}

for isoelectronic species radius increases as

\frac{e^{-}}{p}

ratio increases.

\because \mathrm{O}^{-2} \frac{e}{p} = \frac{10}{8}

\therefore \mathrm{F}^{-} \frac{e}{p} = \frac{10}{9}

So, radius order is

\mathrm{O}^{2-} > \mathrm{F}^{-}

On moving left to right in a period, effective nuclear charge increases as radius decreases. So, order of size is

\mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{O} > \mathrm{F}

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