A) \[{{O}^{2-}}>{{F}^{-}}>O>F\]
B) \[{{O}^{2-}}>{{F}^{-}}>F>O\]
C) \[{{F}^{2-}}>{{O}^{2-}}>F>O\]
D) \[{{O}^{2-}}>O>{{F}^{-}}>F\]
Correct Answer: A
Solution :
Note: In a period from left to right, as the atomic number increases, atomic radii decreases. Therefore, in F and O, oxygen has higher value of atomic radii than fluorine. \[\text{Radii}\propto \frac{\text{1}}{\text{Effective}\,\text{nuclear}\,\text{charge}}\] Hence, anions have bigger size than parent atom as the effective nuclear charge decrease by increasing the -ve charge. So, \[{{\text{O}}^{2-}}\]has greater size than \[{{\text{F}}^{-}}\] Thus, the correct order is: \[{{O}^{2-}}<{{F}^{-}}>O>F\]
You need to login to perform this action.
You will be redirected in
3 sec
