Consider the isoelectronic species, Na^(...

Consider the isoelectronic species, `Na^(+),Mg^(2+),F^(-)` and `O^(2-)`. The correct order of increasing length of their radii is:

`F^(-)ltO^(2-)ltMg^(2+)ltNa^(+)`

`Mg^(2+)ltNa^(+)ltF^(-)ltO^(2-)`

`O^(2-)ltF^(-)ltNa^(+)ltMg^(2+)`

`O^(2-)ltF^(-)ltMg^(2+)ltNa^(+)`

Text Solution

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The correct Answer is:

For isoelectronic species,ionic radii decrease with increase in nuclear charge(I.e.,no. of protons).Thus,the cation with greater +Ve charge will have a smaller radius and the anion with greater -Ve charge will have a larger radius.Thus,the correct order of increasing ionic radii is `Mg^(2+)ltNa^(+)ltF^(-)ltO^(2-)`

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The correct order of increasing radii are

`Be^(2+), Mg^(2+), Na^(o+)`

`K^(o+), Ca^(2+), S^(2-)`

`O^(2-), F^(ɵ), N^(3-)`

`S^(2-), O^(2-), As^(3-)`

Consider the isoelectronic species Na^+,Mg^+2,F^-and O^-2 the correct order of their increasing ionic radii is….

`F^-ltO^-2ltMg^+2ltNa^+`

`Mg^+2ltNa^+ltF^-ltO^-2`

`O^-2ltF^-ltNa^+ltMg^+2`

`O^-2ltF^-ltMg^+2ltNa^+`

For Na^(+),Mg^(2+),F^(-) and O^(2-) , the correct order of increasing ionic radii is :

`Mg^(2+)ltNa^(+)ltF^(-)ltO^(2-)`

`O^(2-)ltF^(-)ltNa^(+)ltMg^(2+)`

`Na^(+)ltMg^(2+)ltF^(-)ltO^(2-)`

`Mg^(2+)ltO^(2-)ltNa^(+)ltF^(-)`