QJEEM04Ch0fec: JEE Main 2004 Chemistry - Class 11 JEEM04Ch0fec Solution: Which one of the following ions has the highest value of ionic…

From a neutral atom where a electron is removed then it become cation and when a electron is added to the atom then it becomes anion.

For a atom $x$ when it loose a electron then it becomes cation $x - {e^ - } \to {x^ + }$

Let atomic no. of $x$ is $z$

then no. of proton in ${x^ + } = z$

and no. of electron in ${x^ + } = z - 1$

So the ratio ${z \over e}$ as no. of electrons decreases. And when no of electron decreases then per electron attraction increases from nucleus and radius of cation deceases.

Similarly when atom $x$ becomes ${x^ - }$ by adding a electron $\left( {x + {e^ - } \to {x^ - }} \right)$ then no. of proton in ${x^ - } = z$ and no. of electron in ${x^ - } = z + 1.$

Now the ratio of ${z \over e}\,\,$ in ${x^ - }$ decreases as no. of electron increases. As electron increases in ${x^ - }$ then attraction from nucleus decreases on per electron and the distance between electron and nucleus increases so the radius also increase in anion.

All $4$ ions are from second period and for ${O^{2 - }}\,\,$ and ${F^ - }$ both have $1s,$ $2s$ and $2p$ shell and for $L{i^ + }\,\,$ and ${B^{3 + }}$ both have $1s$ shell.

So, size of ${O^{2 - }}\,\,\,$ and $\,\,\,{F^ - }\,\,\,$ are more then $L{i^ + }\,\,\,\,$

and $\,\,\,\,{B^{3 + }}$ and among ${O^{2 - }}\,\,\,$ and $\,\,\,\,{F^ - }$

For $\,\,\,$ ${O^{2 - }},\,\,{z \over e} = {8 \over {10}} = 0.8$

For $\,\,\,$ ${F^ - },\,\,\,{z \over e} = {9 \over {10}} = 0.9$

as for ${O^{2 - }},\,\,{z \over e}\,\,\,$ is less so per electron attraction from nucleus decreases and radius increases more than ${F^ - }.$