Question

If the ionic radii of $K^{+}$ of $F^{-}$ are nearly the same (i.e. $1.34A^{\circ }$), then the atomic radii of $K$ and $F$ respectively are:

• A. $1.34A^{\circ },1.34A^{\circ }$
• B. $0.72A^{\circ },1.96A^{\circ }$
• C. $1.96A^{\circ },0.72A^{\circ }$
• D. $1.96A^{\circ },1.34A^{\circ }$

Answer 1

The correct option is C $1.96A^{\circ },0.72A^{\circ }$

Given ionic radii of $K^{+}$ and $F^{-}$ $1.34\stackrel{o}{A}$, Atomic radii of $K$ and $F$ respectively would be $1.96\stackrel{o}{A}$, $0.72\stackrel{o}{A}$. Because $F$ is smaller in size than $K$. And when any atom looses $e^{-}$ its' radius decreases, so as for $K^{+}$. So, $K$ has more radius value than $K^{+}$

When, an $e^{-}$ is added in an atom its' radius increases which increases, which is the case of $F^{-}$. So, $F$ has less radius value than $F^{-}$.