Question

If the ionic radii of $K^{+}$ and $F^{-}$ are $1.34\stackrel{\circ }{\text{A}}$ each, the expected value of atomic radii of $K$ and $F$ should be:

• A. $1.34\stackrel{\circ }{\text{A}}\text{and}1.34\stackrel{\circ }{\text{A}}$
• B. $2.31\stackrel{\circ }{\text{A}}\text{and}0.64\stackrel{\circ }{\text{A}}$
• C. $0.64\stackrel{\circ }{\text{A}}\text{and}2.31\stackrel{\circ }{\text{A}}$
• D. $2.31\stackrel{\circ }{\text{A}}\text{and}1.34\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is B $2.31\stackrel{\circ }{\text{A}}\text{and}0.64\stackrel{\circ }{\text{A}}$

The Cation having a greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Whereas, anion with the greater negative charge will have the larger radius, as the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size.
Therefore, atomic radii of $K$ will be greater than $K^{+}$ and atomic radii of $F$ will be less than $F^{-}$.
So, the correct answer is $\left(b\right)$.