If the ionic radii of K^{+} and F^{-} are nea | vedclass

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Question

(C) The atomic radius of a metal is always larger than its ionic radius because the loss of an electron results in a decrease in the number of shells

Vedclass · Accepted Answer

$1.96 \ \mathring{A}, 0.72 \ \mathring{A}$

Vedclass · Answer

(C) The atomic radius of a metal is always larger than its ionic radius because the loss of an electron results in a decrease in the number of shells or increased effective nuclear charge on the remaining electrons. Thus,the atomic radius of $K$ $(1.96 \ \mathring{A})$ is greater than the ionic radius of $K^{+}$ $(1.34 \ \mathring{A})$.Conversely,the atomic radius of a non-metal is smaller than its ionic radius because the gain of an electron increases inter-electronic repulsion,causing the electron cloud to expand. Thus,the atomic radius of $F$ $(0.72 \ \mathring{A})$ is smaller than the ionic radius of $F^{-}$ $(1.34 \ \mathring{A})$.Therefore,the atomic radii of $K$ and $F$ are $1.96 \ \mathring{A}$ and $0.72 \ \mathring{A}$ respectively.

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

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