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

Similar Questions

Which of the following represents the correct order of ionic radius?

In the isoelectronic species,the ionic radii in $\mathring{A}$ of $N^{3-}$,$O^{2-}$,and $F^{-}$ are respectively given by:

Assertion $(A)$: The ionic radii of $Na^{+}$ and $F^{-}$ are same.
Reason $(R)$: Both $Na^{+}$ and $F^{-}$ are isoelectronic species.
The correct answer is:

The correct sequence which shows the decreasing order of the ionic radii of the elements is:

Which of the following is the correct increasing order of ionic radii for $Na, Rb, K,$ and $Mg$?

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