Since fluorine and potassium form ions to gain the electronic configuration of neon and argon respectively, then shouldn’t $\ce{K+}$ have a greater radius, as argon is larger than neon?

Also which would have a bigger atomic radius, potassium or fluorine?

The original question is

The ionic radii of $\ce{K+}$ and $\ce{F-}$ are nearly the same (1.34), then the atomic radii of $\ce{K}$ and $\ce{F}$ will be

A) 1.34, 1.34

B) 0.72, 1.96

C) 1.96, 0.72

D) 1.96, 1.34

All units in Angstrom ($\pu{10^{-10} m}$).

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    $\begingroup$ "All units in Armstrong." No, those look more like Ångström. And the question is impossible to answer without eliminating the listed choices (as both radii are typically determined experimentally), which makes it ill-posed IMO. $\endgroup$ – andselisk Aug 30 '19 at 15:50
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    $\begingroup$ Relax people! It was the auto-correct. I failed to notice it. Sorry for bother you with a typo. I assume it was obvious that I meant Angstrom. No need to get so triggered over it. $\endgroup$ – Aditya Aug 30 '19 at 16:32
  • $\begingroup$ @andselisk it’s really not advanced chemistry. I think it can found at easily, since the one with the smaller size will have lesser radius. It being an MCQ just makes it painfully obvious. All I have a problem is why K+ is smaller than F- $\endgroup$ – Aditya Aug 30 '19 at 16:35
  • $\begingroup$ "…since the one with the smaller size will have lesser radius" — this statement is both tautologic and sophistic. Nothing really is easy if you look deeper, and potassium cation is not smaller than fluorine anion (in terms of ionic radii). $\ce{K+}$ with C.N. 4 is comparable in size with $\ce{F-},$ and then the values of ionic radius for higher C.N. skyrocket to 1.64 Å for C.N. 12. $\endgroup$ – andselisk Aug 30 '19 at 16:55
  • $\begingroup$ @andselisk I don't think the examiner is really concerned about the real values (pfft, haha, who is? Those weirdos . . . chemists). OP This question does strike me as pretty trivial though. What you're inquiring about is already addressed when a reference point is given: The ionic radii. It's really obvious and if you think about it, you just missed it. There's really nothing to be learned here. $\endgroup$ – It's Over Aug 31 '19 at 15:37

Why do K+ and F− ions have the same ionic radii?

They don't according to wikipedia:

enter image description here

Source: https://commons.wikimedia.org/wiki/File:Atomic_%26_ionic_radii.svg

They do according to "Data taken from John Emsley, The Elements, 3rd edition. Oxford: Clarendon Press, 1998."

enter image description here

Source: https://www.angelo.edu/faculty/kboudrea/periodic/trends_atomicradius.htm

As @andselisk wisely mentions in his comments, radii depend on coordination number (and on how distance between anion and cation are turned into a single radius - you have to set one radius somewhat arbitrarily).

The original question

Irrespective of the actual numbers, one can see nicely in the first figure that monovalent cations are much smaller than the corresponding atom (one less shell filled), and monovalent anions are somewhat larger than the corresponding atom (one last electron placed into the outer shell). Answer c) is the only one that follows the expected pattern.

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    $\begingroup$ Question has been edited, please have a look again, Since the context was missing $\endgroup$ – Aditya Aug 30 '19 at 15:46
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    $\begingroup$ It's a common misconception to assign a single ionic radii to the ion. In general it works OK, but for some ions (such as potassium, actually), ionic radius varies a lot with coordination number. From 97th edition of CRC Handbook of Chemistry and Physics, p. 12-12 $R_\mathrm{i}(\ce{F-}, \mathrm{C.N.}~6) = \pu{1.33 Å}$ and $R_\mathrm{i}(\ce{K+}, \mathrm{C.N.}~4) = \pu{1.37 Å}$ which is indeed pretty close. $\endgroup$ – andselisk Aug 30 '19 at 15:47
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    $\begingroup$ Answer has been edited, please have a look again (@andselisk). $\endgroup$ – Karsten Theis Aug 30 '19 at 16:12
  • $\begingroup$ I would like to know the reason. Why are monovalent cations smaller than the their orignal atom $\endgroup$ – Aditya Aug 30 '19 at 17:18
  • $\begingroup$ @Aditya Look at the electron configuration of the atom, [Ar]4s1, and of the ion, [Ar]. So the atom uses the 4th shell, and the ion has 3rd shell as the highest, with the same number of protons in the nucleus. That explains the big difference. $\endgroup$ – Karsten Theis Aug 30 '19 at 17:27

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