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Which ion has the larger radius? Some sites say $r(\ce{Na+}) > r(\ce{Ca^2+})$. Any explanation would be appreciated.

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The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]:

\begin{array}{cc} \hline \text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \\ \hline \ce{Na+} & 4 & 1.13 & 0.99 \\ & 5 & 1.14 & 1.00 \\ & 6 & 1.16 & 1.02 \\ & 7 & 1.26 & 1.12 \\ & 8 & 1.32 & 1.18 \\ & 9 & 1.38 & 1.24 \\ & 12 & 1.53 & 1.39 \\ \hline \ce{Ca^2+} & 6 & 1.14 & 1.00 \\ & 7 & 1.20 & 1.06 \\ & 8 & 1.26 & 1.12 \\ & 9 & 1.32 & 1.18 \\ & 10 & 1.37 & 1.23 \\ & 12 & 1.48 & 1.34 \\ \hline \end{array}

where C.N. - coordination number; C.R. - crystal radius; I.R. - ionic radius. The data is also available online for free. Comparing similar coordination environments, it's easier to notice that overall $\ce{Na+}$ is indeed larger than $\ce{Ca^2+}$:

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Reference

  1. Shannon, R. D. Acta Cryst A, 1976, 32 (5), 751–767 DOI: 10.1107/S0567739476001551.
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    $\begingroup$ If there is no such thing as a non-integer CN then drawing any lines is strictly speaking wrong... apart from that this is a really useful answer! $\endgroup$ – Floris Oct 17 '17 at 1:33
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According to the diagonal relation, $\ce{Na}$ and $\ce{Ca}$ have similar radii.

$\ce{Na+}$ and $\ce{Ca+}$ have thus similar radii as well but, when one more electron is removed from $\ce{Ca+}$, making it $\ce{Ca^{2+}}$, its effective nuclear charge increases, which will make its radius decrease.

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