They will both have the same number of valence electrons, but $\ce{F-}$ is more electronegative. Why is it not the case that $\ce{Mg^2+}$ has a bigger radius than $\ce{F-}$?
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$\begingroup$ Strongly related if not dupe: chemistry.stackexchange.com/q/64335 $\endgroup$– JanJan 18, 2017 at 0:20
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2 Answers
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These two species are isoelectronic (have the same number of electrons). In fact, they're electronically identical with the except that the magnesium ion has 3 more protons. Since all other electronic factors are equal, the net effect is that the three extra protons pull all of the electrons in closer to the nucleus. I think you understand this part.
Electronegativity is not correctly used in this context. One way to think about it is that you should only be using that concept on a neutral atom. The fact the fluoride anion is negative and magnesium cation is positive means you've already applied electronegativity to the system in order to ionize them.
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$\begingroup$ So because it has more particles in pulling in the e- than Mg+ will have a smaller radius! Thanks I understand this now. $\endgroup$ Jan 17, 2017 at 15:16
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$\begingroup$ so is this also why is the 1st ionisation energy of Na+ bigger than Ne (more protons than Ne so bigger attraction and capability to gain an electron than Ne) $\endgroup$ Jan 17, 2017 at 15:19
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$\begingroup$ Correct with regard to $\ce{Na+}$ vs. $\ce{Ne}$. $\endgroup$– ZheJan 17, 2017 at 15:56
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The number of electrons decreases by 2 in $\ce{Mg^2+}$ compared to neutral Mg atom, but the number of protons inside the nucleus remains the same. The effective nuclear charge hence increases causing the electrons to be pulled towards the nucleus, thus, smaller ionic radius.
But, in case of $\ce{F^-}$ one electron is extra compared to neutral F atom but the number of protons inside the nucleus remains the same. The effective nuclear charge hence decreases causing the electrons to move away from the nucleus, thus, larger ionic radius.
One very important point to be noted is that in case of $\ce{Mg^2+}$, the the outermost orbital n is 2 instead of 3 in case of neutral Mg atom. This absence of the outermost orbital significantly reduces the ionic radius of $\ce{Mg^2+}$.
answered Jan 17, 2017 at 15:20