Exercises — Ionic radius — cation - parent atom, anion - parent atom; isoelectronic series

Rule: lose electrons → smaller cation; gain electrons → larger anion (proton count never moved). (a) ${\text{K}}^{+}$ lost one electron → smaller than $\text{K}$. (b) ${\text{O}}^{2-}$ gained two electrons → larger than $\text{O}$. (c) ${\text{Al}}^{3+}$ lost three electrons (its whole $n=3$ shell) → smaller than $\text{Al}$.

Count electrons (electrons = protons − charge; a $+$ removes electrons, a $-$ adds them). Set A: ${\text{Na}}^{+}=11-1=10$, ${\text{F}}^{-}=9+1=10$, $\text{Ne}=10$. All 10 → isoelectronic ✅ Set B: ${\text{Cl}}^{-}=17+1=18$, ${\text{K}}^{+}=19-1=18$, $\text{Ca}=20$ (neutral, so 20). $18,18,20$ → not isoelectronic (Ca has 20). ❌ Answer: Set A.

All have 10 electrons → isoelectronic → shielding $S$ roughly constant → $Z_{\text{eff}}\approx Z-\text{const}$, so more protons = smaller. Protons: $\text{O}(8)<\text{F}(9)<\text{Na}(11)<\text{Mg}(12)$. Fewest protons pulls weakest → biggest: ${\text{O}}^{2-}>{\text{F}}^{-}>{\text{Na}}^{+}>{\text{Mg}}^{2+}$

Factor $=181/99=1.83$, i.e. it swells to about 1.8× its neutral size. Direction: $\text{Cl}=[\text{Ne}]3s^{2}3p^5$ (17 electrons) gains one electron → $[\text{Ne}]3s^{2}3p^6$ (18 electrons), same $Z=17$. Now 18 electrons share the same 17-proton pull → $Z_{\text{eff}}$ per electron drops, and the extra electron adds repulsion → the cloud expands.

Lost fraction $=(186-102)/186=84/186=0.452$, about 45% — nearly half. Why so large: $\text{Na}=[\text{Ne}]3s^1$ loses its only $n=3$ electron → ${\text{Na}}^{+}=[\text{Ne}]$. An entire outer shell disappears, not just a slight tightening — that is the biggest single kind of jump in ionic sizes.

Electrons: ${\text{S}}^{2-}=16+2=18$, ${\text{Cl}}^{-}=17+1=18$, ${\text{K}}^{+}=19-1=18$, ${\text{Ca}}^{2+}=20-2=18$. All 18 → isoelectronic. Protons: $\text{S}(16)<\text{Cl}(17)<\text{K}(19)<\text{Ca}(20)$. Since $S$ is fixed, $Z_{\text{eff}}\propto Z$, so more protons ⇒ higher $Z_{\text{eff}}$ ⇒ smaller: ${\text{S}}^{2-}>{\text{Cl}}^{-}>{\text{K}}^{+}>{\text{Ca}}^{2+}$ Data check (pm): $184>181>138>100$ ✅

${\text{F}}^{-}=[\text{Ne}]=1s^{2}2s^{2}2p^6$ → outermost electrons in shell $n=2$. ${\text{Cl}}^{-}=[\text{Ar}]=\dots 3s^{2}3p^6$ → outermost electrons in shell $n=3$. Different electron counts ⇒ compare shells first. ${\text{Cl}}^{-}$ occupies a higher shell ($n=3$) which sits physically farther out, so ${\text{Cl}}^{-}>{\text{F}}^{-}.$ Here proton count is a red herring: even though Cl has more protons, its electrons live in a bigger shell, and shell size wins. (Data: ${\text{Cl}}^{-}\approx 181$ pm $>$ ${\text{F}}^{-}\approx 133$ pm.)

Protons: $\text{Ar}(18)<\text{K}(19)$. Pure isoelectronic logic says more protons ⇒ smaller, so ${\text{K}}^{+}$ should be smaller than Ar — the opposite of the numbers. Resolution: the numbers are measured with different definitions of radius. ${\text{K}}^{+}=138$ pm is an ionic radius (from crystals); Ar $\approx 71$ pm is a covalent radius; noble gases have no ordinary ionic radius. You may only apply the isoelectronic rule when comparing radii of the same type. Compare like with like (${\text{S}}^{2-},{\text{Cl}}^{-},{\text{K}}^{+},{\text{Ca}}^{2+}$ — all crystal ionic radii) and the rule holds perfectly.

$\text{X}=\text{Mg}=1s^{2}2s^{2}2p^{6}3s^2$. The $3s$ electrons shield each other by $0.35$ each; the 10 inner electrons ($n=1,2$) shield by $\approx 1.00$ each (crude core rule). For a $3s$ electron before ionizing: $S\approx 10(1.00)+1(0.35)=10.35$, so $Z_{\text{eff}}=12-10.35=\mathrm{1.65.}$ After removing both $3s$ electrons we form ${\text{Mg}}^{2+}=[\text{Ne}]$. The new outermost is a $2p$ electron; the ten electrons rearrange, but the key point is the lost $3s$ shell had $Z_{\text{eff}}\approx 1.65$, whereas a $2p$ electron in Ne-like ${\text{Mg}}^{2+}$ feels a far larger $Z_{\text{eff}}$ (roughly $12-4.15\approx 7.85$ using Slater for $n=2$). $\Delta Z_{\text{eff}}\approx 7.85-1.65=6.2(\text{much stronger pull}).$ Prediction: the outer electrons of ${\text{X}}^{2+}$ feel a hugely larger $Z_{\text{eff}}$ and a whole shell vanished → ${\text{X}}^{2+}$ is dramatically smaller than $\text{X}$. (Real data: $\text{Mg}\approx 160$ pm, ${\text{Mg}}^{2+}\approx 72$ pm.) ✅

Need 18 electrons. Choose ions whose (protons − charge) $=18$:

• ${\text{P}}^{3-}$: $15+3=18$
• ${\text{S}}^{2-}$: $16+2=18$
• ${\text{K}}^{+}$: $19-1=18$
• ${\text{Ca}}^{2+}$: $20-2=18$ Protons: $\text{P}(15)<\text{S}(16)<\text{K}(19)<\text{Ca}(20)$, so by the rule (fewer protons = bigger): ${\text{P}}^{3-}>{\text{S}}^{2-}>{\text{K}}^{+}>{\text{Ca}}^{2+}.$

First separate isoelectronic ($=10$ e⁻) from the odd one out.

• 10-electron group: ${\text{N}}^{3-}(7\text{p}),{\text{O}}^{2-}(8),{\text{F}}^{-}(9),{\text{Na}}^{+}(11),{\text{Mg}}^{2+}(12),{\text{Al}}^{3+}(13)$ → order by protons (fewer = bigger): ${\text{N}}^{3-}>{\text{O}}^{2-}>{\text{F}}^{-}>{\text{Na}}^{+}>{\text{Mg}}^{2+}>{\text{Al}}^{3+}.$
• Neutral $\text{F}$ has 9 electrons (not 10) — remove one electron from ${\text{F}}^{-}$. Fewer electrons than ${\text{F}}^{-}$ with the same 9 protons → higher $Z_{\text{eff}}$ per electron → smaller than ${\text{F}}^{-}$. Where does it sit? $\text{F}\approx 72$ pm; the 10-electron ions run from ${\text{N}}^{3-}(146)$ down to ${\text{Al}}^{3+}(53)$. So $\text{F}$ (72 pm) sits between ${\text{Na}}^{+}(102)$ and ${\text{Mg}}^{2+}(72\text{–}86)$ region — comparable to ${\text{Mg}}^{2+}$. Final ranking (using standard pm values): ${\text{N}}^{3-}(146)>{\text{O}}^{2-}(140)>{\text{F}}^{-}(133)>{\text{Na}}^{+}(102)>\text{F}(72)\approx {\text{Mg}}^{2+}(72)>{\text{Al}}^{3+}(53).$ Boundary reasons: within the ion group each step is +1 proton, same 10 electrons ⇒ tighter; the ${\text{F}}^{-}\to {\text{Na}}^{+}$ jump is large because we cross from anions to cations (removing the extra electrons and gaining protons together); neutral $\text{F}$ falls far below its own ${\text{F}}^{-}$ because it has one fewer electron over the same 9 protons.

Contact distance $\approx r_{\text{cation}}+r_{\text{anion}}$. ${\text{Na}}^{+}\approx 102$ pm in both. Anions: ${\text{F}}^{-}\approx 133$ pm (electrons in $n=2$), ${\text{Cl}}^{-}\approx 181$ pm (electrons in $n=3$, higher shell). Since ${\text{Cl}}^{-}>{\text{F}}^{-}$: $d_{\text{NaCl}}=102+181=283\text{ pm}>d_{\text{NaF}}=102+133=235\text{ pm}.$ Link to lattice energy: Coulomb attraction $\propto 1/d$, so the smaller separation in NaF means a stronger lattice → NaF has the larger (more exothermic) lattice energy. Smaller ions ⇒ tighter packing ⇒ stronger ionic bonding.

Give the two-question checklist that solves every ranking problem on this page ::: (1) Are the species isoelectronic (same electron count)? If yes and same radius type → more protons = smaller. (2) If not isoelectronic → compare occupied shells first, then $Z_{\text{eff}}$.