Question

which ion in each pair has the smaller atomic radius?

1. k+ or o2-
2. ba2+ or i-
3. al3+ or p3-
4. k+ or cs+
5. fe2+ or fe3+
6. f- or s2-

Explanation:

9. \( \boldsymbol{K^+} \) or \( \boldsymbol{O^{2-}} \)

Step1: Determine electron configurations

\( K^+ \) has 18 electrons (K: 19, lose 1), \( O^{2-} \) has 10 electrons (O: 8, gain 2). Wait, no: \( K^+ \): K (19) → \( K^+ \) (18 e⁻, [Ar] configuration). \( O^{2-} \): O (8) → \( O^{2-} \) (10 e⁻, [Ne] configuration). Wait, actually, \( K^+ \) and \( O^{2-} \) have different electron shells? Wait, no, \( K^+ \) is in period 4 (lost 1 e⁻, so electron configuration [Ar]), \( O^{2-} \) is [Ne]. Wait, but when comparing isoelectronic species (same number of electrons), but here they are not. Wait, no, \( K^+ \) has 18 electrons, \( O^{2-} \) has 10? No, wait O is atomic number 8, gain 2 e⁻: 10 e⁻. K is atomic number 19, lose 1 e⁻: 18 e⁻. Wait, maybe I made a mistake. Wait, \( K^+ \) and \( O^{2-} \): actually, \( K^+ \) has more protons (19) than \( O^{2-} \) (8). The number of electrons: \( K^+ \) has 18, \( O^{2-} \) has 10? No, that can't be. Wait, no, \( O^{2-} \) is 8 + 2 = 10 e⁻, \( K^+ \) is 19 - 1 = 18 e⁻. Wait, but the key is: for ions, when comparing, if they are isoelectronic (same e⁻), the one with more protons has smaller radius. If not isoelectronic, we look at the principal quantum number (n) of the outermost electrons. Wait, \( K^+ \) has outermost electrons in n=3 (since [Ar] is 1s²2s²2p⁶3s²3p⁶), \( O^{2-} \) has outermost in n=2 ([Ne] is 1s²2s²2p⁶). Wait, but \( K^+ \) has more protons (19) than \( O^{2-} \) (8), but also higher n? Wait, no, maybe I messed up. Wait, no, \( K^+ \) is a cation from period 4, \( O^{2-} \) is an anion from period 2. Wait, actually, the correct approach: for ions, when comparing, if they have the same number of electrons (isoelectronic), radius decreases with increasing nuclear charge. If not, the one with smaller n (principal quantum number) for the outermost shell is smaller, but also nuclear charge. Wait, let's recast: \( K^+ \): atomic number 19, electrons 18 (n=3 for outermost, since [Ar] is 1s²2s²2p⁶3s²3p⁶). \( O^{2-} \): atomic number 8, electrons 10 (n=2 for outermost, [Ne] 1s²2s²2p⁶). Wait, but \( K^+ \) has a larger n (3 vs 2), but more protons. Wait, this is confusing. Wait, no, actually, \( K^+ \) and \( O^{2-} \): let's check their electron configurations. \( K^+ \): 1s²2s²2p⁶3s²3p⁶ (18 e⁻, n=3 for outermost). \( O^{2-} \): 1s²2s²2p⁶ (10 e⁻, n=2 for outermost). Wait, but n=2 is smaller than n=3, so the outermost shell of \( O^{2-} \) is closer to the nucleus. But \( K^+ \) has more protons (19 vs 8). Wait, maybe I made a mistake in electron count. Wait, K is 19, lose 1 e⁻: 18 e⁻ (correct). O is 8, gain 2 e⁻: 10 e⁻ (correct). So \( K^+ \) has electrons in n=3, \( O^{2-} \) in n=2. But the radius of an ion also depends on the number of electron shells. Wait, no, \( O^{2-} \) has 2 electron shells (n=1, n=2), \( K^+ \) has 3 electron shells (n=1, n=2, n=3). Wait, but the outermost shell of \( K^+ \) is n=3, which is further from the nucleus than n=2 of \( O^{2-} \)? But \( K^+ \) has more protons, which pulls the electrons closer. Wait, this is a conflict. Wait, actually, the correct way: when comparing ions, if they are isoelectronic (same number of electrons), the one with more protons has smaller radius. If not isoelectronic, the one with fewer electron shells (lower n) is smaller, but also nuclear charge. Wait, let's take another approach. The atomic radius (ionic radius) generally decreases across a period and increases down a group. But for ions, cations are smaller than their parent atoms, anions larger. Now, \( K^+ \) and \( O^{2-} \): \( K^+ \) is a cation from period 4, \( O^{2-} \) is an anion from period 2. Wait, but \( K…

Step1: Determine electron count

\( Ba^{2+} \): Ba (atomic number 56) loses 2 e⁻ → 54 e⁻ (noble gas Xe configuration). \( I^- \): I (atomic number 53) gains 1 e⁻ → 54 e⁻ (Xe configuration). So they are isoelectronic (same number of electrons, 54).

Step2: Compare nuclear charge

Nuclear charge (Z) for \( Ba^{2+} \) is 56, for \( I^- \) is 53. In isoelectronic species, higher Z (nuclear charge) means smaller radius (since electrons are pulled more tightly).

Step3: Conclusion

Since \( Ba^{2+} \) has higher nuclear charge (56) than \( I^- \) (53) and same electrons, \( Ba^{2+} \) has smaller radius.

Step1: Electron count

\( Al^{3+} \): Al (13) loses 3 e⁻ → 10 e⁻ (Ne configuration). \( P^{3-} \): P (15) gains 3 e⁻ → 18 e⁻? Wait, no: P is 15, gain 3 e⁻ → 18 e⁻? Wait, no, 15 + 3 = 18? Wait, Al is 13, lose 3 → 10. Wait, no, I messed up. Wait, \( Al^{3+} \): 13 - 3 = 10 e⁻ (Ne: 10 e⁻). \( P^{3-} \): 15 + 3 = 18 e⁻ (Ar: 18 e⁻). Wait, no, that can't be. Wait, no, \( P^{3-} \): P is in group 15, gains 3 e⁻ to reach noble gas (Ar), so 15 + 3 = 18 e⁻. \( Al^{3+} \): Al in group 13, loses 3 e⁻ to reach Ne (10 e⁻). Wait, so they are not isoelectronic. Wait, no, wait: \( Al^{3+} \) (10 e⁻) and \( Na^+ \) (10 e⁻) are isoelectronic. \( P^{3-} \) (18 e⁻) and \( S^{2-} \), \( Cl^- \), \( K^+ \), etc., are isoelectronic. Wait, I made a mistake. Wait, \( Al^{3+} \) has 10 e⁻, \( P^{3-} \) has 18 e⁻. So different electron counts. Wait, no, wait: Al is 13, lose 3 → 10 (Ne). P is 15, gain 3 → 18 (Ar). So \( Al^{3+} \) has 10 e⁻ (n=2 shell), \( P^{3-} \) has 18 e⁻ (n=3 shell). Now, nuclear charge: \( Al^{3+} \) has Z=13, \( P^{3-} \) has Z=15. Wait, but \( Al^{3+} \) has fewer electron shells (n=2 vs n=3 for \( P^{3-} \)). Let's check Zeff. For \( Al^{3+} \): electrons in n=2, inner shell 1s² (2 e⁻), so S=2, Zeff=13-2=11. For \( P^{3-} \): electrons in n=3, inner shells 1s²2s²2p⁶ (10 e⁻), so S=10, Zeff=15-10=5. Now, radius is proportional to \( \frac{n^2}{Zeff} \). For \( Al^{3+} \): \( \frac{2^2}{11} ≈ 0.36 \). For \( P^{3-} \): \( \frac{3^2}{5} = 1.8 \). So \( Al^{3+} \) has smaller radius because n is smaller and Zeff is higher. Alternatively, since \( Al^{3+} \) is in n=2 (closer to nucleus) and \( P^{3-} \) in n=3 (farther), and even with lower Z (13 vs 15), the smaller n dominates. So \( Al^{3+} \) is smaller.

Step2: Conclusion

\( Al^{3+} \) has fewer electron shells (n=2) and higher effective nuclear charge (Zeff=11) than \( P^{3-} \) (n=3, Zeff=5), so smaller radius.

Answer:

\( K^+ \)

10. \( \boldsymbol{Ba^{2+}} \) or \( \boldsymbol{I^-} \)