2.3 · Periodic TrendsPractice · 元素周期律

The ionization energy jumps about 6.8 times from I₃ (3660) to I₄ (25026), so X has 3 valence electrons and the 4th would come from the core. X forms $\ce{X^3+}$, which combines with $\ce{O^2-}$ so the charges cancel: 2(+3) + 3(−2) = 0, giving $\ce{X2O3}$ (X is boron).

From Al to Ga the nuclear charge rises by 18, ten of them matched by newly filled 3d electrons. d electrons are diffuse and shield poorly compared with s and p electrons, so Ga’s 4p electron feels a larger effective nuclear charge than the trend predicts, and the ionization energy doesn’t drop. Ga (122 pm) is about the same size as Al (121 pm); the extra shell would lower the energy, and that is exactly what gets cancelled; and Ga has a single 4p electron, so nothing is paired.

Ionic radii (Shannon): $\ce{Li+}$ 76 pm, $\ce{Na+}$ 102 pm, $\ce{Mg^2+}$ 72 pm, $\ce{Al^3+}$ 54 pm. $\ce{Na+}$, $\ce{Mg^2+}$, and $\ce{Al^3+}$ all have 10 electrons, so the higher the nuclear charge, the smaller the ion. $\ce{Li+}$ has only 2 electrons (one shell fewer), so it is smaller than $\ce{Na+}$; but with only a +1 charge it ends up about the size of $\ce{Mg^2+}$, slightly larger (the diagonal relationship).

Data: first ionization energy S 1000, Cl 1251 kJ/mol; electronegativity S 2.58, Cl 3.16; atomic radius S 105, Cl 102 pm; ionic radius $\ce{S^2-}$ 184, $\ce{Cl-}$ 181 pm; electron affinity S 200, Cl 349 kJ/mol. Cl is to the right of S, with more nuclear charge and a smaller radius, so its ionization energy, electronegativity, and electron affinity are all larger. $\ce{S^2-}$ and $\ce{Cl-}$ have the same 18 electrons, but S has the smaller nuclear charge, so $\ce{S^2-}$ is larger.

Ionization energy is per mole of gaseous atoms, so find moles first: n = 1.00 ÷ 6.94 = 0.1441 mol. Energy = 0.1441 × 520 = 74.9 kJ.

After losing 3s¹, Na becomes $\ce{Na+}$ (1s²2s²2p⁶); the next electron must come from 2p, much closer to the nucleus and poorly shielded, so the energy shoots up. Mg has two 3s electrons, so its second is still a valence electron; its jump comes at $I_3$ (7733 kJ/mol). $\ce{Na+}$ and $\ce{Mg+}$ both carry +1, Na is larger than Mg, and both have the same number of shells.

C is 2p², so the new electron can enter an empty 2p orbital. N is 2p³, with one electron in each 2p orbital, so the new one must pair up, and the pairing repulsion outweighs the larger nuclear charge. The same half-filled stability is why N’s first ionization energy (1402) exceeds O’s (1314). O (2p⁴) is back to an electron affinity of 141 kJ/mol.

Cl is 1s²2s²2p⁶3s²3p⁵. Shielding S: the other 6 electrons in shell 3, 6 × 0.35 = 2.10; the 8 in shell 2, 8 × 0.85 = 6.80; the 2 in shell 1, 2 × 1.00 = 2.00; S = 10.90. $Z_{eff}$ = 17 − 10.90 = 6.1. Na’s 3s electron feels only 2.2, which is why Cl is much smaller and much harder to ionize.

Write the configurations (unpaired electrons in parentheses): V [Ar]4s²3d³ (3); Cr [Ar]4s¹3d⁵ (6); Co [Ar]4s²3d⁷ (3); As [Ar]4s²3d¹⁰4p³ (3); Ga [Ar]4s²3d¹⁰4p¹ (1). V’s 3d³, Co’s 3d⁷ (two pairs and three singles in five orbitals), and As’s 4p³ each give 3; Cr, 4s¹3d⁵, has 6; Ga has a single 4p electron.

Oxidizing strength falls down the group: $\ce{F2 > Cl2 > Br2 > I2}$. A halogen can oxidize the halide ions below it: $\ce{Cl2 + 2KI -> 2KCl + I2}$. $\ce{Br2}$ can’t displace $\ce{Cl-}$, and $\ce{I2}$ can’t displace $\ce{Br-}$ or $\ce{Cl-}$. Electronegativities: Cl 3.16 > Br 2.96 > I 2.66.