2.3 · Periodic TrendsPractice · 元素周期律

The block is set by the sublevel filled last, not just the outermost one. 4s fills before 3d, so this element is filling 3d and sits in the d block: period 4, group 5, vanadium (V, element 23). The 4s² can tempt you into choosing the s block.

Estimating with “nuclear charge − core electrons,” both valence electrons feel about +1 (Na: 11 − 10; K: 19 − 18), so that isn’t the difference. K has one more shell, so its valence electron is in n = 4, farther from the nucleus (radius K 203 pm, Na 166 pm). By Coulomb’s law a greater distance means a weaker attraction, so it’s easier to remove. K’s nuclear charge is actually larger, and both have just 1 valence electron.

All four ions have 18 electrons, arranged like argon’s; they differ only in nuclear charge. From S (16) to Ca (20) the nuclear charge rises, pulling the same 18 electrons in more tightly, so the radius shrinks. Data (pm): $\ce{S^2-}$ 184, $\ce{Cl-}$ 181, $\ce{K+}$ 138, $\ce{Ca^2+}$ 100.

Find the big jump: 1577 ÷ 787 ≈ 2.0, 3232 ÷ 1577 ≈ 2.0, 4356 ÷ 3232 ≈ 1.3, 16091 ÷ 4356 ≈ 3.7, 19805 ÷ 16091 ≈ 1.2. The biggest jump is between the 4th and 5th, so the first 4 electrons are valence electrons and the 5th comes from an inner shell. A period 3 element with 4 valence electrons is in group 14: silicon.

Ionization energy usually rises across a period, and S is an exception, for the same reason as N and O in period 2. P is [Ne]3s²3p³ with one electron in each 3p orbital; S is [Ne]3s²3p⁴, so one orbital holds a pair, and the repulsion between paired electrons makes one of them easier to remove. A, B, and D are false: S has the larger nuclear charge (16), a smaller radius (105 pm vs 107 pm for P), and the same number of shells.

Bond polarity depends on the electronegativity difference. H is 2.20; F, Cl, O, and N are 3.98, 3.16, 3.44, 3.04, giving differences of 1.78, 0.96, 1.24, 0.84. Fluorine is the most electronegative, so H–F is the most polar.

Electronegativity differences: Li–F 3.00, Al–Cl 1.55, Ca–O 2.44, P–Cl 0.97, Mg–Cl 1.85. Li–F, Ca–O, and Mg–Cl reach 1.7. Al–Cl is a metal and a nonmetal, but its difference is only 1.55, and aluminum chloride really is covalent. The rule is only a guide; cases near the cutoff need a closer look.

From F to I the melting point rises (F₂ −219.6 °C, Cl₂ −101.5 °C, Br₂ −7.2 °C, I₂ 113.7 °C), going from gas to solid; atoms get larger (F 57, Cl 102, Br 120, I 139 pm) and less electronegative (3.98, 3.16, 2.96, 2.66). Extending the trends, astatine is a solid, larger than iodine, and less electronegative. Nonmetal character weakens down the group: oxidizing strength goes $\ce{Cl2 > Br2 > I2}$, so astatine is weaker still and can’t displace iodine, and the hydrogen halides get less stable, so HAt is less stable than HI.

From Li to Cs the atoms gain shells, so the radius increases (Li 128, Na 166, K 203, Rb 220, Cs 244 pm). The valence electron is farther out and easier to lose, so the first ionization energy (520, 496, 419, 403, 376 kJ/mol) and electronegativity fall, and the metals get more reactive. Metallic bonding weakens, so the melting point drops (180.5, 97.8, 63.5, 39.3, 28.4 °C).

496 kJ/mol is per mole of atoms; convert to joules and divide by the number of atoms in a mole: $\dfrac{496 \times 10^{3}\ \text{J}}{6.022 \times 10^{23}} = 8.24 \times 10^{-19}$ J.