U1.7 · Periodic TrendsPractice · 周期趋势

Data (kJ/mol): Na 496, Al 578, Mg 738, Si 787. Ionization energy generally rises across a period, but Al is below Mg: Al loses a 3p electron, which is higher in energy than 3s and a bit farther out, so it’s easier to remove. Ordering by atomic number (A) misses this exception.

Compare successive values: 1757 ÷ 900 ≈ 2.0, 14849 ÷ 1757 ≈ 8.5, 21007 ÷ 14849 ≈ 1.4. The big jump is between the 2nd and 3rd, so X has 2 valence electrons and forms X²⁺. Combined with O²⁻, the charges balance as XO (X is beryllium).

Energy released (kJ/mol): O 141, F 328, S 200, Cl 349. Halogens release more than the group 16 element in the same period. F is the most electronegative, but it’s so small that an added electron is crowded by the other 2p electrons; the extra repulsion makes its electron affinity smaller than Cl’s.

F and Cl both have 7 valence electrons and similar effective nuclear charge (about +7). F has two shells (radius 57 pm) and Cl three (102 pm). The shorter distance means a stronger Coulombic attraction, so F pulls a shared pair toward itself more strongly. Cl actually has the larger nuclear charge, and F is the smaller atom.

Mg is [Ne]3s² and $\ce{Mg^2+}$ is 1s²2s²2p⁶: both 3s electrons are gone, so the outermost shell is now n = 2. The nuclear charge is unchanged (still 12), but fewer electrons share it. Anions are larger than their atoms, so “ions are always smaller” is false.

Look for where the bars suddenly jump: 1451 kJ/mol for the 2nd and 7733 kJ/mol for the 3rd, a factor of 5.3, while every other step is less than a factor of 2. The first 2 electrons are valence electrons (3s²); the 3rd would come from the inner n = 2 shell. So the element is Mg, in group 2. Na would jump after the 1st, Al after the 3rd, and Si after the 4th.

First identify J and K: across period 2 (Z = 3–10) the values rise overall, with dips at Z = 5 and Z = 8. J is N (1402 kJ/mol) and K is O (1314 kJ/mol). O is 1s²2s²2p⁴, so one 2p orbital holds a pair of electrons that repel each other, which makes one easier to remove. Both electrons come from the n = 2 shell, and O has the larger nuclear charge (8) and the smaller radius, so the other three choices are wrong.