Group 2 (A2)
A-Level Chemistry Topic 27 8:06 English narration · English + 中文 subtitles burned in
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Here is a puzzle.
先看一个谜题。
Go down Group 2, from magnesium to barium, and the hydroxides get more soluble. Go down the same group, and the sulfates get less soluble.
沿第二族向下，从镁到钡，氢氧化物越来越易溶； 沿同一族向下，硫酸盐却越来越难溶。
Same group, same two plus charge — yet the trends point opposite ways.
同一族、同样的二价正电荷， 两条趋势却指向相反。
At AS you learned that these trends exist.
在AS阶段你学过这两条趋势的存在； 今天我们回答为什么。
Today we answer why: it is a balance between two energies.
答案是两种能量之间的一场较量。
Welcome to the A2 half of Group 2.
欢迎来到第二族的A2部分。
Two jobs only. Explain the thermal stability of the carbonates and the nitrates, using ionic radius and polarisation. And explain the solubility of the hydroxides and the sulfates, using lattice energy and hydration.
任务只有两个：用离子半径和极化， 解释碳酸盐和硝酸盐的热稳定性；再用晶格能和水合， 解释氢氧化物和硫酸盐的溶解度。
Let's begin.
让我们开始吧。
Everything today runs on two energies.
今天的一切都建立在两种能量上。
The lattice energy is the energy change when gaseous ions come together to form one mole of the solid.
晶格能是气态离子结合成一摩尔固体时的能量变化， 它永远是放热的。
It is always exothermic. The enthalpy change of hydration is the energy change when one mole of gaseous ions dissolves in water.
水合焓变是一摩尔气态离子溶于水时的能量变化，它同样是放热的， 因为水分子会被离子吸引。
It is exothermic too, because water molecules are attracted to the ions. One energy holds the solid together. The other pays you to pull it apart.
一种能量把固体抓在一起，另一种则让你有"本钱"把它拆开。
To combine those two energies we need Hess's law.
要把这两种能量结合起来，我们需要盖斯定律。
An enthalpy change depends only on where you start and where you finish, never on the route.
焓变只取决于起点和终点，与所走的路径无关。
So dissolving a solid in one step must cost the same as going the long way round.
所以一步溶解固体，与绕远路走一圈，代价必须完全相同。
Here is the cycle.
这就是那个循环。
Along the top, the solid dissolves in water: the enthalpy change of solution, the change you actually want.
上面一条路：固体溶解在水中，那就是溶解焓变， 也是我们真正想要的那个量。
Now the long way round.
现在走远路。
Go down and pull the lattice apart into separate gaseous ions. That costs energy, so it is the lattice energy with its sign reversed.
先向下：把晶格拆成一个个气态离子， 这需要吸收能量，所以它等于晶格能取相反的符号。
Then go up on the right: wrap every gaseous ion in water and add the hydration energies. Same start, same finish — the routes are equal.
再从右边向上： 给每一个气态离子裹上水分子，把所有水合焓变加起来。
So the enthalpy change of solution equals the total hydration energy minus the lattice energy.
起点相同、终点相同，两条路必然相等。 所以溶解焓变等于总水合焓变减去晶格能。
What makes each one big?
是什么决定它们的大小？
The same two things, and the exam asks for both by name: ionic charge and ionic radius.
是同样的两个因素，而且考试会点名要这两个词： 离子电荷和离子半径。
The lattice energy depends on the two charges divided by the sum of the two radii, so bigger charges, or smaller ions, make it more exothermic.
晶格能取决于两个电荷之积除以两个半径之和， 所以电荷越大、离子越小，它就越放热。
The enthalpy change of hydration depends on the charge divided by the radius of one ion.
水合焓变取决于某一个离子的电荷除以它的半径。
Again: bigger charge, or smaller ion, more exothermic.
同样：电荷越大、离子越小，越放热。
Now real exam data.
我们用真题数据来走一遍这个循环。
Magnesium fluoride.
氟化镁。
Its lattice energy is minus two thousand nine hundred and fifty-seven.
它的晶格能是负二九五七千焦每摩尔。
The hydration energy of the magnesium ion is minus one thousand nine hundred and twenty-six, and of the fluoride ion, minus five hundred and five.
镁离子的水合焓变是负一九二六，氟离子的水合焓变是负五零五。
Find the enthalpy change of solution.
求溶解焓变。
Pause here and try it.
先暂停，自己算一算。
Ready?
好了吗？
Put the numbers into the cycle.
把数字代入循环。
Add the hydration energies first — there are two fluoride ions, so that value counts twice.
先把水合焓变加起来—— 注意有两个氟离子，所以那个数值要算两次。
Then subtract the lattice energy.
然后减去晶格能。
That gives plus twenty-one kilojoules per mole.
结果是正二十一千焦每摩尔。
Positive, so magnesium fluoride is only sparingly soluble.
它是正的，所以氟化镁只是微溶。
Now go down the group.
现在沿族向下。
The charge stays at two plus, but the cation gets bigger at every step.
电荷始终是二价正电，但每往下一步阳离子就更大一些。
A bigger cation weakens both energies: both of them become less exothermic.
阳离子变大会削弱这两种能量：两者都变得不那么放热。
The lattice energy falls, and so does the hydration energy.
晶格能下降，水合焓变也下降。
That is the whole difficulty. They both fall, so the trend depends on which one falls faster.
难点正在于此：既然两者都在下降， 趋势就取决于哪一个下降得更快。
Compare the two anions.
来比较这两种阴离子。
With a small anion, like hydroxide, the two ions sit close, so the lattice energy depends strongly on the size of the cation.
对于小的阴离子，比如氢氧根，两个离子靠得很近， 所以晶格能强烈依赖阳离子的大小。
Grow the cation and the lattice energy falls faster than the hydration energy. The enthalpy change of solution becomes more exothermic: the hydroxides get more soluble — solubility rises.
阳离子变大时，晶格能下降得比水合焓变更快， 溶解焓变因此变得更放热：氢氧化物越来越易溶——溶解度上升。
With a large anion, like sulfate, the big anion already sets the distance, so the lattice energy hardly depends on the cation. Grow the cation and it barely moves, while the hydration energy still falls.
对于大的阴离子，比如硫酸根，离子间距主要由这个大阴离子决定， 晶格能几乎不受阳离子影响；阳离子变大时晶格能几乎不动，而水合焓变仍在下降。
Now the enthalpy change of solution becomes less exothermic: the sulfates get less soluble.
这时溶解焓变变得不那么放热：硫酸盐越来越难溶。
Always compare the changes, not the sizes.
所以永远要比较变化量，而不是比较大小。
Here is a real four-mark question. The fluorides of calcium, strontium and barium.
这就是最近一份真题里的原题：钙、锶、钡的氟化物。
Describe the trend in solubility, and explain it.
描述它们溶解度的变化趋势，并加以解释，共四分。
Pause and try it. Ready.
先暂停，自己写一写。
Mark one is the order: least soluble is calcium fluoride, then strontium fluoride, then barium fluoride.
好了吗？ 第一分是顺序：最难溶的是氟化钙，然后是氟化锶，氟化钡最易溶。
Mark two: going down, both energies become less exothermic.
第二分：沿族向下，两种能量都变得不那么放热。
Mark three: the fluoride ion is small, so the lattice energy changes more.
第三分：氟离子很小，所以晶格能的变化更大。
Mark four: therefore the enthalpy change of solution becomes more exothermic, and the salt is more soluble.
第四分：因此溶解焓变变得更放热，这种盐也就更易溶。
Four sentences, four marks.
四句话，四分。
Now the other half: heat.
现在看这个专题的另一半：加热。
Every Group 2 cation pulls on the electrons of the anion beside it.
每一个第二族阳离子都会拉扯旁边阴离子的电子。
How hard it pulls is what we call its polarising power, and that is the charge divided by the radius.
它拉得有多用力，就是我们所说的极化能力，等于电荷除以半径。
To pull hard the cation must be small and highly charged. And the anion must be large, because a big electron cloud is easy to distort.
要拉得用力，阳离子必须又小、电荷又高； 而阴离子必须够大，因为大的电子云容易被扭曲。
Carbonate and nitrate are both large anions.
碳酸根和硝酸根都是大阴离子。
A small magnesium ion distorts the carbonate a lot, and one bond inside it is weakened. A big barium ion barely distorts it at all.
小小的镁离子会把碳酸根扭曲得很厉害，被扭曲的碳酸根内部就有一条键被削弱了； 而大大的钡离子几乎不会扭曲它。
So it needs more heat to break, and stability increases down the group.
所以它需要更多热量才能分解，热稳定性沿族向下增大。
Give the decomposition products.
给出分解产物。
Get the equations right.
把方程式写对。
A Group 2 carbonate gives the metal oxide and carbon dioxide.
第二族的碳酸盐分解生成金属氧化物和二氧化碳。
A Group 2 nitrate gives the metal oxide, nitrogen dioxide and oxygen — and that one needs care when you balance it: two of the nitrate, two of the oxide, four of the brown gas, one of oxygen.
第二族的硝酸盐分解生成金属氧化物、二氧化氮和氧气——这一条要小心配平： 硝酸盐两份，氧化物两份，棕色气体四份，氧气一份。
A recent paper also asked for the hydrogencarbonate, which gives the carbonate, carbon dioxide and water.
最近的一份真题还考了碳酸氢盐，它分解生成碳酸盐、二氧化碳和水。
And if the question asks what you see, the brown gas is nitrogen dioxide.
如果题目问你看到什么现象，那种棕色气体就是二氧化氮。
Two marks, straight from the paper.
两分，直接来自真题。
The thermal stability of the Group 2 carbonates increases down the group.
第二族碳酸盐的热稳定性沿族向下增大，请解释这个趋势。
Explain this trend. Pause and try it. Ready.
先暂停，自己写一写。
Mark one is the cation: its radius increases down the group, so its charge density falls.
好了吗？ 第一分讲阳离子：它的半径沿族向下增大， 所以电荷密度下降。
Mark two is the anion: it is polarised less, so the carbonate ion is distorted less and needs more heat to break.
第二分讲阴离子：它受到的极化变小， 所以碳酸根离子被扭曲得更少，需要更多的热量才能分解。
Two marks, two ideas: one sentence for the cation, one for the anion.
两分，两个要点：一句写阳离子，一句写阴离子。
Never write only that it is more stable.
千万不要只写"更稳定"。
Put both answers side by side, because the examiner marks the steps, not the conclusion.
我们把两个答案并排放在一起，因为阅卷老师给分给的是步骤，不是结论。
The thermal stability chain: bigger cation, lower charge density, less polarisation of the large anion, more heat needed.
热稳定性这条链：阳离子更大、电荷密度更低、对大阴离子的极化更弱、需要更多热量。
The solubility chain: both energies less exothermic; then decide which one changes more — with a small anion it is the lattice energy, with a large anion it is the hydration; then say whether the enthalpy change of solution becomes more or less exothermic.
溶解度这条链：两种能量都变得不那么放热；再判断哪一个变化更大—— 小阴离子看晶格能，大阴离子看水合焓变； 最后说出溶解焓变是变得更放热还是更不放热。
Three marks students throw away.
三个学生常丢的分。
First, signs.
第一，符号。
In the cycle you subtract the lattice energy, and it is already negative, so subtracting it adds a positive number.
在循环里你要减去晶格能， 而晶格能本身是负的，所以减去它相当于加上一个正数。
Second, never write only that the ion is bigger.
第二，绝不要只写"离子更大"。
Say what bigger does: lower charge density, less polarisation of the anion.
要说出变大带来了什么： 电荷密度更低，所以对阴离子的极化更弱。
Third, compare the changes, not the sizes: the mark is for saying which energy changes more.
第三，比较变化量，而不是比较大小。 得分点是说出哪一种能量变化更大。
Get those right, and this topic is yours.
做对这三点，这个专题就是你的了。