Group 17
A-Level Chemistry Topic 11 7:28 English narration · English + 中文 subtitles burned in
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Transcript
Here are three colourless solutions.
这里有三份无色溶液。
They look exactly the same.
它们看上去一模一样。
One is a chloride, one is a bromide, one is an iodide — but you cannot tell which is which by looking.
一份是氯化物，一份是溴化物， 一份是碘化物——但只靠眼睛看，你分不出哪个是哪个。
Now add a few drops of silver nitrate.
现在滴入几滴硝酸银。
Within seconds, each one shows its own colour: white, cream and pale yellow.
几秒钟内，每一份都显出自己的颜色：白色、奶油色和浅黄色。
One family, clear trends — and every trend is a mark in the exam.
同一个家族，清晰的规律——而每一条规律都是考试的分数。
Welcome to Group 17, the halogens.
欢迎来到第十七族——卤素。
We will follow four trends down the group: colour and volatility, oxidising power, the halide tests, and finally the two faces of chlorine.
我们将沿着这一族向下追踪四条规律： 颜色与挥发性、氧化能力、卤离子的检验，最后是氯的两副面孔。
Let's begin.
让我们开始吧。
First, what you can see.
先看你能看到的部分。
The halogens exist as pairs of atoms.
卤素以双原子分子存在。
Chlorine is a pale green gas.
氯是淡绿色气体。
Bromine is a red-brown liquid.
溴是红棕色液体。
Iodine is a grey-black solid that gives off a purple vapour.
碘是灰黑色固体，会放出紫色蒸气。
So volatility falls as you go down the group.
所以沿着这一族向下，挥发性下降。
Why?
为什么？
Going down, the molecules are bigger and hold more electrons.
向下走，分子更大，电子更多。
That makes the instantaneous dipole, induced dipole forces between the molecules stronger, and stronger forces are harder to pull apart.
这使分子之间的瞬时偶极—诱导偶极作用力更强， 而作用力更强就更难被拉开。
Now look inside the molecule.
现在看分子内部。
The bond energy of the halogen molecules falls as you go down.
卤素分子的键能沿着这一族向下降低。
For chlorine it is about two hundred and forty kilojoules per mole, then one hundred and ninety for bromine, then one hundred and fifty for iodine.
氯大约是二百四十千焦每摩尔， 溴是一百九十，碘是一百五十。
The reason is size.
原因在于尺寸。
In a bigger atom the shared pair of electrons sits further from the two nuclei, so the pull on it is weaker and the bond is easier to break.
原子越大，共用电子对离两个原子核越远， 受到的吸引就越弱，键也就更容易断。
That pattern explains the next two trends as well.
这个规律同样能解释接下来的两条趋势。
Before we use those trends, a quick reminder.
在使用这些规律之前，先快速复习一点。
Redox is simply a transfer of electrons.
氧化还原本质上就是电子的转移。
Here zinc gives two electrons away, so zinc is oxidised, and the copper ion takes them, so it is reduced.
这里锌给出两个电子，所以锌被氧化；铜离子接过电子，所以它被还原。
The substance that takes electrons is the oxidising agent; the one that gives them away is the reducing agent.
接受电子的物质是氧化剂，给出电子的物质是还原剂。
The whole of Group 17 turns on those two words.
整个第十七族都建立在这两个词上。
Every halogen wants one more electron, to fill its outer shell.
每种卤素都想再得到一个电子，把最外层填满。
Taking that electron makes a one-minus ion, so the halogen acts as an oxidising agent.
得到这个电子就形成一价负离子， 所以卤素起氧化剂的作用。
But this power gets weaker as you go down the group — the halogens get less reactive down the group.
但这种能力沿着这一族向下逐渐减弱——卤素沿族向下反应性变弱。
The atom is bigger, so the extra electron would sit further from the nucleus with more inner shells in the way, and it is held less strongly.
原子更大，多出来的电子会离原子核更远，中间还隔着更多内层电子， 因此被吸引得没那么牢。
That gives you displacement: chlorine is a stronger oxidising agent than bromine, so chlorine will push bromine out of a bromide solution, and the colourless solution turns orange.
这就带来了置换：氯是比溴更强的氧化剂， 所以氯会把溴从溴化物溶液中置换出来，无色溶液变成橙色。
Each halogen also reacts with hydrogen to make a hydrogen halide, and the same size trend shows up again.
每种卤素也能与氢气反应生成卤化氢，这里又出现了同样的尺寸规律。
Chlorine and hydrogen explode in sunlight.
氯与氢在阳光下会爆炸。
Bromine needs heat.
溴需要加热。
Iodine reacts slowly, and never completely.
碘反应缓慢，而且永远不完全。
Now look at the hydrogen halide itself.
再看卤化氢本身。
Its thermal stability also falls down the group, because the bond from hydrogen to the halogen gets longer and weaker.
它的热稳定性同样沿着这一族向下降低， 因为氢与卤素之间的键更长、更弱。
Hydrogen chloride is very stable; hydrogen iodide breaks apart on gentle heating, and purple iodine vapour appears.
氯化氢非常稳定； 碘化氢在温和加热下就会分解，并出现紫色的碘蒸气。
Now turn the whole thing around and look at the halide ions.
现在把整件事反过来，看卤离子。
A halide ion already holds that extra electron, so it can give one away.
卤离子已经拥有那个多出来的电子， 所以它可以把电子给出去。
That makes it a reducing agent, and this power runs the opposite way: it increases down the group.
这使它成为还原剂，而这种能力的方向正好相反： 沿着这一族向下增强。
A larger ion holds its outer electron more loosely, so iodide gives its electron up most easily.
离子越大，对最外层电子的束缚就越松， 所以碘离子最容易给出电子。
Two arrows, pointing in opposite directions. For the halogens, oxidising power falls; for the halide ions, reducing power rises.
两个箭头，方向相反：对卤素来说氧化能力下降， 对卤离子来说还原能力上升。
So how do you identify an unknown halide?
那么怎样鉴别一个未知的卤离子？
Add dilute nitric acid first, to remove anything else that could give a precipitate.
先加稀硝酸，除掉其他可能生成沉淀的物质。
Then add aqueous silver nitrate.
然后加入硝酸银溶液。
Silver chloride is a white precipitate. Silver bromide is cream. Silver iodide is yellow.
氯化银是白色沉淀，溴化银是奶油色，碘化银是黄色。
Cream and yellow are hard to tell apart, so confirm with ammonia.
奶油色和黄色不容易区分，所以要用氨水来确认。
Silver chloride dissolves in dilute ammonia; silver bromide needs concentrated ammonia; silver iodide will not dissolve in either.
氯化银溶于稀氨水； 溴化银需要浓氨水；碘化银在两者中都不溶解。
Colour, then ammonia — that is the full answer.
先看颜色，再用氨水—— 这才是完整的答案。
Here is the reaction examiners love: a solid halide warmed with concentrated sulfuric acid.
这是考官最爱考的反应：固体卤化物与浓硫酸共热。
Two things can happen.
会发生两件事。
First, every halide gives the hydrogen halide — that step is just an acid handing a proton to the halide ion, and no electrons move.
第一，每一种卤化物都会先生成卤化氢——这一步只是酸把一个质子交给卤离子， 没有电子转移。
Second, if the halide is a strong enough reducing agent, it attacks the sulfur in the acid and can pull it down from plus six.
第二，如果这个卤离子是足够强的还原剂， 它还会攻击酸中的硫，把硫从正六价往下拉。
To plus four gives sulfur dioxide. To zero gives yellow sulfur. To minus two gives hydrogen sulfide, the rotten-egg gas.
拉到正四价，生成二氧化硫； 拉到零价，生成黄色的硫；拉到负二价，生成有臭鸡蛋味的硫化氢。
Let's do the classic question.
我们来做这道经典题。
Solid sodium chloride, sodium bromide and sodium iodide are each warmed with concentrated sulfuric acid. Predict what you would see.
把固体氯化钠、溴化钠和碘化钠分别与浓硫酸共热， 预测你会观察到什么。
Pause the video and try it. Ready?
先暂停视频，自己试一试。
Chloride is too weak to reduce anything, so you only get misty fumes of hydrogen chloride.
好了吗？ 氯离子太弱， 什么都还原不了，所以只得到氯化氢的白雾。
Bromide reduces the acid a little: hydrogen bromide, plus brown bromine and choking sulfur dioxide.
溴离子能把酸还原一点： 生成溴化氢，还有棕色的溴和有刺激性的二氧化硫。
Iodide is the strongest reducing agent, so it goes furthest: hydrogen iodide, purple iodine, and hydrogen sulfide with its bad-egg smell.
碘离子是最强的还原剂， 所以走得最远：生成碘化氢、紫色的碘，以及有臭鸡蛋味的硫化氢。
Now chlorine itself, with sodium hydroxide.
现在看氯本身与氢氧化钠的反应。
With cold, dilute sodium hydroxide you get sodium chloride and sodium chlorate one.
与冷的稀氢氧化钠反应，得到氯化钠和次氯酸钠。
With hot, concentrated sodium hydroxide you get sodium chloride and sodium chlorate five.
与热的浓氢氧化钠反应，得到氯化钠和氯酸钠。
Now track the oxidation number of chlorine.
现在追踪氯的氧化数。
It starts at zero. In the chloride it has dropped to minus one. In the chlorate it has risen to plus one, or to plus five.
它从零开始； 在氯化物中降到了负一；在含氧酸盐中升到了正一，或者正五。
So the same element goes up and down at the same time. That is disproportionation.
同一种元素同时向上和向下变化——这就是歧化。
Finally, chlorine in water purification — the reason chlorine is in your tap water.
最后，氯在水净化中的用途——说说为什么自来水里要加氯。
A little chlorine reacts with water in a reversible reaction, giving hydrochloric acid and chloric one acid.
少量氯与水发生可逆反应， 生成盐酸和次氯酸。
Look at the numbers again: chlorine goes both up and down, so this is disproportionation too.
再看一次氧化数：氯同时上升和下降，所以这也是歧化反应。
The two active species are shown on screen — that acid and the chlorate one ion.
两种有效物种显示在屏幕上——次氯酸和次氯酸根离子。
It is these that kill the bacteria.
正是它们杀死细菌。
A very small amount makes water safe to drink.
极少的量就能让水安全可饮。
Three marks students give away.
三个学生常送掉的分。
First, never just write that reactivity decreases down the group.
第一，不要只写"沿着这一族向下反应性减弱"。
Say why: the atom is bigger, there is more shielding, so the extra electron is held less strongly.
要说明原因： 原子更大，屏蔽更多，所以多出来的电子被吸引得更弱。
Second, in the silver nitrate test, always give the colour and the ammonia step; colour on its own is not a full identification.
第二，做硝酸银检验时， 一定要同时给出颜色和氨水这一步，只写颜色不算完整的鉴别。
Third, whenever chlorine reacts with water or with cold sodium hydroxide, use the word disproportionation and show the oxidation numbers.
第三， 只要氯与水或与冷的氢氧化钠反应，就要用"歧化"这个词，并写出氧化数的变化。