Chemistry of transition elements
A-Level Chemistry Topic 28 8:24 English narration · English + 中文 subtitles burned in
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Look at these six test tubes: pink, orange, yellow, green, blue, violet.
看看这六支试管：粉色、橙色、黄色、绿色、蓝色、紫色。
Every one holds a d-block metal in water, and every metal comes from the same short block in the middle of the periodic table.
每一支里都含有一种 d 区金属， 而这些金属都来自元素周期表中间那一小块。
A ruby is red for the same reason.
红宝石之所以是红色，原因也一样。
One idea explains the colour of a transition metal complex — a d sub-shell that is only partly full.
一个想法就能解释过渡金属配合物的颜色——一个只被部分填满的 d 亚层。
Welcome to the chemistry of the transition elements.
欢迎来到过渡元素的化学。
We will get the definition right, build complexes from ligands, explain the colour, meet the isomers, and finish with a titration calculation.
我们会先把定义讲清楚，再用配体搭出配合物， 解释颜色是怎么来的，认识各种异构体，最后完成一道完整的滴定计算。
The definition is worth a mark on its own.
先从定义开始——它本身就值一分。
A transition element is a d-block element that forms at least one stable ion with an incomplete d sub-shell.
过渡元素是指能形成至少一种 d 亚层未填满的稳定离子的 d 区元素。
The word ion is the key, not the atom.
关键词是"离子"，不是原子。
Scandium and zinc are the test: scandium's ion has an empty d sub-shell, zinc's is completely full, so neither is a transition element.
钪和锌就是检验：钪的离子 d 亚层是空的， 锌的离子则完全填满，所以两者都不是过渡元素。
And when these atoms form ions, the outer s electrons always leave first, which is why a copper two plus ion is d nine.
另外，这些原子形成离子时， 总是外层的 s 电子先离开，所以铜二价离子是 3d9。
That incomplete d sub-shell gives four properties, and the exam wants the reason as well as the name.
d 亚层未填满带来四条性质，而考试要的是原因，不只是名称。
Variable oxidation states, because the d sub-shell and the s sub-shell beside it are very close in energy.
第一，可变氧化态，因为 d 亚层和旁边的 s 亚层能量非常接近。
They act as catalysts, because they have more than one stable oxidation state and empty d orbitals that are easy to reach.
第二，它们能作催化剂，因为它们有不止一种稳定氧化态，还有容易被利用的空 d 轨道。
They form complex ions, because those same empty orbitals accept lone pairs.
第三，它们能形成配离子，因为这些空轨道可以接受孤对电子。
And they form coloured compounds — the biggest idea in this whole topic.
第四，它们能形成有色化合物——这是整个专题里最重要的一点。
A ligand is a species that has a lone pair of electrons and uses it to form a dative covalent bond to a central metal ion.
配体是含有孤对电子、并用这对电子与中心金属离子形成配位键的物种。
Learn that sentence.
把这句话背下来。
Ligands are sorted by how many bonds each one can make.
配体按一个配体能形成几条键来分类。
Monodentate ligands make one — water, ammonia, chloride, cyanide.
单齿配体形成一条—— 水、氨、氯离子、氰离子。
Bidentate ligands make two — E N, and the ethanedioate ion.
双齿配体形成两条——乙二胺（en）和乙二酸根离子。
A polydentate ligand makes many: E D T A uses six on its own.
多齿配体能形成很多条：EDTA 一个就能用上六条。
A complex is a metal ion surrounded by ligands, and one number fixes its shape.
配合物是被配体包围的中心金属离子，而它的形状由一个数字决定。
The coordination number is how many dative bonds reach the metal.
配位数就是有多少条配位键连到金属上。
Six gives an octahedral complex, at ninety degrees.
六是八面体形，键角九十度。
Four gives either tetrahedral, at one hundred and nine point five degrees, or square planar, at ninety.
四是四面体形，键角一百零九点五度，或者平面正方形，键角九十度。
Two gives linear, at one hundred and eighty.
二是直线形，键角一百八十度。
Careful — a bidentate ligand makes two bonds, so three of them still give six.
注意——一个双齿配体形成两条键，所以三个仍然是六。
Try this one.
来做一道。
A nickel two plus ion bonds to three ethanedioate ions, each one bidentate and each one two minus.
一个镍二价离子与三个乙二酸根离子结合。 乙二酸根是双齿的，带两个负电荷。
Give the formula, the charge and the shape.
写出配合物的化学式、电荷和形状。
Pause the video now. Ready?
先暂停视频试一试。
For the charge, add up: two plus from the metal, three lots of two minus from the ligands, so the complex is four minus.
好了吗？ 算电荷就是把电荷加起来：金属是正二，三个配体各负二共是负六， 所以整个配合物带四个负电荷。
Now the coordination number: each ligand makes two bonds, so three make six — and six means octahedral.
再看配位数：每个配体形成两条键，三个配体就是六， 而六就意味着八面体形。
So the answer is nickel with three ethanedioates, four minus, octahedral.
所以答案是：镍配三个乙二酸根，带四个负电荷，八面体形。
Ligands can swap.
配体是可以交换的。
Copper two plus in water is a pale blue octahedral complex with six water ligands.
铜二价在水中是一个浅蓝色的八面体配合物，带六个水配体。
Add ammonia and four waters are replaced: the solution turns deep royal blue.
加入氨水，其中四个水被取代，溶液变成深蓝色。
Add concentrated hydrochloric acid instead and chloride takes over: the colour goes yellow. Chloride is large, so only four of them fit, and the complex becomes tetrahedral.
改加浓盐酸，则由氯离子取代： 颜色变黄，而且因为氯离子体积大，只能容下四个，所以配合物变成四面体形。
Cobalt does the same — pink in water, blue with concentrated acid.
钴也一样——在水中是粉色，加浓盐酸后变蓝色。
In a free metal ion the five d orbitals have the same energy — they are degenerate.
在自由的金属离子中，五个 d 轨道能量完全相同——它们是简并的。
Bring the ligands close, and their lone pairs repel the d electrons, but not equally, so the orbitals split into two sets with an energy gap between them.
把配体拉近，配体的孤对电子会排斥 d 电子，但排斥程度并不相同， 于是五个轨道分裂成两组，中间隔着一个能量差。
In an octahedral complex, three go lower and two go higher.
在八面体配合物中，三个轨道降低、两个升高。
In a tetrahedral complex it is two lower and three higher, and the gap is smaller.
在四面体配合物中正好相反：两个低、三个高，而且能量差更小。
That gap is everything.
这个能量差就是关键。
Light whose frequency matches the gap is absorbed, and its energy promotes one electron from a lower d orbital to a higher one.
频率与能量差相匹配的光会被吸收， 它的能量把一个电子从较低的 d 轨道激发到较高的 d 轨道。
That colour is taken out of the white light, so what reaches your eye is the complementary colour of the light absorbed.
这些颜色从白光中被拿走了，所以进入你眼睛的是被吸收光的互补色。
Change the ligand and you change the gap, so you change the colour — which is exactly why ligand exchange changes colour.
换一个配体，能量差的大小就变了，颜色也就变了——这正是配体交换会改变颜色的原因。
Complexes form isomers too, and one kind you have already met.
配合物也会形成异构体，其中一种你已经学过。
A carbon with four different groups is not the same as its mirror image — like your left and right hand, no turning will ever line them up.
一个连着四个不同基团的碳原子， 和它的镜像并不相同——就像你的左手和右手，无论怎么转都无法重合。
That is optical isomerism, and a complex ion can do exactly the same.
这就是旋光异构。 记住它，因为配离子也能做到完全一样的事。
Complexes show two kinds of stereoisomerism.
配合物有两类立体异构。
First, geometrical.
第一类是几何异构。
Take the square planar platinum complex with two ammonia and two chloride ligands.
以带两个氨和两个氯配体的平面正方形铂配合物为例。
With the ammonias next to each other you get the cis isomer — cisplatin, a real anticancer drug.
两个氨相邻，就是顺式异构体——顺铂，一种真正在用的抗癌药。
Opposite each other gives the trans isomer, no use as a drug.
两个氨相对，就是反式异构体， 而它完全没有药效。
It also decides polarity: in the trans form the dipoles cancel, so it is non-polar; the cis form is polar.
这也决定了极性：反式中两个偶极相互抵消，所以是非极性的； 顺式中则不抵消。
Second, optical isomerism.
第二类是旋光异构。
An octahedral complex with three bidentate ligands has two mirror images that never superimpose.
带三个双齿配体的八面体配合物， 有两个永远无法重合的镜像。
Which complex is more stable?
哪一个配合物更稳定？
The stability constant answers that.
稳定常数给出答案。
It is the equilibrium constant for forming the complex ion in solution from the metal ion and its ligands — and water is never included: it is in huge excess.
它是在溶液中，由金属离子和它的配体 形成配离子这一过程的平衡常数——而且表达式中绝不写水，因为溶剂是大大过量的。
A large constant means a very stable complex, so ligand exchange moves towards the larger one.
稳定常数大，说明配合物很稳定，所以配体交换总是向稳定常数更大的一方进行。
Silver with two cyanides has an enormous value, so cyanide pushes ammonia out.
银与两个氰配体的稳定常数极大，所以氰离子会直接把氨挤出去。
A bidentate ligand beats the monodentate one it replaces.
而双齿配体总是胜过被它取代的单齿配体。
Transition ions change oxidation state easily, so they are the ions you meet in redox titrations.
过渡金属离子很容易改变氧化态，所以氧化还原滴定几乎都用它们。
First, check feasibility: find the cell potential from the two standard electrode potentials — positive means it can happen.
计算之前先判断可行性： 用两个标准电极电势算出电池电动势，为正就说明反应可以发生。
Then learn three systems.
然后记住三个体系。
Manganate seven with iron two in acid, purple to colourless.
酸性条件下高锰酸根与亚铁离子，紫色变无色。
Manganate seven with ethanedioate in acid, warmed to start it.
酸性条件下高锰酸根与乙二酸根，需要加热引发。
And copper two with iodide, which sets iodine free — titrate that with thiosulfate.
还有铜二价与碘离子，会放出碘——再用硫代硫酸盐滴定这些碘。
Now the calculation.
现在做计算。
Twenty-five point zero cubic centimetres of iron two solution needs twenty-two point zero cubic centimetres of manganate seven at zero point zero two zero zero moles per cubic decimetre.
二十五点零立方厘米的亚铁溶液，需要二十二点零立方厘米、 浓度为零点零二零零摩尔每立方分米的高锰酸钾溶液才到终点。
Find the iron two concentration.
求亚铁离子的浓度。
Pause and try it.
先暂停试一试。
Ready?
好了吗？
Step one, the ratio: one manganate seven ion oxidises five iron two ions.
第一步，比例：一个高锰酸根离子氧化五个亚铁离子。
Step two, moles of manganate seven — concentration times volume gives four point four zero times ten to the minus four.
第二步，高锰酸根的物质的量等于浓度乘以体积：四点四零乘以十的负四次方。
Step three, multiply by five: two point two zero times ten to the minus three.
第三步，乘以五得到亚铁离子的物质的量：二点二零乘以十的负三次方。
Step four, divide by the volume of the iron solution, and the answer is zero point zero eight eight zero.
第四步，除以亚铁溶液的体积，答案是零点零八八零摩尔每立方分米。
Three marks students throw away.
三个学生常丢的分。
First, the definition: it is the ion, not the atom, that must have a partially filled d sub-shell — which is why scandium and zinc are out.
第一，定义：必须是"离子"而不是原子的 d 亚层部分填充—— 这正是钪和锌被排除的原因。
Second, colour: give all three steps — the d orbitals split, light matching the gap is absorbed and promotes an electron, and you see the complementary colour.
第二，颜色：三步都要写——d 轨道分裂、 吸收与能量差匹配的光并激发一个电子、然后看到互补色。
Third, balance the equation for a ligand exchange: count the ligands, and check the charge on both sides.
第三，配体交换的方程式要配平： 先数配体，再检查两边的总电荷。