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LEARNING OBJECTIVES  学习目标

Be able to explain: 能够解释:
the trends in melting and boiling temperatures of the elements of Periods 2 and 3 of the Periodic Table in terms of the structure of the element and the bonding between its atoms or molecules;
元素周期表 2 和 3 周期元素的熔化和沸腾温度趋势,包括元素的结构及其原子或分子之间的键合;
  • the specific trends in ionisation energy of the elements across Periods 2 and 3 of the Periodic Table.
    元素周期表第 2 周期和第 3 周期中元素电离能的具体趋势。

WHAT ARE PERIODIC PROPERTIES?
什么是周期性属性?

The elements in a period exhibit periodic properties (also sometimes called periodicity).
周期中的元素表现出周期性 (有时也称为周期性) 。

We can illustrate the idea of periodic properties by looking at the elements in Periods 2 and 3. We have already seen one example of periodic properties in Periods 2 and 3 - the regular repeating pattern of electronic configurations from ns 1 ns 1 ns^(1)\mathrm{ns}^{1} through to ns 2 np 6 ns 2 np 6 ns^(2)np^(6)\mathrm{ns}^{2} \mathrm{np}{ }^{6}. Other examples are the trends in atomic radii, melting and boiling temperatures, and first ionisation energies.
我们可以通过查看周期 2 和 3 中的元素来说明周期性的概念。我们已经在周期 2 和 3 中看到了周期性特性的一个例子 - 从 ns 1 ns 1 ns^(1)\mathrm{ns}^{1} 到 的 ns 2 np 6 ns 2 np 6 ns^(2)np^(6)\mathrm{ns}^{2} \mathrm{np}{ }^{6} 电子构型的规则重复模式。其他例子是原子半径、熔化和沸腾温度以及第一电离能的趋势。

ATOMIC RADII 原子半径

The atomic radius of an element is a measurement of the size of its atoms. It is the distance from the centre of the nucleus to the boundary of the electron cloud. Since the atom does not have a well-defined boundary, we can find the atomic radius by determining the distance between the two nuclei and dividing it by two.
元素的原子半径是其原子大小的量度。它是从原子核中心到电子云边界的距离。由于原子没有明确定义的边界,我们可以通过确定两个原子核之间的距离并将其除以 2 来找到原子半径。


fig A Measuring (a) the covalent radius and (b) the van der Waals radius.
图 A 测量 (a) 共价半径和 (b) 范德华半径。

Diagram (a) in fig A shows two bonded atoms. The atoms are closer together than they are in diagram (b) (where they are only just touching).
图 A 中的图 (a) 显示了两个键合原子。原子比图 (b) 中的原子靠得更近(其中它们只是刚刚接触)。
In diagram (a), we are measuring the covalent radius.
在图 (a) 中,我们正在测量共价半径。

The radius we are measuring in diagram (b) is called the van der Waals radius. This is the only radius that we can determine for neon and argon, because they do not bond with other elements.
我们在图 (b) 中测量的半径称为范德华半径。这是我们唯一可以确定的氖和氩的半径,因为它们不与其他元素键合。
There is a third radius that is used for metals. It is called the metallic radius.
还有第三个半径用于金属。它被称为金属半径。

Fig B shows the trend in covalent radii across the second period (lithium to fluorine, or Li to F) and third period (sodium to chlorine, or Na to Cl ) ) )) measured in nanometres ( 10 9 m ) 10 9 m (10^(-9)(m))\left(10^{-9} \mathrm{~m}\right).
图 B 显示了第二周期(锂到氟,或锂到氟)和第三周期(钠到氯,或钠到氯,以纳米 ) ) )) 为单位)的共价半径趋势 ( 10 9 m ) 10 9 m (10^(-9)(m))\left(10^{-9} \mathrm{~m}\right)

DID YOU KNOW? 您是否了解?

Different types of radii give different measurements for the same element. The van der Waals radius is always larger. So always compare like with like when you are looking at the trends in atomic radii.
不同类型的半径对同一单元给出不同的测量值。范德华半径总是更大。因此,当您查看原子半径的趋势时,请始终进行同类比较。


fig B Trend in covalent radii across the second and third periods.
图 B 第二和第三周期共价半径的趋势。

You can see that the radius decreases across each period. This is because as the number of protons in the nucleus increases, so does the nuclear charge. This results in an increase in the attractive force between the nucleus and the outer electrons. This increase in attractive force offsets (counterbalances) the increase in electron-electron repulsion as the number of electrons in the outer quantum shell increases.
您可以看到半径在每个周期中都会减小。这是因为随着原子核中质子数量的增加,核电荷也会增加。这导致原子核和外电子之间的吸引力增加。随着外量子壳中电子数量的增加,这种吸引力的增加抵消(抵消了)电子-电子排斥的增加。

MELTING AND BOILING TEMPERATURES
熔化和沸腾温度

Table A below illustrates the changes in melting and boiling temperatures for the elements in Periods 2 and 3.
下面的表 A 说明了周期 2 和 3 中元素的熔化和沸腾温度变化。
PERIOD 2 第 2 期 Li  Be  B C (DIAMOND) C (钻石) N 0 0 0\mathbf{0} F
melting temperature / C / C //^(@)C/{ }^{\circ} \mathrm{C}
熔融温度 / C / C //^(@)C/{ }^{\circ} \mathrm{C}
181 1278 2300 3550 -210 -218 -220
boiling temperature / C / C //^(@)C/{ }^{\circ} \mathrm{C}
沸腾温度 / C / C //^(@)C/{ }^{\circ} \mathrm{C}
1342 2970 3927 4827 -196 -183 -188
type of bonding 粘合类型 metallic 金属 metallic 金属 covalent 共价的 covalent 共价的 covalent 共价的 covalent 共价的 covalent 共价的
structure 结构 giant lattice 巨型晶格 giant lattice 巨型晶格 giant lattice 巨型晶格 giant lattice 巨型晶格
 简单分子
simple
molecular
simple molecular| simple | | :---: | | molecular |
 简单分子
simple
molecular
simple molecular| simple | | :---: | | molecular |
 简单分子
simple
molecular
simple molecular| simple | | :---: | | molecular |
PERIOD 2 Li Be B C (DIAMOND) N 0 F melting temperature //^(@)C 181 1278 2300 3550 -210 -218 -220 boiling temperature //^(@)C 1342 2970 3927 4827 -196 -183 -188 type of bonding metallic metallic covalent covalent covalent covalent covalent structure giant lattice giant lattice giant lattice giant lattice "simple molecular" "simple molecular" "simple molecular"| PERIOD 2 | Li | Be | B | C (DIAMOND) | N | $\mathbf{0}$ | F | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | melting temperature $/{ }^{\circ} \mathrm{C}$ | 181 | 1278 | 2300 | 3550 | -210 | -218 | -220 | | boiling temperature $/{ }^{\circ} \mathrm{C}$ | 1342 | 2970 | 3927 | 4827 | -196 | -183 | -188 | | type of bonding | metallic | metallic | covalent | covalent | covalent | covalent | covalent | | structure | giant lattice | giant lattice | giant lattice | giant lattice | simple <br> molecular | simple <br> molecular | simple <br> molecular |
PERIOD 3 第 3 期 Na  Mg 毫克 Al  Si  P P P\mathbf{P} S S S\mathbf{S} CI
melting temperature / C / C //^(@)C/{ }^{\circ} \mathrm{C}
熔融温度 / C / C //^(@)C/{ }^{\circ} \mathrm{C}
98 649 660 1440 44 113 -101
boiling temperature / C / C //^(@)C/{ }^{\circ} \mathrm{C}
沸腾温度 / C / C //^(@)C/{ }^{\circ} \mathrm{C}
883 1107 2467 2355 280 445 -35
type of bonding 粘合类型 metallic 金属 metallic 金属 metallic 金属 covalent 共价的 covalent 共价的 covalent 共价的 covalent 共价的
structure 结构 giant lattice 巨型晶格 giant lattice 巨型晶格 giant lattice 巨型晶格 giant lattice 巨型晶格
 简单分子
simple
molecular
simple molecular| simple | | :---: | | molecular |
 简单分子
simple
molecular
simple molecular| simple | | :---: | | molecular |
 简单分子
simple
molecular
simple molecular| simple | | :---: | | molecular |
PERIOD 3 Na Mg Al Si P S CI melting temperature //^(@)C 98 649 660 1440 44 113 -101 boiling temperature //^(@)C 883 1107 2467 2355 280 445 -35 type of bonding metallic metallic metallic covalent covalent covalent covalent structure giant lattice giant lattice giant lattice giant lattice "simple molecular" "simple molecular" "simple molecular"| PERIOD 3 | Na | Mg | Al | Si | $\mathbf{P}$ | $\mathbf{S}$ | CI | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | melting temperature $/{ }^{\circ} \mathrm{C}$ | 98 | 649 | 660 | 1440 | 44 | 113 | -101 | | boiling temperature $/{ }^{\circ} \mathrm{C}$ | 883 | 1107 | 2467 | 2355 | 280 | 445 | -35 | | type of bonding | metallic | metallic | metallic | covalent | covalent | covalent | covalent | | structure | giant lattice | giant lattice | giant lattice | giant lattice | simple <br> molecular | simple <br> molecular | simple <br> molecular |
table A A A\mathbf{A} Changes in melting and boiling temperatures for Period 2 and 3 elements.
A A A\mathbf{A} 第 2 周期和第 3 周期元素的熔化和沸腾温度变化。

EXAM HINT 考试提示

If you are asked to explain trends in melting points, you do not need to be able to remember specific melting points of elements. You only need to be able to explain the trends in terms of bond type.
如果要求您解释熔点的趋势,您不需要能够记住元素的特定熔点。您只需要能够解释债券类型的趋势。
You may have noticed that the elements with giant lattice structures have high melting and boiling temperatures, and those with simple molecular structures have low melting and boiling temperatures. We will explain the reason for this in Topic 3.
您可能已经注意到,具有巨大晶格结构的元素具有较高的熔化和沸腾温度,而具有简单分子结构的元素具有较低的熔化和沸腾温度。我们将在主题 3 中解释其原因。

FIRST IONISATION ENERGIES
第一电离能

Fig C C C\mathbf{C} is a plot of first ionisation energy for the first three Periods.
C C C\mathbf{C} 是前三个周期的第一电离能图。


Δ Δ Delta\Delta fig C First ionisation energy for Periods 1-3.
Δ Δ Delta\Delta 图 C 第 1-3 周期的第一电离能。

HYDROGEN AND HELIUM 氢气和氦气

The electronic configurations of hydrogen and helium are 1 s 1 1 s 1 1s^(1)1 \mathrm{~s}^{1} and 1 s 2 1 s 2 1s^(2)1 s^{2}, respectively.
氢和氦的电子构型分别为 1 s 1 1 s 1 1s^(1)1 \mathrm{~s}^{1} 1 s 2 1 s 2 1s^(2)1 s^{2}

We can explain the increase in first ionisation energy from hydrogen to helium by the increase in nuclear charge from 1 to 2 as an extra proton is added. This increase in nuclear charge more than offsets the increase in electron-electron repulsion in the 1 s orbital as a second electron is added.
我们可以解释为什么随着额外质子的增加,核电荷从 1 增加到 2 而从氢到氦的第一电离能增加。随着第二个电子的添加,核电荷的增加超过了 1 s 轨道中电子-电子排斥的增加。

LEARNING TIP 学习小贴士

You may hear or read about a suggestion that the first ionisation energy of boron is less than that of beryllium, because by losing an electron the boron atom will acquire a full 2 s orbital and so become more stable. In fact, because energy has to be supplied to the boron atom in order to remove an electron, the ion formed is energetically less stable than the atom because it has a higher energy value.
您可能听说或读到过这样一种说法,即硼的第一电离能小于铍的第一电离能,因为通过失去一个电子,硼原子将获得完整的 2 s 轨道,从而变得更加稳定。事实上,由于必须向硼原子提供能量才能去除电子,因此形成的离子在能量上不如原子稳定,因为它具有更高的能量值。

ANOMALIES 异常

We have already explained the general increase in first ionisation energy across the period from lithium to neon ( Li to Ne ) and from sodium to argon (Na to Ar), in terms of the increasing nuclear charge. However, you will notice that there are two anomalies in each case. The first ionisation energy of the Group 3 element is less than that of the Group 2 element, and the first ionisation energy of the Group 6 element is less than that of the Group 5 element.
我们已经解释了从锂到氖(Li 到 Ne)和从钠到氩(Na 到 Ar)期间第一电离能的普遍增加,就核电荷的增加而言。但是,您会注意到每种情况下都有两个异常。第 3 族元素的第一电离能小于第 2 族元素的第一电离能,第 6 族元素的第一电离能小于第 5 族元素的第一电离能。

First, consider the case of beryllium (Be) and boron (B).
首先,考虑铍 (Be) 和硼 (B) 的情况。

The electronic configurations are:
电子配置为:
Be: 1 s 2 2 s 2 B: 1 s 2 2 s 2 2 p 1  Be:  1 s 2 2 s 2  B:  1 s 2 2 s 2 2 p 1 {:[" Be: ",1s^(2)2s^(2)],[" B: ",1s^(2)2s^(2)2p^(1)]:}\begin{array}{ll} \text { Be: } & 1 s^{2} 2 s^{2} \\ \text { B: } & 1 s^{2} 2 s^{2} 2 p^{1} \end{array}
Although the nuclear charge of the boron atom is greater than that of the beryllium atom, the outer electron of boron has more energy, since it is in a 2 p orbital as opposed to the 2 s orbital for beryllium. For this reason, the energy required to remove 2 p 2 p 2p2 p electron in boron is less than the energy required to remove a 2 s electron from a beryllium atom. In addition, the 2 p 2 p 2p2 p electron in boron experiences greater electron-electron repulsion (i.e. greater shielding) because there are two inner electron sub-shells as opposed to only one in the beryllium atom.
尽管硼原子的核电荷大于铍原子的核电荷,但硼的外电子具有更多的能量,因为它位于 2 p 轨道上,而铍的轨道为 2 s 轨道。因此,去除 2 p 2 p 2p2 p 硼中电子所需的能量小于从铍原子中去除 2 s 电子所需的能量。此外,硼中的 2 p 2 p 2p2 p 电子经历更大的电子-电子排斥(即更大的屏蔽),因为有两个内部电子子壳层,而铍原子中只有一个。
A similar argument applies to magnesium and aluminium (Mg and Al ), except that in this case it involves the 3 s and 3 p electrons.
类似的论点适用于镁和铝(Mg 和 Al),只是在这种情况下它涉及 3 s 和 3 p 电子。

LEARNING TIP 学习小贴士

You may hear or read about an explanation stating that the outer electron in the boron atom is further from the nucleus than the outer electron in the beryllium atom. This is not correct. The boron atom is smaller than the beryllium atom as shown in fig B.
您可能会听到或读到这样一种解释,即硼原子中的外电子比铍原子中的外电子离原子核更远。这是不正确的。硼原子比铍原子小,如图 B 所示。

NITROGEN AND OXYGEN 氮气和氧气

Now consider nitrogen ( N ) and oxygen ( O ).
现在考虑氮 ( N ) 和氧 ( O )。

The electronic configurations are:
电子配置为:
N : 1 s 2 2 s 2 2 p x 1 2 p y 1 2 p z 1 O : 1 s 2 2 s 2 2 p x 2 2 p y 1 2 p z 1 N : 1 s 2 2 s 2 2 p x 1 2 p y 1 2 p z 1 O : 1 s 2 2 s 2 2 p x 2 2 p y 1 2 p z 1 {:[N:,1s^(2)2s^(2)2p_(x)^(1)2p_(y)^(1)2p_(z)^(1)],[O:,1s^(2)2s^(2)2p_(x)^(2)2p_(y)^(1)2p_(z)^(1)]:}\begin{array}{ll} \mathrm{N}: & 1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}_{\mathrm{x}}{ }^{1} 2 p_{\mathrm{y}}{ }^{1} 2 \mathrm{p}_{\mathrm{z}}{ }^{1} \\ \mathrm{O}: & 1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}_{\mathrm{x}}{ }^{2} 2 \mathrm{p}_{\mathrm{y}}{ }^{1} 2 \mathrm{p}_{\mathrm{z}}{ }^{1} \end{array}
The first electron removed from the oxygen atom is one of the two paired electrons in the 2 p x 2 p x 2p_(x)2 p_{x} orbital.
从氧原子中去除的第一个电子是 2 p x 2 p x 2p_(x)2 p_{x} 轨道中两个成对电子之一。

The presence of two electrons in a single orbital increases the electron-electron repulsion in this orbital. So less energy is required to remove one of these electrons than is required to remove a 2 p electron from a nitrogen atom, despite the larger nuclear charge of the oxygen atom.
单个轨道中存在两个电子会增加该轨道中的电子-电子排斥。因此,尽管氧原子的核电荷较大,但去除其中一个电子所需的能量比从氮原子中去除 2 p 电子所需的能量要少。

LEARNING TIP 学习小贴士

You may see an explanation stating that the first ionisation energy of oxygen is less than that of nitrogen, saying this is because when the oxygen atom loses an electron, it acquires a stable half-full 2 p 2 p 2p2 p subshell. However, there is no special stability associated with a half-full sub-shell. Also, as with the boron atom, energy must be supplied to the oxygen atom in order to remove an electron, so the ion formed is energetically less stable than the atom.
你可能会看到一个解释,说氧的第一电离能小于氮的第一电离能,说这是因为当氧原子失去一个电子时,它会获得一个稳定的半满 2 p 2 p 2p2 p 亚壳层。但是,半满子壳没有特殊的稳定性。此外,与硼原子一样,必须向氧原子提供能量才能去除电子,因此形成的离子在能量上不如原子稳定。

You may also hear or read about an explanation stating that the outer electron in the oxygen atom is further from the nucleus than the outer electron in the nitrogen atom. This is not correct. The oxygen atom is smaller than the nitrogen atom as shown in fig B.
您可能还会听到或读到一种解释,指出氧原子中的外电子比氮原子中的外电子离原子核更远。这是不正确的。氧原子比氮原子小,如图 B 所示。

CHECKPOINT 检查站

SKILLS REASONING 技能推理

  1. An element has very high melting and boiling temperatures.
    元素具有非常高的熔化和沸腾温度。

    (a) What type of structure is it likely to have?
    (a) 它可能具有什么类型的结构?

    (b) Which physical property would help you determine whether the bonding was metallic or covalent? Explain your answer.
    (b) 哪种物理特性可以帮助您确定键合是金属键还是共价键?解释你的答案。
  2. Explain why the first ionisation energy of helium is higher than that of hydrogen.
    解释为什么氦的第一电离能高于氢的第一电离能。
  3. Explain why the first ionisation energy of lithium is much lower than that of helium, even though the lithium atom has a greater nuclear charge.
    解释为什么锂的第一电离能远低于氦,即使锂原子具有更大的核电荷。
  4. Would you expect the first ionisation energy of gallium (Ga) to be higher or lower than that of calcium (Ca)? Explain your answer.
    您预计镓 (Ga) 的第一电离能会高于还是低于钙 (Ca) 的第一电离能?解释你的答案。
  5. Why does neon have the highest first ionisation energy of all the elements in Period 2?
    为什么氖在周期 2 的所有元素中具有最高的第一电离能?

SUBJECT VOCABULARY 学科词汇

periodic properties (periodicity) regularly repeating patterns of atomic, physical and chemical properties, which can be predicted using the Periodic Table and explained using the electron configurations of the elements
周期性 (Periodicity) 定期重复原子、物理和化学性质的模式,可以使用元素周期表进行预测,并使用元素的电子构型进行解释

2
THINKING BIGGER
2 放眼更远

ELEMENTAL FINGERPRINTS 元素指纹

Is it possible to know what stars in distant galaxies are made from?
有可能知道遥远星系中的恒星是由什么组成的吗?

ELEMENTAL ‘FINGERPRINTS' 元素“指纹”

In 1825, the French philosopher Auguste Comte said that there are some things we will never know, among these the chemical composition of the stars. This was an unfortunate example. We do not need actual material from the stars in order to analyse them. What we do need, of course, is information, and this they send us plentifully, in the form of light. To reveal this information, we must separate the light into its different wavelengths or colours.
1825 年,法国哲学家奥古斯特·孔德 (Auguste Comte) 说,有些事情我们永远不会知道,其中包括恒星的化学成分。这是一个不幸的例子。我们不需要来自星星的实际材料来分析它们。当然,我们确实需要的是信息,而这些信息以光的形式大量发送给我们。为了揭示这些信息,我们必须将光分成不同的波长或颜色。

Emission spectrum 发射光谱


fig A 图 A
From From Stars to Stalagmites: How Everything Connects by Paul Braterman
从星星到石笋:万物如何连接 保罗·布拉特曼

THINKING BIGGER TIPS 思考更宏大的技巧

An emission spectrum shows the frequencies of electromagnetic radiation that are emitted by a substance. An absorption spectrum shows the frequencies of electromagnetic radiation that are absorbed by the substance. This means that the absorption spectrum is essentially the ‘negative’ of the emission spectrum.
发射光谱显示物质发射的电磁辐射的频率。吸收光谱显示了被物质吸收的电磁辐射的频率。这意味着吸收光谱本质上是发射光谱的“负”值。
Many years earlier Isaac Newton had passed sunlight through a glass prism, and found that this gave him all the colours of the rainbow. In 1802 William Wollaston in England … repeated the experiment, using an up-to-date high quality glass prism, and discovered that the continuous spectrum was interrupted by narrow dark lines. (Rather unfairly these are now known as Frauenhofer lines, after the German physicist Joseph von Frauenhofer, who confirmed and extended Wollaston’s observations). These lines later proved to match exactly the light given out by different elements when heated or in electric discharges. A now familiar example is provided by the element sodium, whose yellow emission is used in street lamps. The lines can be used as a kind of fingerprint for each element. We can take this fingerprint using electrical discharges here on Earth and compare it with the lines found in the spectrum of the Sun. In this way we can get a good chemical analysis of the surface layer of the Sun or of any other star whose light we can collect and astronomers now extend the same process to the most distant galaxies.
许多年前,艾萨克·牛顿 (Isaac Newton) 将阳光穿过玻璃棱镜,发现这给了他彩虹的所有颜色。1802 年,威廉·沃拉斯顿 (William Wollaston) 在英格兰......使用最新的高质量玻璃棱镜重复实验,发现连续光谱被狭窄的黑线打断。(相当不公平的是,这些现在被称为弗劳恩霍夫线,以德国物理学家约瑟夫·冯·弗劳恩霍夫 (Joseph von Frauenhofer) 的名字命名,他证实并扩展了沃拉斯顿的观察结果)。这些线路后来证明与不同元件在加热或放电时发出的光完全匹配。现在熟悉的一个例子是钠元素,其黄色发射用于路灯。这些线条可以用作每个元素的一种指纹。我们可以使用地球上的放电来获取这个指纹,并将其与太阳光谱中发现的线进行比较。通过这种方式,我们可以对太阳或任何其他恒星的表层进行良好的化学分析,天文学家现在将相同的过程扩展到最遥远的星系。

DID YOU KNOW? 您是否了解?

Many images of galaxies taken by the Hubble telescope use the line emission spectra of elements such as hydrogen, sulfur and oxygen to build up a colour image of the galaxy. The different elements are assigned red, green and blue frequencies to build up a full colour image.
哈勃望远镜拍摄的许多星系图像都使用氢、硫和氧等元素的线发射光谱来构建星系的彩色图像。不同的元素被分配了红色、绿色和蓝色频率,以构建全彩色图像。

SCIENGE COMMUNIGATION 科学通讯

The extract is taken from the book From Stars to Stalagmites: How Everything Connects by Paul Braterman (2012).
摘录自保罗·布拉特曼 (Paul Braterman) (2012) 的《从星星到石笋:万物如何连接》一书。
  1. (a) Do you think this article is aimed at scientists, the general public, or people who are not scientists but have an interest in science? Look back through the extract and find examples to support your answer.
    (a) 你认为这篇文章是针对科学家、公众还是不是科学家但对科学感兴趣的人?回顾摘录并找到示例来支持您的答案。

    (b) Explain how the book’s subtitle How Everything Connects is supported by the extract.
    (b) 解释该摘录如何支持该书的副标题“万物如何连接”。

CHEMISTRY IN DETAIL 化学细节

  1. (a) Describe the structure of the sodium atom in terms of protons, neutrons and electrons.
    (a) 用质子、中子和电子来描述钠原子的结构。

    (b) Give the electronic structure of a sodium atom using s, p, d notation.
    (b) 用 s、p、d 符号给出钠原子的电子结构。

    © What is giving rise to (i) the absorption and (ii) the emission spectrum of sodium?
    © 是什么导致了钠的 (i) 吸收和 (ii) 发射光谱?

    (d) Can you suggest why sodium vapour is used in preference to potassium vapour in street lamps?
    (d) 您能否说明为什么路灯中首选使用钠蒸气而不是钾蒸气?
  2. Would you expect heavier isotopes of sodium to give a different ‘fingerprint’? Explain your answer.
    您会期望较重的钠同位素产生不同的“指纹”吗?解释你的答案。
  3. The element helium was first discovered in the outer layer of the Sun. Suggest why it was discovered there before it was discovered here on Earth. Think back to work you have done on identifying alkali metals using flame tests.
    氦元素最早是在太阳的外层发现的。请说明为什么它在地球上被发现之前就在那里被发现。回想一下您使用火焰测试识别碱金属所做的工作。

ACTIVITY 活动

Today, the elemental fingerprints in the light collected from distant galaxies have provided evidence for an expanding universe. The elemental fingerprints are subject to a phenomenon called ‘red shift’.
今天,从遥远星系收集的光中的元素指纹为宇宙膨胀提供了证据。元素指纹会受到一种称为“红移”的现象的影响。

Explain in 200-300 words how red shift has provided evidence for an expanding universe. Try to structure your explanation so that the concepts can be understood by an audience of 16 -year-old students.
用 200-300 字解释红移如何为膨胀的宇宙提供证据。尝试构建您的解释,以便 16 岁的学生可以理解这些概念。

fig B A picture taken by the Hubble Telescope showing over 10000 galaxies!
图 B 哈勃望远镜拍摄的照片显示了超过 10000 个星系!

THINKING BIGGER TIP 考虑更宏大的建议

As well as considering the complexity of the scientific ideas, think about the level of detail in which they are discussed. Notice how many different branches of natural science are mentioned in the extract. Are all of the scientists mentioned chemists or did they work in different scientific disciplines?
除了考虑科学思想的复杂性外,还要考虑讨论它们的细节程度。请注意摘录中提到了多少个不同的自然科学分支。是所有科学家都提到了化学家,还是他们在不同的科学学科工作?

WRITING SCIENTIFICALLY 科学写作

You need to use reasoning and maybe examples to give an explanation of your answer and support your point. Consider how the particles involved (e.g. atoms, ions, molecules, electrons, etc.) affect your answer.
你需要使用推理,也许还有例子来解释你的答案并支持你的观点。考虑所涉及的粒子(例如原子、离子、分子、电子等)如何影响您的答案。

THINKING BIGGER TIP 考虑更宏大的建议

The Activity provides opportunity for creative explanation of advanced concept to younger students.
该活动为低年级学生提供了创造性地解释高级概念的机会。
1 The relative atomic mass of boron is 10.8 .
1 硼的相对原子质量为 10.8 。

A sample of boron contains the isotopes 5 10 B 5 10 B _(5)^(10)B{ }_{5}^{10} \mathrm{~B} and 5 11 B 5 11 B _(5)^(11)B{ }_{5}^{11} \mathrm{~B}.
硼样品包含同位素 5 10 B 5 10 B _(5)^(10)B{ }_{5}^{10} \mathrm{~B} 5 11 B 5 11 B _(5)^(11)B{ }_{5}^{11} \mathrm{~B}

What is the percentage of 5 11 B 5 11 B _(5)^(11)B{ }_{5}^{11} \mathrm{~B} atoms in the isotopic mixture of this sample?
该样品的同位素混合物中 5 11 B 5 11 B _(5)^(11)B{ }_{5}^{11} \mathrm{~B} 原子的百分比是多少?

A 0.8 % 0.8 % 0.8%0.8 \% 一个 0.8 % 0.8 % 0.8%0.8 \%
B 8.0 % 8.0 % 8.0%8.0 \%  8.0 % 8.0 % 8.0%8.0 \%
C 20 % 20 % 20%20 \%  20 % 20 % 20%20 \%
D 80 % 80 % 80%80 \%
[1]
(Total for Question 1 = 1 mark)
(问题 1 的总分 = 1 分)

2 Which of the following elements has no paired p p pp electrons in a single uncombined atom of the element?
2 以下哪种元素在该元素的单个未组合原子中没有成对 p p pp 电子?

A carbon 
B oxygen B 氧
C fluorine D neon C 氟 D 霓虹灯
[1]
(Total for Question 2 = 1 2 = 1 2=12=1 mark)
(问 2 = 1 2 = 1 2=12=1 号总计)

3. Which of the following electronic configurations is that of an atom of an element which forms a simple ion with a charge of -3 ?
3. 以下哪种电子构型是形成电荷为 -3 的简单离子的元素的原子构型?

A 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1} 一个 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 1s^(2)2s^(2)2p^(6)3s^(2)3p^(1)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}
B 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}  1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 1 s 2 2 s 2 2 p 6 3 s 2 3 p 3 1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}
C 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 1 4 s 2 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 1 4 s 2 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)4s^(2)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{1} 4 s^{2}  1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 1 4 s 2 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 1 4 s 2 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)4s^(2)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{1} 4 s^{2}
D 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 3 4 s 2 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 3 4 s 2 1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(3)4s^(2)1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3} 4 s^{2}
(Total for Question 3 = 1 3 = 1 3=13=1 mark)
(问 3 = 1 3 = 1 3=13=1 号总计)

4 A sample of chlorine contains isotopes of mass numbers 35 and 37.
4 氯样品含有质量数为 35 和 37 的同位素。

The sample is analysed in a mass spectrometer. How many peaks corresponding to Cl 2 + Cl 2 + Cl_(2)^(+)\mathrm{Cl}_{2}{ }^{+}are recorded?
样品在质谱仪中进行分析。记录了多少个对应的 Cl 2 + Cl 2 + Cl_(2)^(+)\mathrm{Cl}_{2}{ }^{+} 峰?

A 1 答 1
B 2 乙 2
C 3
D 4
(Total for Question 4 = 1 4 = 1 4=14=1 mark)
(问 4 = 1 4 = 1 4=14=1 号总计)
5 What is the atomic number of an element that contains atoms which have four unpaired electrons in their ground state?
5 包含基态有四个不成对电子的原子的元素的原子序数是多少?

A 6 一个 6
B 16 C 乙 16 C
22
D 26 第 26 条
[1]
(Total for Question 5 = 1 5 = 1 5=15=1 mark)
(问 5 = 1 5 = 1 5=15=1 号总计)

6 Which of the following ions has more electrons than protons, and also has more protons than neutrons?
6 以下哪个离子的电子数比质子多,质子数也比中子多?

( H = 1 1 H H = 1 1 H H=_(1)^(1)H\mathrm{H}={ }_{1}^{1} \mathrm{H}  H = 1 1 H H = 1 1 H H=_(1)^(1)H\mathrm{H}={ }_{1}^{1} \mathrm{H}
D = 1 2 H D = 1 2 H D=_(1)^(2)H\mathrm{D}={ }_{1}^{2} \mathrm{H}
He = 2 4 He O = 8 16 O ) He = 2 4 He O = 8 16 O {:He=_(2)^(4)HequadO=_(8)^(16)O)\left.\mathrm{He}={ }_{2}^{4} \mathrm{He} \quad \mathrm{O}={ }_{8}^{16} \mathrm{O}\right)
A OD A OD AOD^(-)\mathrm{A} \mathrm{OD}^{-}
B D 3 O + D 3 O + D_(3)O^(+)\mathrm{D}_{3} \mathrm{O}^{+}  D 3 O + D 3 O + D_(3)O^(+)\mathrm{D}_{3} \mathrm{O}^{+}
C He + D OH He + D OH He^(+)quadDOH^(-)\mathrm{He}^{+} \quad \mathrm{D} \mathrm{OH}^{-}  He + D OH He + D OH He^(+)quadDOH^(-)\mathrm{He}^{+} \quad \mathrm{D} \mathrm{OH}^{-}
(Total for Question 6 = 1 6 = 1 6=16=1 mark)
(问 6 = 1 6 = 1 6=16=1 号总计)

7 A sample of helium from a rock was found to contain two isotopes with the following composition by mass: 3 He , 0.992 % 3 He , 0.992 % ^(3)He,0.992%{ }^{3} \mathrm{He}, 0.992 \%; 4 He , 99.008 % 4 He , 99.008 % ^(4)He,99.008%{ }^{4} \mathrm{He}, 99.008 \%.
7 发现岩石中的氦样品含有两种同位素,按质量计算,其成分如下: 3 He , 0.992 % 3 He , 0.992 % ^(3)He,0.992%{ }^{3} \mathrm{He}, 0.992 \% ; 4 He , 99.008 % 4 He , 99.008 % ^(4)He,99.008%{ }^{4} \mathrm{He}, 99.008 \%

(a) State what is meant by isotopes.
(a) 说明同位素的含义。

(b) State the difference in the atomic structures of 3 He 3 He ^(3)He{ }^{3} \mathrm{He} and 4 He 4 He ^(4)He{ }^{4} \mathrm{He}.
(b) 说明 和 4 He 4 He ^(4)He{ }^{4} \mathrm{He} 3 He 3 He ^(3)He{ }^{3} \mathrm{He} 原子结构的差异。

© (i) Which isotope is used as the basis for relative atomic mass measurements?
© (i) 哪种同位素用作相对原子质量测量的基础?

(ii) Calculate the relative atomic mass of helium in the rock sample.
(ii) 计算岩石样品中氦的相对原子质量。

(d) Helium has the largest first ionisation energy of all the elements.
(d) 氦在所有元素中具有最大的第一电离能。

(i) State what is meant by first ionisation energy.
(i) 说明第一电离能的含义。

(ii) Write an equation, including state symbols, to
(ii) 写一个方程式,包括状态符号,以

represent the first ionisation energy of helium.
代表氦的第一电离能。

(iii) Explain why the first ionisation energy of helium is larger than that of hydrogen.
(iii) 解释为什么氦的第一电离能大于氢的第一电离能。

(Total for Question 7= 12 marks)
(问题 7 的总分 = 12 分)

8 The five ionisation energies of boron are:
8 硼的五种电离能是:
801 2427 3660 25026 32828 801 2427 3660 25026 32828 {:[801,2427,3660,25026,32828]:}\begin{array}{lllll} 801 & 2427 & 3660 & 25026 & 32828 \end{array}
(a) State and justify the group in the Periodic Table in which boron is placed.
(a) 说明并证明元素周期表中放置硼的族。

(b) Which of the following represents the second ionisation energy of boron?
(b) 以下哪项代表硼的第二电离能?

A B ( g ) B 2 + ( g ) + 2 e Δ H = + 2427 kJ mol 1 B ( g ) B 2 + ( g ) + 2 e Δ H = + 2427 kJ mol 1 quadB(g)rarrB^(2+)(g)+2e^(-)Delta H=+2427kJmol^(-1)\quad \mathrm{B}(\mathrm{g}) \rightarrow \mathrm{B}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} \Delta H=+2427 \mathrm{~kJ} \mathrm{~mol}^{-1} 一个 B ( g ) B 2 + ( g ) + 2 e Δ H = + 2427 kJ mol 1 B ( g ) B 2 + ( g ) + 2 e Δ H = + 2427 kJ mol 1 quadB(g)rarrB^(2+)(g)+2e^(-)Delta H=+2427kJmol^(-1)\quad \mathrm{B}(\mathrm{g}) \rightarrow \mathrm{B}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} \Delta H=+2427 \mathrm{~kJ} \mathrm{~mol}^{-1}
B B + ( g ) B 2 + ( g ) + e Δ H = + 2427 kJ mol 1 B + ( g ) B 2 + ( g ) + e Δ H = + 2427 kJ mol 1 quadB^(+)(g)rarrB^(2+)(g)+e^(-)Delta H=+2427kJmol^(-1)\quad \mathrm{B}^{+}(\mathrm{g}) \rightarrow \mathrm{B}^{2+}(\mathrm{g})+\mathrm{e}^{-} \Delta H=+2427 \mathrm{~kJ} \mathrm{~mol}^{-1}  B + ( g ) B 2 + ( g ) + e Δ H = + 2427 kJ mol 1 B + ( g ) B 2 + ( g ) + e Δ H = + 2427 kJ mol 1 quadB^(+)(g)rarrB^(2+)(g)+e^(-)Delta H=+2427kJmol^(-1)\quad \mathrm{B}^{+}(\mathrm{g}) \rightarrow \mathrm{B}^{2+}(\mathrm{g})+\mathrm{e}^{-} \Delta H=+2427 \mathrm{~kJ} \mathrm{~mol}^{-1}
C B ( g ) B 2 + ( g ) + 2 e Δ H = 2427 kJ mol 1 B ( g ) B 2 + ( g ) + 2 e Δ H = 2427 kJ mol 1 B(g)rarrB^(2+)(g)+2e^(-)Delta H=-2427kJmol^(-1)\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{B}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} \Delta H=-2427 \mathrm{~kJ} \mathrm{~mol}^{-1}