Chapter 4: Structure of Atom

In the early years of the twentieth century, most scientists pictured the atom as a soft, uniform ball of positive matter with electrons dotted through it. If that picture were right, a fast, heavy particle fired at a thin sheet of metal should pass straight through, perhaps nudged very slightly off course — rather like a bullet through a bank of fog.

In Ernest Rutherford’s laboratory, that idea was put to the test. A beam of α-particles was directed at a piece of gold foil so thin it was only a few thousand atoms thick, and a screen was used to record where the particles landed. Nearly all of them went through undeflected, exactly as expected. But a very small number came back on the same side they had entered from. Rutherford was astonished; he later compared the result to firing a heavy shell at a sheet of tissue paper and having it bounce back.

Only one arrangement of matter could explain it. The positive charge and almost the entire mass of an atom had to be crushed into a tiny central core — the nucleus — with the rest of the atom being mostly empty space. In 1911 Rutherford proposed exactly that, and the modern picture of the atom began.

Diagram

Fig. 4.0 – Split illustration: on the left, the old uniform positive-sphere model with a beam passing straight through; on the right, Rutherford’s nuclear atom with most α-particles passing through and a few deflected sharply back from a tiny dense nucleus. Label nucleus, α-particle beam, gold foil, deflected paths.

NoteCuriosity Corner

Q1. If an atom is mostly empty space, why does a gold ring feel completely solid and not hollow? Q2. Rutherford’s atom looks like a miniature solar system, with electrons circling the nucleus like planets. Why should such an atom collapse in a tiny fraction of a second — and why do real atoms not collapse? Q3. When hydrogen gas is made to glow in a discharge tube, why does it give only a few sharp coloured lines instead of a smooth rainbow of all colours? Q4. The atomic mass of chlorine is given as 35.5 u, yet no single chlorine atom has a mass of 35.5 u. Where does the fraction come from? Q5. Electrons fill orbitals in a neat, predictable order for element after element — so why do chromium and copper break that order?

Keep these questions in mind as you read. This chapter will give you everything you need to reason out each one for yourself.

4.1 Subatomic Particles

Before you begin, recall the following: - What is the smallest unit of matter? - What is the difference between molecules of an element and those of a compound? - Does an atom have any internal structure, or is it indivisible? - Which particle was identified by J. J. Thomson in the cathode ray tube experiment? - Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil?

Where Dalton’s Model Fell Short

John Dalton’s atomic theory was a remarkable success. It explained the laws of chemical combination — the fixed mass ratios in which elements combine — very neatly, because it treated every atom as a tiny, solid, indivisible sphere.

However, the theory could not explain every property of matter. One familiar example is the electrical behaviour of substances. When glass or ebonite is rubbed with silk or fur, it generates electricity, and a model of hard, featureless spheres offers no reason why this should happen.

The answer had to lie inside the atom itself. If rubbing can pull electric charge out of a material, then the atom must contain charged pieces that can be separated from it — which means the atom is not indivisible after all.

The discovery of subatomic particles in the late nineteenth and early twentieth centuries therefore struck a blow to Dalton’s model of the atom as a hard sphere. Three important subatomic particles that concern Chemistry — the proton, the electron and the neutron — were discovered in this period.

Q. If atoms are electrically neutral, how can rubbing two objects together produce charge?

A. Rubbing transfers some electrons from one surface to the other. The object that gains electrons becomes negatively charged and the one that loses them becomes positively charged. Neither object has “created” charge — the charge was inside the atoms all along, in particles loose enough to be moved.

Where the Particles Sit

The proton and the neutron are present in the atomic nucleus, the extremely small and dense core at the centre of an atom. Because they share this location, protons and neutrons together are given a single collective name — nucleons.

NoteDefinition

Nucleons are the particles present in the nucleus of an atom, that is, protons and neutrons taken together.

Electrons are present in the extranuclear part of the atom — the far larger region surrounding the nucleus. The word simply means “outside the nucleus,” and this is where all of an atom’s chemistry happens, since it is the outer electrons that meet the electrons of a neighbouring atom.

Properties of the Three Particles

The measured properties of the electron, proton and neutron are summarised in Table 4.1. Read the table in pairs of columns: the absolute charge is the charge in coulomb (C), the SI unit of electric charge, while the relative charge simply compares that value with the charge on one proton and so has no unit.

Table 4.1 : Properties of subatomic particles

Name Symbol Absolute charge/C Relative charge Symbol for charge Mass/kg Mass/u Approximate mass/u
Electron \(\mathrm{e^-}\) \(-1.6022\times10^{-19}\) \(-1\) \(-e\) \(9.10938\times10^{-31}\) \(0.00054\) \(0\ \text{u}\)
Proton \(\mathrm{p}\) \(+1.6022\times10^{-19}\) \(+1\) \(+e\) \(1.6726\times10^{-27}\) \(1.00727\) \(1\ \text{u}\)
Neutron \(\mathrm{n}\) \(0\) \(0\) — \(1.67493\times10^{-27}\) \(1.00867\) \(1\ \text{u}\)

Two features of this table deserve attention. First, the electron and the proton carry charges of exactly equal size but opposite sign, which is why an atom containing equal numbers of both is electrically neutral overall. Second, masses in kilogram are clumsy at this scale, so a smaller unit called the unified atomic mass unit (symbol u, also called the dalton) is used; on this scale a proton and a neutron each weigh almost exactly 1 u, while the electron is so light that its mass is approximated as 0 u.

charge on a subatomic particle (SI unit: coulomb, C. Sign convention — the electron is assigned a negative charge, \(-e\); the proton an equal and opposite positive charge, \(+e\); the neutron carries no charge at all. Relative charge is a pure number with no unit, obtained by dividing the absolute charge by \(1.6022\times10^{-19}\ \text{C}\).)

Using the masses given in Table 4.1, calculate how many times heavier a proton is than an electron.

Solution: The required quantity is the ratio of the two masses. Since both are given in the same unit (kilogram), the unit cancels and the answer is a pure number.

\[\frac{m_{\mathrm{p}}}{m_{\mathrm{e}}} = \frac{1.6726\times10^{-27}\ \text{kg}}{9.10938\times10^{-31}\ \text{kg}}\]

\[= \frac{1.6726}{9.10938}\times10^{(-27)-(-31)} = 0.18361\times10^{4}\]

\[\frac{m_{\mathrm{p}}}{m_{\mathrm{e}}} \approx 1836\]

A proton is therefore about 1836 times heavier than an electron. This is why, when we later add up the mass of an atom, the electrons can safely be ignored.

The word electron comes from the Greek ēlektron, meaning amber. Amber was the material the ancient Greeks rubbed to attract light objects — the very same static-electricity effect that Dalton’s model could not explain. The name was coined by the Irish physicist George Johnstone Stoney for the fundamental unit of electric charge, before anyone had actually detected the particle. Proton has an equally fitting root: it comes from the Greek protos, meaning first — the first and most basic building block found inside every nucleus.

4.1.1 Discovery of Electron

In the year 1897, J. J. Thomson investigated cathode rays. Cathode rays are the rays produced when a high voltage is applied across two metal electrodes sealed inside a glass tube from which most of the air has been pumped out; they travel from the cathode (the negative electrode) towards the anode (the positive electrode). The arrangement is shown in Fig. 4.1.

Diagram

Fig. 4.1 – Cathode ray tube experiment: evacuated glass tube with cathode and anode, a beam travelling to a fluorescent screen, charged plates above and below the beam; label cathode, anode, original path, deflected path (bending towards the positive plate).

Thomson found that the cathode rays are a stream of very small, negatively charged particles which are 1837 times lighter than a hydrogen atom, and that they are present in all atoms. He established this by studying how the beam bent when electric and magnetic fields were applied across its path — a beam that bends towards a positive plate must itself be negatively charged.

The last point was the crucial one. Whatever metal was used for the cathode, the same particles appeared, so these particles could not belong to one element alone; they had to be a component of every atom. Later these particles were named as electrons.

The cathode ray tube was not just a laboratory curiosity. The same principle — a beam of electrons steered by electric and magnetic fields onto a fluorescent screen — was the working heart of television sets, computer monitors and oscilloscopes for most of the twentieth century, before flat-panel screens replaced them. In the laboratory these tubes must be handled with care, as they operate at high voltage and their evacuated glass envelope can implode if cracked.

(question only — no solution) Using Table 4.1, calculate the total charge, in coulomb, carried by all the electrons in a neutral atom of an element whose nucleus contains 8 protons. State the sign of your answer.

4.1.2 Discovery of Proton

In the year 1911, Ernest Rutherford found, in the experiment of scattering of α-particles by thin gold foil, that a few α-particles bounce back. An α-particle is a fast-moving, positively charged particle emitted by certain radioactive substances; it is heavy compared with an electron, so anything that can turn it around must be both massive and strongly charged. The experimental arrangement is shown in Fig. 4.2.

Diagram

Fig. 4.2 – Rutherford’s scattering experiment: source of α-particles in a lead block, narrow beam passing through a lead plate, striking a thin gold foil, a circular fluorescent/photographic detecting screen around it; show most particles passing straight, a few deflected at large angles and one bouncing back. Label source of α-particles, lead plate, gold foil, scintillation, photographic plate.

From this Rutherford inferred the presence of a massive and positively charged nucleus inside the atom. A positive core repels an incoming positive α-particle; if that core is also heavy, it does not move aside, and a particle that happens to head straight for it is thrown back the way it came.

From Nucleus to Proton

Following the discovery of the nucleus in an atom, Rutherford found (1919) that fast moving α-particles transmuted nitrogen into oxygen with simultaneous liberation of hydrogen.

\[^{14}_{7}\mathrm{N} + {}^{4}_{2}\alpha \longrightarrow {}^{17}_{8}\mathrm{O} + {}^{1}_{1}\mathrm{H}\]

Read this equation the way you read a balanced chemical equation, but with two totals to check instead of atoms: the superscripts (mass numbers) add to 18 on both sides, and the subscripts (charges on the nuclei) add to 9 on both sides. Nothing is lost — one element has simply been converted into another, which is what transmutation means.

He further showed that other elements could also be transmuted, but hydrogen was always emitted. That last word — always — carries the whole argument. If a hydrogen nucleus can be knocked out of nitrogen, and out of other elements too, then a hydrogen nucleus must already be sitting inside all of them.

On this basis Rutherford proposed that the hydrogen nucleus must be contained inside nuclei of all the elements. Hence, the hydrogen nucleus was renamed as proton.

Q. Why was a new name needed at all — why not keep calling it a hydrogen nucleus?

A. Because it had stopped being a property of one element. Once the same particle was shown to be a building block of every nucleus, it deserved a name of its own rather than one borrowed from hydrogen.

(question only — no solution) In a transmutation, an α-particle strikes an aluminium nucleus, \(^{27}_{13}\mathrm{Al}\), and a neutron is emitted along with a single new nucleus. Work out the mass number and the atomic number of that new nucleus.

4.1.3 Discovery of Neutron

The existence of an electrically neutral and massive particle in the nucleus was predicted by Ernest Rutherford in 1920 to account for the disparity in atomic number and atomic mass of an element. In simple terms, the protons alone could not supply enough mass: a nucleus was regularly found to be about twice as heavy as its total proton charge suggested, so some other heavy but uncharged particle had to be sitting there as well.

In the year 1932, James Chadwick measured the velocity of protons knocked out from paraffin by an unidentified radiation from beryllium. The set-up is shown in Fig. 4.3, and the logic of the experiment is laid out step by step in FC 4.1.

Diagram

Fig. 4.3 – Discovery of neutron: polonium source emitting α-particles onto a beryllium target, the unidentified neutral radiation travelling on to a block of paraffin, protons ejected from the paraffin and detected. Label polonium source, beryllium, α-particles, neutrons, paraffin, protons.

Flowchart

FC 4.1 – Chadwick’s neutron experiment: polonium source → α-particles → strike beryllium → unidentified radiation given off (undeflected by electric/magnetic fields, so uncharged) → strikes paraffin → protons knocked out → velocity of ejected protons measured → mass of the unknown particle calculated ≈ mass of a proton → particle named the neutron

Paraffin was a shrewd choice of target, because it is rich in hydrogen and therefore full of protons. From the velocity of the protons that were knocked out, Chadwick determined the mass of the particles of the unidentified neutral radiation, which came out to be almost the same as that of a proton. This is exactly what a head-on collision between two particles of equal mass predicts.

He named this particle as the ‘neutron’, which was predicted by Rutherford earlier.

NoteDefinition

A neutron is a subatomic particle present in the nucleus that carries no electric charge and has a mass almost equal to that of a proton.

the three discovery experiments | Feature | Thomson’s experiment | Rutherford’s experiment | Chadwick’s experiment | |—|—|—|—| | Year | 1897 | 1911 (nucleus); 1919 (transmutation) | 1932 | | What was studied | Cathode rays in a discharge tube | Scattering of α-particles by thin gold foil | Radiation from beryllium falling on paraffin | | Key observation | The rays bend, showing negatively charged particles lighter than a hydrogen atom | A few α-particles bounce back | Protons are knocked out of the paraffin | | Particle established | Electron | Proton (after the nucleus) | Neutron | | Charge of that particle | Negative | Positive | Zero |

Chadwick’s identification of the neutron in 1932 was recognised with the Nobel Prize in Physics in 1935. The particle mattered far beyond bookkeeping of mass: because a neutron carries no charge, it is not repelled by a positively charged nucleus and can walk straight into one. That single property opened the way to nuclear fission and, with it, to nuclear power — a theme you will meet again in Chapter 13.

Table 4.1, the mass and charge columns (Compare the mass of the electron with those of the proton and the neutron. (a) Roughly what fraction of a proton’s mass is one electron? (b) A neutral atom of an element has 20 protons, 20 neutrons and 20 electrons — estimate what percentage of its total mass is contributed by its electrons. (c) On the basis of your answer, is it reasonable to say that the mass of an atom is concentrated in its nucleus?)

4.2 Atomic Number and Atomic Mass Number

The Number That Names an Element

The number of protons in the nucleus is the chemical identity of an element. Change that number and you no longer have the same element at all — which is precisely why Rutherford’s nitrogen turned into oxygen once a proton was added to its nucleus.

This number is called the atomic number (Z) of the element. Every carbon atom in the universe has 6 protons; every gold atom has 79. Nothing else about the atom is so fixed.

NoteDefinition

The atomic number (Z) of an element is the number of protons present in the nucleus of its atom.

The positive charge on the nucleus is due to the protons present in it, and equals \(+Ze\), where \(e\) is the magnitude of the charge on one proton. An atom being electrically neutral, it contains the same number of extranuclear electrons in it as its atomic number. Therefore the total electronic charge on an atom is \(-Ze\), which exactly cancels the nuclear charge. Thus in any atom,

\[\text{Atomic number }(Z) = \text{Number of protons} = \text{Number of electrons}\]

Note carefully what this last equality does and does not say. The equality of protons and electrons holds because the atom as a whole is neutral; the identity of the element rests on the proton count alone.

Q. If the number of electrons also equals Z, why do we define the element by its protons rather than by its electrons?

A. Because electrons can be gained or lost. When a sodium atom loses one electron it becomes a sodium ion, but it is still sodium — the nucleus is untouched. Protons are not removed in ordinary chemical change, so the proton number is the one reliable label.

For a long time chemists arranged the elements by atomic mass, and a few of them stubbornly refused to fit. In the early 1910s the young English physicist Henry Moseley studied the X-rays given off by different metals and found that their frequencies varied in a regular way with a whole number that increased by one from element to element. That whole number was the nuclear charge — the atomic number. Moseley’s work showed that the true order of the elements is set by Z, not by atomic mass. He was killed in the First World War a few years later, at the age of twenty-seven.

Counting the Mass of an Atom

As can be seen from Table 4.1, the mass of an electron is negligibly small compared to that of the nucleons. As a result the mass of an atom can be considered to be concentrated in its nucleus.

The approximate mass of one proton or one neutron is 1 u. Therefore the approximate atomic mass in daltons is numerically equal to the number of nucleons in the atom. In other words, counting the heavy particles in the nucleus is the same as reading off the atom’s approximate mass.

The number of neutrons in the nucleus is designated by the symbol \(N\); and the total number of protons and neutrons, that is nucleons, in an atom is called its atomic mass number (A).

\[\text{Mass number }(A) = \text{Proton number }(Z) + \text{Neutron number }(N)\]

\[\text{Therefore}\quad A = Z + N, \qquad N = A - Z\]

NoteDefinition

The atomic mass number (A) of an atom is the total number of nucleons — protons plus neutrons — present in its nucleus.

Both \(Z\) and \(A\) are counts of particles, so both are whole numbers with no unit. The second relation, \(N = A - Z\), is simply the first one rearranged, and it is the form you will use most often, because a nuclide symbol hands you \(A\) and \(Z\) and expects you to work out \(N\) for yourself.

atomic number vs neutron number vs mass number | Feature | Atomic number, \(Z\) | Neutron number, \(N\) | Mass number, \(A\) | |—|—|—|—| | What it counts | Protons in the nucleus | Neutrons in the nucleus | All nucleons (protons + neutrons) | | Where it is written in the symbol | Subscript on the left | Not written; it is calculated | Superscript on the left | | Relation to the others | \(Z = A - N\) | \(N = A - Z\) | \(A = Z + N\) | | Fixes the identity of the element? | Yes | No | No | | Also equals, in a neutral atom | Number of electrons | — | — |

Writing the Composition of an Atom

The composition of any atom is represented by the element symbol (X) with the atomic mass number (\(A\)) as superscript on the left and the atomic number (\(Z\)) as subscript on the left:

\[^{A}_{Z}\mathrm{X}\]

The layout is worth memorising by its logic rather than by rote: the larger number sits on top, because \(A\) can never be smaller than \(Z\). The arrangement of the symbol is shown in Fig. 4.2a.

Diagram

Fig. 4.2a – Nuclide notation: a large element symbol X with the mass number A as a left superscript and the atomic number Z as a left subscript; arrows labelling “A = total nucleons (protons + neutrons)” and “Z = protons = electrons in a neutral atom”, with a worked tag showing ⁴⁰₁₈Ar → 18 p, 18 e⁻, 22 n.

The atom or nucleus having a unique composition as specified by \(^{A}_{Z}\mathrm{X}\) is called a nuclide.

NoteDefinition

A nuclide is an atom or nucleus of a definite composition, that is, with a specific number of protons and a specific number of neutrons, represented as \(^{A}_{Z}\mathrm{X}\).

Reading a nuclide symbol is a two-step habit, set out in FC 4.2, and every question of this type in an examination follows the same route.

Flowchart

FC 4.2 – Reading a nuclide symbol ᴬ_ZX: read A from the top left → read Z from the bottom left → number of protons = Z → number of electrons = Z (neutral atom only) → number of neutrons N = A − Z → state the composition. Branch: if the species carries a charge (an ion), electrons ≠ Z; subtract the positive charge from Z, or add the negative charge to Z.

Find out the number of protons, electrons and neutrons in the nuclide \(^{40}_{18}\mathrm{Ar}\).

Solution: In case of the nuclide \(^{40}_{18}\mathrm{Ar}\),

\[A = 40 \quad \text{and} \quad Z = 18\]

Number of protons = number of electrons = \(Z\) = 18 and number of neutrons

\[N = A - Z\]

\[= 40 - 18 = 22\]

The number of electrons could be written down directly only because argon here is a neutral atom; the electron count is a consequence of neutrality, not of the symbol itself.

An atom of an element contains 26 electrons and 30 neutrons. Write the nuclide symbol of this atom.

Solution: The atom is neutral, so its number of electrons equals its number of protons.

\[Z = \text{number of protons} = \text{number of electrons} = 26\]

An atomic number of 26 identifies the element as iron, symbol Fe.

The mass number is the total number of nucleons:

\[A = Z + N = 26 + 30 = 56\]

Writing \(A\) as the left superscript and \(Z\) as the left subscript, the nuclide symbol is

\[^{56}_{26}\mathrm{Fe}\]

Nuclide notation is everyday shorthand in a hospital, not only in a chemistry class. In nuclear medicine, a patient may be given a compound tagged with a particular nuclide so that its path through the body can be tracked by a detector outside. Technetium-99m — a nuclide of technetium, \(Z = 43\) — is the workhorse of this field and is used in millions of imaging scans worldwide each year, because it emits radiation that leaves the body quickly and is easy to detect. The dose given is carefully limited, and handling of such material is strictly regulated.

(question only — no solution) An atom of an element has a mass number of 39 and contains 20 neutrons. (a) Find its atomic number. (b) Write its nuclide symbol. (c) How many electrons will the unipositive ion of this element contain?

(question only — no solution) For the nuclide \(^{31}_{15}\mathrm{P}\), calculate (a) the number of nucleons, (b) the number of neutrons, and (c) the total charge on the nucleus in coulomb, using \(e = 1.6022\times10^{-19}\ \text{C}\).

4.3 Isotopes, Isobars and Isotones

Every nuclide is fixed by two numbers, \(Z\) and \(A\). Once you have two nuclides in front of you, there are only a few ways in which they can resemble each other — they may share \(Z\), they may share \(A\), or they may share \(N\). Similarities in composition of nuclides result in exactly these three types of relationships, and each has its own name.

Isotopes: Same Element, Different Mass

Some elements exist as a single natural nuclide. For example, fluorine occurs in nature only as \(^{19}_{9}\mathrm{F}\), so every fluorine atom you will ever handle has 9 protons and 10 neutrons.

However, many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element.

All the isotopes of an element have the same number of protons but different number of neutrons in their nuclei. The proton count is what makes them the same element; the neutron count is what makes them different nuclides.

NoteDefinition

Isotopes are the nuclides of the same element, having the same number of protons but different numbers of neutrons in their nuclei.

As the proton number is the atomic number, all the isotopes of an element have the same position in the modern periodic table and exhibit similar chemical properties. All the natural isotopes of an element coexist and have a definite natural abundance — that is, the proportion in which they occur is not random but a fixed characteristic of the element. Table 4.2 shows various features of the three common isotopes of carbon, whose nuclei are compared in Fig. 4.3a.

Diagram

Fig. 4.3a – Three carbon nuclei side by side, each drawn with 6 protons but with 6, 7 and 8 neutrons respectively, each surrounded by 6 electrons; label C-12, C-13, C-14 and mark “Z the same, N different”.

Table 4.2 : Isotopes of carbon

Symbol Atomic number \(Z\) Atomic mass number \(A\) Neutron number \(N = A - Z\) % Abundance Stability
\(^{12}_{6}\mathrm{C}\) or C-12 6 12 6 98.9 % Stable
\(^{13}_{6}\mathrm{C}\) or C-13 6 13 7 1.1 % Stable
\(^{14}_{6}\mathrm{C}\) or C-14 6 14 8 < 0.00017 % Radioactive

Q. If C-12, C-13 and C-14 have different masses, why do all three behave the same way in a chemical reaction?

A. Because chemical behaviour is decided by the electrons, and the number of electrons in a neutral atom equals \(Z\). All three carbon nuclides have \(Z = 6\), so all three carry 6 electrons arranged in exactly the same way. The extra neutrons add mass but no charge, and so they leave the chemistry untouched.

The word isotope was introduced by the English chemist Frederick Soddy, who was studying the products of radioactive decay and kept finding substances that were chemically inseparable yet differed in mass. The name is built from the Greek isos (equal) and topos (place) — “same place” — because such nuclides must all be given the very same box in the periodic table. It is said to have been suggested to him by Margaret Todd, a physician, during a conversation about the problem. Soddy received the Nobel Prize in Chemistry in 1921 for this work on isotopes.

Average atomic mass

Since an element’s isotopes coexist in fixed proportions, the atomic mass we quote for an element is not the mass of any single atom. It is the average mass of its atoms, weighted according to how abundant each isotope is. This is why several atomic masses in the periodic table are not whole numbers.

The two natural isotopes of chlorine, viz. \(^{35}_{17}\mathrm{Cl}\) and \(^{37}_{17}\mathrm{Cl}\), exist in relative abundance of 3:1. Find out the average atomic mass of chlorine.

Solution: From the relative abundance 3:1, it is understood that out of 4 chlorine atoms, 3 atoms have mass 35 and 1 has mass 37.

Therefore, the average atomic mass of chlorine

\[= \frac{3\times35 + 1\times37}{4} = 35.5\]

Notice how the ratio was used: a 3:1 ratio means 4 parts in all, so the total mass of 4 atoms is divided by 4. The heavier isotope pulls the average up only a little, because it is present in only one part out of four.

Q4. The atomic mass of chlorine is given as 35.5 u, yet no single chlorine atom has a mass of 35.5 u. Where does the fraction come from?

A. Natural chlorine is a mixture of two isotopes, \(^{35}_{17}\mathrm{Cl}\) and \(^{37}_{17}\mathrm{Cl}\), present in a fixed 3:1 proportion. Every individual atom has a mass of either 35 or 37 — never 35.5. The figure 35.5 is the weighted average of the two, and since 3 out of every 4 atoms are the lighter kind, the average lies close to 35 rather than midway between the two. A fractional atomic mass is therefore a signal that the element occurs as a mixture of isotopes.

Using the percentage abundances given in Table 4.2 for the two stable isotopes of carbon, calculate the average atomic mass of carbon. (The abundance of C-14 is so small that it may be neglected.)

Solution: When abundances are given as percentages, each isotopic mass is multiplied by its percentage and the total is divided by 100.

\[\text{Average atomic mass} = \frac{(98.9\times12) + (1.1\times13)}{100}\]

\[= \frac{1186.8 + 14.3}{100} = \frac{1201.1}{100}\]

\[= 12.011\ \text{u}\]

The answer lies very close to 12 because almost 99 out of every 100 carbon atoms are C-12. This is the general rule for weighted averages: the value is always pulled towards the most abundant isotope.

The tiny radioactive share of carbon shown in Table 4.2 is the basis of radiocarbon dating. Carbon-14 is continually formed in the upper atmosphere and is taken up by living plants and, through them, by animals, so a living organism carries the same small proportion of C-14 as its surroundings. Once the organism dies it stops taking in fresh carbon, and the C-14 already present slowly decays away. By measuring how much C-14 is left in a piece of wood, bone or cloth, archaeologists can estimate how long ago it stopped living — a technique used to date everything from ancient timber to preserved textiles. You will study the mathematics of such decay in Chapter 13.

(question only — no solution) Boron occurs in nature as two isotopes, \(^{10}_{5}\mathrm{B}\) and \(^{11}_{5}\mathrm{B}\), with abundances of 19.9 % and 80.1 % respectively. Calculate the average atomic mass of boron and state which isotope the answer lies closer to.

Isobars: Same Mass, Different Elements

Isobars are the atoms of different elements having the same mass number but different atomic numbers.

NoteDefinition

Isobars are atoms of different elements that have the same mass number (A) but different atomic numbers (Z).

Isobars are different elements. They have different chemical properties and occupy different positions in the modern periodic table. This follows directly from the definition: their proton numbers differ, so their electron numbers differ, and it is the electrons that decide chemistry.

Table 4.3 shows an illustration of isobars. Both nuclides listed there have 14 nucleons in total, but they distribute those 14 nucleons differently between protons and neutrons — 6 and 8 in one case, 7 and 7 in the other.

Table 4.3 : Isobars

Isobars Atomic number \(Z\) Mass number \(A\) Number of protons \(Z\) Number of neutrons \(N\)
\(^{14}_{6}\mathrm{C}\) 6 14 6 8
\(^{14}_{7}\mathrm{N}\) 7 14 7 7

Three elements Q, R and T have mass number 40. Their atoms contain 22, 21 and 20 neutrons, respectively. Represent their atomic composition with appropriate symbol.

Solution: \(A = Z + N \quad \therefore\ Z = A - N\)

For the given three elements \(A = 40\). Values of their atomic numbers \(Z\), are obtained from the given values of neutron numbers, \(N\), using the above expression.

for Q : \(Z = A - N = 40 - 22 = 18\)

for R : \(Z = A - N = 40 - 21 = 19\)

for T : \(Z = A - N = 40 - 20 = 20\)

\(\therefore\) The atomic composition of the three elements is written as follows :

\[^{40}_{18}\mathrm{Q},\quad ^{40}_{19}\mathrm{R},\quad ^{40}_{20}\mathrm{T}\]

Since all three share the mass number 40 but differ in atomic number, Q, R and T are isobars of one another.

Isotones: The Same Neutron Count

Isotones are the atoms of different elements having the same number of neutrons in their nuclei.

NoteDefinition

Isotones are atoms of different elements whose nuclei contain the same number of neutrons (N), although their atomic numbers and mass numbers differ.

Notice that \(N\) is never written in a nuclide symbol; it must be worked out as \(A - Z\) before two nuclides can be tested for this relationship. Table 4.4 shows examples of isotones, and in both cases the subtraction gives 6.

Table 4.4 : Isotones

Isotones Atomic number \(Z\) Mass number \(A\) Number of Neutrons \(N = A - Z\)
\(^{11}_{5}\mathrm{B}\) 5 11 6
\(^{12}_{6}\mathrm{C}\) 6 12 6

isotopes vs isobars vs isotones | Feature | Isotopes | Isobars | Isotones | |—|—|—|—| | What is the same | Atomic number \(Z\) (protons) | Mass number \(A\) (nucleons) | Neutron number \(N\) | | What is different | Mass number \(A\), neutron number \(N\) | Atomic number \(Z\), neutron number \(N\) | Atomic number \(Z\), mass number \(A\) | | Same element or different? | Same element | Different elements | Different elements | | Position in the periodic table | Same position | Different positions | Different positions | | Chemical properties | Similar | Different | Different | | Example from this section | \(^{12}_{6}\mathrm{C}\) and \(^{14}_{6}\mathrm{C}\) | \(^{14}_{6}\mathrm{C}\) and \(^{14}_{7}\mathrm{N}\) | \(^{11}_{5}\mathrm{B}\) and \(^{12}_{6}\mathrm{C}\) |

A useful way to keep the three apart is to ask which quantity is being held fixed. Fix the protons and you have isotopes; fix the total nucleons and you have isobars; fix the neutrons and you have isotones. The routine for deciding is set out in FC 4.3.

Flowchart

FC 4.3 – Classifying a pair of nuclides: read A and Z for both → calculate N = A − Z for both → is Z the same? if yes → isotopes; if no → is A the same? if yes → isobars; if no → is N the same? if yes → isotones; if none of the three match → no special relationship.

(question only — no solution) Consider the nuclides \(^{40}_{18}\mathrm{Ar}\), \(^{40}_{20}\mathrm{Ca}\), \(^{39}_{19}\mathrm{K}\) and \(^{38}_{18}\mathrm{Ar}\). Calculate \(N\) for each, then pick out (a) a pair of isotopes, (b) a pair of isobars and (c) a pair of isotones, giving a reason in each case.

We will consider some more aspects of nuclides — particularly why certain nuclides such as C-14 are radioactive — in Chapter 13.

4.4 Drawbacks of Rutherford Atomic Model

The Atom That Should Collapse

Let us now go back to the point of time when Rutherford put forth his nuclear model of the atom. It is akin to a miniature of the solar system, the nucleus playing the role of the massive sun and the electrons being the lighter planets.

Electrons in this model could not be stationary. If an electron simply stood still somewhere outside the nucleus, the electrostatic force of attraction exerted on it would pull it towards the nucleus, and the whole atom would collapse into a single blob of charge — a miniature version of Thomson’s atomic model, which the scattering experiment had already ruled out.

So the electrons must be moving. However, the electrons revolving about the nucleus, as described by Rutherford, also pose a problem.

Electrons in the Rutherford model are negatively charged particles in orbital motion. Such orbital motion is an accelerated motion, accompanied by a continuous change in the velocity of the electron as noticed from its continuously changing direction.

Q. The electron goes round its orbit at a steady speed. How can it then be said to be accelerating?

A. Because velocity includes direction, not just speed. An electron going round a circle is turning at every instant, so the direction of its velocity changes continuously even if the speed does not. A changing velocity is exactly what acceleration means, so circular motion is always accelerated motion.

According to the electromagnetic theory of Maxwell, accelerated charged particles would emit electromagnetic radiation. This is a general result of classical physics: whenever a charge is made to change its velocity, it sheds energy into the space around it as radiation.

NotePrinciple / Law

(Maxwell’s electromagnetic theory) A charged particle in accelerated motion continuously emits electromagnetic radiation, and in doing so loses energy.

An electron in an orbit would therefore emit radiation, equivalent to the energy possessed by the radiation associated with the electronic motion. Energy that leaves the electron cannot be replaced from anywhere, so the electron must fall closer to the nucleus. The orbit would, therefore, shrink continuously.

Thus, an electron orbiting about the nucleus would follow a spiral path to the nucleus, as drawn in Fig. 4.4a. It can be seen that the Rutherford atomic model has an intrinsic instability of the atom. However, real atoms are stable.

Diagram

Fig. 4.4a – Predicted collapse in Rutherford’s model: a nucleus at the centre with an electron spiralling inwards along a decreasing path, wavy arrows leaving the electron to show radiation being emitted continuously; label “electron loses energy”, “orbit shrinks”, “spiral path to the nucleus”.

That last sentence is the heart of the objection. This is not a small numerical error in the model — the calculation predicts that every atom in existence should have collapsed almost the instant it formed, whereas the matter around us is perfectly stable. A model whose central prediction is contradicted by the very existence of matter cannot be the final word.

The theory that condemned Rutherford’s atom was already several decades old. In the 1860s the Scottish physicist James Clerk Maxwell united electricity, magnetism and light in a single set of equations, from which it followed that accelerating charges must radiate energy as electromagnetic waves. In the 1880s Heinrich Hertz confirmed this in the laboratory by generating and detecting such waves — the beginning of radio. Maxwell’s theory was among the great triumphs of nineteenth-century physics, which is exactly why its verdict on the nuclear atom was so uncomfortable to ignore.

Maxwell’s prediction is not merely an obstacle for atomic models; it is put to work every day. In a synchrotron, electrons are kept moving at very high speed around a large ring, and because they are constantly being pushed into a curved path, they radiate — exactly as the theory says an accelerated charge must. The intense X-rays produced this way are used by chemists and biologists to determine the detailed structures of crystals, catalysts and protein molecules. What ruined the Rutherford model turned out to be one of the most powerful tools available for studying matter.

Thomson’s model vs Rutherford’s model | Feature | Thomson’s model | Rutherford’s model | |—|—|—| | Picture of the atom | Positive charge spread through the whole atom, with electrons embedded in it | Almost all the mass and all the positive charge in a tiny central nucleus | | Position of the electrons | Fixed within the positive matter | Revolving around the nucleus in orbits | | Nucleus | No nucleus proposed | Nucleus proposed and central to the model | | Supported by | — | Large-angle scattering of α-particles by gold foil | | Main failure | Cannot explain why a few α-particles bounce back | Predicts that the orbiting electron must radiate and spiral into the nucleus |

No Account of Electron Energies

The second serious drawback of the Rutherford model is that it does not describe the distribution of electrons around the nucleus and their energies.

This is a drawback of silence rather than of contradiction. The model tells us where the positive charge is, but not how the electrons are arranged in the space around it, nor how much energy each of them has. Since chemistry is entirely the behaviour of outer electrons, a model that says nothing about their arrangement can say nothing about why sodium reacts as it does and argon does not.

The drawbacks of the Rutherford model were overcome in the Bohr atomic model, which is taken up in the sections that follow.

Rutherford’s atom is often compared with the solar system. State one respect in which this comparison holds well, and one respect in which it breaks down completely.

Solution:

Where it holds: In both, almost the entire mass is concentrated in a small central body — the sun in one case, the nucleus in the other — while much lighter bodies revolve around it in orbits, held by an attractive force directed towards the centre.

Where it breaks down: Planets are electrically neutral, so their accelerated motion costs them no energy and their orbits are stable. An electron, however, is charged. By Maxwell’s theory an accelerated charge must continuously radiate energy, so an orbiting electron would lose energy, spiral inwards and fall into the nucleus. The analogy therefore fails on the single point that matters most — stability.

(question only — no solution) Read the two statements below and decide whether each is true, and whether the second correctly explains the first.

****Assertion:** According to Rutherford’s model, an atom should not be able to exist for any appreciable length of time.**

****Reason:** An electron revolving in a circular orbit is in accelerated motion.**

4.5 Developments Leading to the Bohr’s Atomic Model

At the time when different models of atomic structure were being put forth, some results obtained from the studies of interactions of radiation with matter required to be correlated to atomic structure. In other words, physicists had learnt a great deal about what happens when light falls on matter, or is given out by it, but nobody had yet linked those findings to the question of how an atom is built.

Niels Bohr utilized these results to get over the drawbacks of the Rutherford atomic model. These results were: (1) wave particle duality of electromagnetic radiation and (2) line emission spectra of hydrogen.

Both of these must be understood before Bohr’s model can make sense, so we take them one at a time.

4.5.1 Wave Particle Duality of Electromagnetic Radiation

A dilemma was posed by electromagnetic radiation in the world of science. Light appeared to be two contradictory things at once, depending on which experiment was performed.

Phenomena such as diffraction and interference of light could be explained by treating light as an electromagnetic wave. Diffraction is the bending of light around an obstacle or through a narrow slit, and interference is the way two light beams can add up to give bright bands or cancel to give dark ones. Only waves behave in these ways.

On the other hand, the black-body radiation or photoelectric effect could not be explained by the wave nature of light, and could be accounted for by considering the particle nature of light. Black-body radiation is the pattern of light emitted by a hot object as its temperature rises; the photoelectric effect is the ejection of electrons from a metal surface when light of a suitable kind falls on it. In both cases the wave picture gave predictions that disagreed with what was actually measured.

The only way to resolve the dilemma was to accept that light has dual behaviour. When light interacts with matter it behaves as a stream of particles called photons; when light propagates, it behaves as an electromagnetic wave.

NoteDefinition

Wave–particle duality of electromagnetic radiation means that radiation behaves as a wave while it travels through space, and as a stream of particles (photons) when it interacts with matter.

Q. Is light “really” a wave or “really” a stream of particles?

A. Neither picture alone is complete. Each is a model that works in its own domain — the wave model when light travels, the particle model when light is absorbed or emitted. Nature is not obliged to match either of the two familiar pictures we borrow from everyday objects, and light is our first clear warning of this.

wave nature vs particle nature of electromagnetic radiation | Feature | Wave nature | Particle nature | |—|—|—| | When it shows itself | While the radiation propagates | When the radiation interacts with matter | | Phenomena explained | Diffraction, interference | Black-body radiation, photoelectric effect | | Radiation is treated as | A continuous electromagnetic wave | A stream of photons | | Quantity that matters | Wavelength, frequency, amplitude | Energy of each photon, \(E = h\nu\) |

Characteristics of Electromagnetic Wave

Figure 4.4 shows a schematic representation of an electromagnetic wave. Notice in Fig. 4.4 that the electric field and the magnetic field oscillate at right angles to each other, and that both are at right angles to the direction in which the wave travels.

Diagram

Fig. 4.4 – Electromagnetic wave: an oscillating electric field in one plane and an oscillating magnetic field in the perpendicular plane, travelling to the right; label electric field, magnetic field, crest, trough, amplitude, wavelength λ, direction of propagation.

Various parameters used to describe the different types of electromagnetic radiation are wavelength, frequency, wavenumber, amplitude, and velocity.

Wavelength (\(\lambda\)) is the distance between two consecutive crests or troughs. It is represented by the symbol \(\lambda\), which is a Greek letter (lambda). The SI unit for wavelength is metre (m). Since atomic-scale wavelengths are very small, they are often quoted in nanometres, where \(1\ \text{nm} = 10^{-9}\ \text{m}\).

NoteDefinition

Wavelength is the distance between two consecutive crests or two consecutive troughs of a wave.

Frequency (\(\nu\)) is the number of waves that pass a given point in one second. It is represented by the Greek letter nu, (\(\nu\)). The SI unit of frequency is hertz (Hz or \(\text{s}^{-1}\)). A frequency of one hertz simply means one complete wave passing per second.

NoteDefinition

Frequency is the number of waves passing a given point in one second.

Wavenumber (\(\bar{\nu}\)) is the number of wavelengths per unit length. Wavenumber is represented by the symbol nu bar (\(\bar{\nu}\)). The commonly used unit for wavenumber is reciprocal centimetre (\(\text{cm}^{-1}\)), while the SI unit is \(\text{m}^{-1}\). Wavenumber is related to the wavelength by the expression

\[\bar{\nu} = \frac{1}{\lambda}\]

NoteDefinition

Wavenumber is the number of complete wavelengths present in unit length, and is the reciprocal of the wavelength.

Amplitude (\(A\)) of a wave is the height of the crest. The square of the amplitude denotes the intensity of the radiation. This is why a brighter beam and a dimmer beam of the same colour differ in amplitude but not in wavelength or frequency.

NoteDefinition

Amplitude is the height of a crest of the wave, and the square of the amplitude gives the intensity of the radiation.

the wave parameters (Wavelength \(\lambda\) — SI unit metre, m; often quoted in nm, where \(1\ \text{nm} = 10^{-9}\ \text{m}\). Frequency \(\nu\) — SI unit hertz, Hz, which is the same as \(\text{s}^{-1}\). Wavenumber \(\bar{\nu}\) — SI unit \(\text{m}^{-1}\), commonly used unit \(\text{cm}^{-1}\). All three are positive quantities. Take care to convert the wavelength into the same length unit as the one demanded by the answer before substituting.)

Q. Why is wavenumber measured in \(\text{cm}^{-1}\) when the SI unit is \(\text{m}^{-1}\)?

A. Convenience. Visible and infrared radiation have wavelengths of the order of \(10^{-5}\) cm, so their wavenumbers come out as manageable numbers of a few thousand \(\text{cm}^{-1}\) rather than awkward powers of ten in \(\text{m}^{-1}\). Spectroscopists adopted the smaller unit and it has stayed.

Different regions of electromagnetic radiation have different values of frequency or wavelengths, as set out in Fig. 4.5a. Thus the radiofrequency region is around \(10^{6}\) Hz, the microwave region is around \(10^{10}\) Hz, the infrared region is around \(10^{13}\) Hz, and the ultraviolet region is around \(10^{16}\) Hz.

Diagram

Fig. 4.5a – The electromagnetic spectrum drawn as a horizontal bar with frequency increasing to the right; mark the four regions given in the text at their approximate frequencies — radiofrequency ~10⁶ Hz, microwave ~10¹⁰ Hz, infrared ~10¹³ Hz, ultraviolet ~10¹⁶ Hz — with the narrow visible band placed between infrared and ultraviolet.

Fig. 4.5a, the electromagnetic spectrum ((a) Moving from the radiofrequency end towards the ultraviolet end, does the wavelength increase or decrease? Justify your answer using \(c = \nu\lambda\). (b) Which of the four named regions consists of photons of the highest energy, and why? (c) Roughly how many times greater is the frequency of ultraviolet radiation than that of infrared radiation?)

In vacuum, the speed of all the types of electromagnetic radiation is the same, which is \(3.0\times10^{8}\ \text{m s}^{-1}\) (\(2.997925\times10^{8}\ \text{m s}^{-1}\) to be accurate). This is called the speed of light, and is denoted by the symbol ‘\(c\)’.

The parameters wavelength (\(\lambda\)), frequency (\(\nu\)) and the speed of light (\(c\)) are related by the expression:

\[c = \nu\lambda\]

This single relation carries an important message. Since \(c\) is a fixed number, \(\nu\) and \(\lambda\) must move in opposite directions: radiation of short wavelength necessarily has high frequency, and radiation of long wavelength has low frequency.

Particle Nature of Electromagnetic Radiation

In the year 1900, Max Planck put forth his quantum theory to explain black-body radiation. According to this theory, the energy of electromagnetic radiation depends upon the frequency and not the amplitude.

This was a sharp break with the wave picture, in which a brighter beam (larger amplitude) simply carries more energy. Planck’s claim was that the energy comes in packets whose size is set by the frequency alone, and that a brighter beam merely delivers more of those packets.

Planck gave the name ‘quantum’ to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation.

NoteDefinition

A quantum is the smallest possible packet of energy that can be emitted or absorbed as electromagnetic radiation.

The energy (\(E\)) of one quantum of radiation is proportional to its frequency (\(\nu\)) and given by

\[E = h\nu \qquad \text{......(4.1)}\]

The proportionality constant ‘\(h\)’ is called Planck’s Constant. Later its value was found out to be \(6.626\times10^{-34}\ \text{J s}\).

NotePrinciple / Law

(Planck’s quantum theory) Electromagnetic radiation is emitted or absorbed not continuously but in whole packets called quanta, the energy of each quantum being proportional to the frequency of the radiation, \(E = h\nu\), where \(h\) is Planck’s constant.

Because \(c = \nu\lambda\), the frequency in Eq. (4.1) may be replaced by \(c/\lambda\), giving a form that is often more convenient when the wavelength is the quantity supplied:

\[E = \frac{hc}{\lambda}\]

The extremely small value of \(h\) explains why the graininess of energy went unnoticed for so long. One quantum of ordinary light carries so little energy that light of everyday brightness appears perfectly smooth and continuous.

In the year 1905, Albert Einstein explained the photoelectric effect using Planck’s quantum theory. In doing so he considered electromagnetic radiation as a stream of photons of energy \(h\nu\). A photon has zero rest mass.

NoteDefinition

A photon is a particle of electromagnetic radiation carrying one quantum of energy, \(E = h\nu\), and having zero rest mass.

Planck introduced the quantum in 1900 with considerable reluctance. He had been trying for years to find a formula that matched the measured black-body curve, and the assumption that energy comes in discrete packets was, in his own view, an act of desperation rather than a picture of reality. It was Einstein who took the idea seriously as physics rather than as a mathematical device, and used it in 1905 to explain the photoelectric effect. Both were later honoured with the Nobel Prize in Physics — and in Einstein’s case, the prize awarded in 1921 cited this work on the photoelectric effect, not the theory of relativity for which he is more famous.

The photoelectric effect is not a laboratory curiosity. Light striking a suitable material and knocking electrons loose is the working principle of the photodiodes and light sensors that switch on street lights at dusk, of the light meters in cameras, and — in a related form — of the solar cells that convert sunlight into electricity. In each case the crucial point is Planck’s and Einstein’s: what matters is the energy of each individual photon, so radiation below a certain frequency produces no effect at all, however intense the beam is made.

Visible light has wavelengths ranging from 400 nm (violet) to 750 nm (red). Express these wavelengths in terms of frequency (Hz). (\(1\ \text{nm} = 10^{-9}\ \text{m}\))

Solution:

\[c = \nu\lambda \quad \therefore\ \nu = \frac{c}{\lambda}\]

The wavelengths are first converted from nanometre to metre, because the speed of light is given in \(\text{m s}^{-1}\) and the two units must match.

\(\therefore\) frequency of violet light =

\[\frac{3\times10^{8}\ \text{m s}^{-1}}{400\times10^{-9}\ \text{m}} = 7.50\times10^{14}\ \text{Hz}\]

and frequency of red light =

\[\frac{3\times10^{8}\ \text{m s}^{-1}}{750\times10^{-9}\ \text{m}} = 4.00\times10^{14}\ \text{Hz}\]

The metre cancels between numerator and denominator, leaving \(\text{s}^{-1}\), which is the hertz.

Parameters of blue and red light are 400 nm and 750 nm respectively. Which of the two is of higher energy?

Solution: 400 nm and 750 nm are the wavelengths of blue and red light, respectively. Energy of radiation is given by the expression \(E = h\nu\) and \(\nu\), the frequency, of radiation is related to the wavelength by the expression.

\[\nu = \frac{c}{\lambda}\]

Therefore, shorter the wavelength, \(\lambda\), larger the frequency, \(\nu\), and higher the energy, \(E\). Thus, blue light which has shorter \(\lambda\) (400 nm) than red light (750 nm) has higher energy.

Note that no arithmetic was needed here. Once the chain “shorter \(\lambda\) → larger \(\nu\) → larger \(E\)” is understood, comparison questions of this type can be answered by inspection — a habit worth building, since examinations often ask you to order radiations by energy rather than to calculate.

Calculate the energy of one photon of light whose wavelength is 600 nm. (\(h = 6.626\times10^{-34}\ \text{J s}\), \(c = 3.0\times10^{8}\ \text{m s}^{-1}\))

Solution: The wavelength is given, so the convenient form of the relation is

\[E = \frac{hc}{\lambda}\]

Converting the wavelength to metre:

\[\lambda = 600\ \text{nm} = 600\times10^{-9}\ \text{m} = 6.00\times10^{-7}\ \text{m}\]

Substituting the values:

\[E = \frac{(6.626\times10^{-34}\ \text{J s})\times(3.0\times10^{8}\ \text{m s}^{-1})}{6.00\times10^{-7}\ \text{m}}\]

\[E = \frac{1.9878\times10^{-25}\ \text{J m}}{6.00\times10^{-7}\ \text{m}}\]

\[E = 3.313\times10^{-19}\ \text{J}\]

Check the units: \(\text{J s}\times\text{m s}^{-1} = \text{J m}\), and dividing by metre leaves joule, as it must for an energy.

(question only — no solution) Radiation used in an infrared spectrometer has a wavelength of \(5.0\times10^{-6}\ \text{m}\). Calculate (a) its frequency in Hz, (b) its wavenumber in \(\text{cm}^{-1}\), and (c) the energy of one of its photons in joule.

4.5.2 Line Emission Spectrum of Hydrogen

When a substance is irradiated with light it absorbs energy. Atoms, molecules or ions, which have absorbed radiation are said to be ‘excited’. Heating can also result in an excited state.

When an excited species gives away the absorbed energy in the form of radiation, the process is called emission of radiation. The recorded spectrum of this emitted radiation is called an ‘emission spectrum’.

NoteDefinition

An emission spectrum is the record of the radiation given out by an excited atom, molecule or ion as it releases the energy it had absorbed.

Of all the elements, hydrogen has the simplest emission spectrum. This is not an accident — hydrogen has only one electron, so there is only one electron whose behaviour has to be accounted for, and its spectrum became the natural test case for every atomic model.

The fluorescent tube, the sodium vapour lamp, the neon sign and the halogen lamp are all examples of atomic emission. In each of them, atoms of a gas are excited by an electric discharge and then give the energy back as light. The characteristic yellow glow of a sodium street lamp and the red-orange of a neon sign are simply the emission spectra of those atoms, seen by eye rather than through an instrument. The same principle is used analytically: because every element emits its own fixed set of wavelengths, a laboratory can identify which metals are present in a sample from the light its atoms give out — the basis of the flame tests you may have seen performed, and of atomic emission spectroscopy in industry.

The five series of hydrogen lines

When electric discharge is passed through gaseous hydrogen, it emits radiation. The hydrogen emission spectrum was studied by five scientists in the late nineteenth century. They recorded the hydrogen emission spectra in different regions of electromagnetic radiation.

The spectra were not continuous and comprised of a series of lines corresponding to different frequencies, as shown in Fig. 4.5. This was the striking result. Sunlight passed through a prism gives an unbroken band of colour, but excited hydrogen gives only a handful of sharp bright lines separated by darkness — as though the atom were allowed to emit some frequencies and forbidden to emit all the others.

Diagram

Fig. 4.5 – Emission line spectrum of hydrogen: a horizontal wavelength scale from 0 to 3.5, with groups of sharp vertical lines marked as the Paschen series (infrared), Balmer series (partly visible) and Lyman series (ultraviolet); note that within each group the lines get closer and closer together and eventually reach the series limit.

In the year 1885 Balmer expressed the wavenumbers of the emission lines in the visible region of electromagnetic radiation by the formula

\[\bar{\nu} = 109677\left[\frac{1}{2^{2}} - \frac{1}{n^{2}}\right]\ \text{cm}^{-1} \qquad \text{......(4.2)}\]

where \(n = 3, 4, 5.......\)

The lines described by this formula are called the Balmer series. Substituting \(n = 3\) gives one observed line, \(n = 4\) the next, and so on — a single expression reproducing a whole family of measured wavelengths.

J. Rydberg found that all the five series of lines could be described by the following general expression.

\[\bar{\nu} = 109677\left[\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right]\ \text{cm}^{-1} \qquad \text{......(4.3)}\]

where \(n_1 = 1, 2, 3\ .....................\)

\[n_2 = [n_1+1],\ [n_1+2],\ [n_1+3],.........\]

Read the two integers carefully, because they are not interchangeable. For any one series \(n_1\) is held fixed while \(n_2\) runs through all the larger whole numbers, generating the successive lines of that series. Balmer’s Eq. (4.2) is simply Eq. (4.3) with \(n_1\) set equal to 2. Since \(n_2 > n_1\) always, the bracket stays positive and the wavenumber comes out positive, as it must.

The value \(109{,}677\ \text{cm}^{-1}\) is called the Rydberg constant for hydrogen ‘\(R_H\)’. Table 4.5 shows the distinctive features of these five series.

Table 4.5 : Series of emission spectral lines for hydrogen

Series \(n_1\) \(n_2\) Spectral region
Lyman 1 2,3,….. Ultraviolet
Balmer 2 3,4,….. Visible
Paschen 3 4,5,….. Infrared
Bracket 4 5,6,….. Infrared
Pfund 5 6,7,….. Infrared

Notice the pattern running down the table. As \(n_1\) increases, the series shifts from ultraviolet through visible to infrared — that is, the emitted radiation becomes progressively lower in frequency and therefore lower in energy. Only one series, the Balmer series with \(n_1 = 2\), happens to fall in the region our eyes can detect, which is why it was the first to be measured and fitted.

Johann Balmer was a Swiss schoolteacher of mathematics, not a physicist, and he was almost sixty when he took up the problem. He had before him just four numbers — the measured wavelengths of the four visible hydrogen lines — and he searched for a formula that would reproduce them. The expression he found in 1885 fitted all four, and it also predicted further lines that were later found exactly where it said they would be. Balmer had no explanation whatever for why his formula worked; that had to wait nearly thirty years for Bohr.

Q. Balmer’s formula fitted the hydrogen lines perfectly. Why was it still not a theory of the atom?

A. Because it described without explaining. The formula says which wavelengths appear, but not why whole numbers should appear in an atom at all, nor what physically distinguishes \(n = 2\) from \(n = 3\). A relation that fits the data is a valuable clue; it becomes a theory only when something in the structure of the atom is shown to require it — which is precisely what Bohr set out to do.

(question only — no solution) Using Eq. (4.3) with \(R_H = 109677\ \text{cm}^{-1}\), calculate the wavenumber and the wavelength (in nm) of the first line of the Lyman series, that is, the line for which \(n_1 = 1\) and \(n_2 = 2\). State the region of the electromagnetic spectrum in which this line falls.

4.6 Bohr’s Model for Hydrogen Atom

Niels Bohr (1913) put forth his postulates about the atomic model for hydrogen. While doing so he used the quantum theory, the wave particle duality of electromagnetic radiation and the emission line spectra of hydrogen.

Bohr’s strategy is worth noticing before we look at the postulates themselves. Classical physics predicted a collapsing atom and could not explain the sharp spectral lines. Rather than patch the old rules, Bohr simply declared that certain classical conclusions do not apply inside an atom, and then checked whether the resulting model reproduced the measured hydrogen spectrum. It did — with remarkable precision.

4.6.1 Postulates of Bohr Atomic Theory

Bohr’s model of the hydrogen atom is based on the following postulates.

Postulate 1. The electron in the hydrogen atom can move around the nucleus in one of the many possible circular paths of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus in an increasing order of energy.

The word stationary here does not mean that the electron stands still. It means that the state itself is unchanging — an electron sitting in such an orbit keeps the same radius and the same energy indefinitely.

NotePrinciple / Law

(Bohr’s first postulate) The electron in a hydrogen atom revolves around the nucleus only in certain allowed circular orbits of fixed radius and fixed energy, called stationary states, arranged concentrically in increasing order of energy.

Postulate 2. The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state if and when the required amount of energy is absorbed by the electron. Energy is emitted when the electron moves from a higher stationary state to a lower one. The energy change does not take place in a continuous manner.

The phrase the required amount is the crucial one. An electron cannot accept a little energy and rise a little; it either receives exactly the amount needed to reach an allowed state above, or it stays where it is.

NotePrinciple / Law

(Bohr’s second postulate) An electron in a stationary state neither gains nor loses energy with time. It changes state only by absorbing the exact energy needed to move to a higher state, or by emitting the exact energy released in falling to a lower one, and such changes are not continuous.

This postulate answers the objection raised in Section 4.4 head-on. Bohr’s electron does not radiate while it revolves; it radiates only while changing from one allowed state to another.

Q2. Rutherford’s atom looks like a miniature solar system, with electrons circling the nucleus like planets. Why should such an atom collapse in a tiny fraction of a second — and why do real atoms not collapse?

A. It should collapse because an electron moving in a circle is accelerating, and Maxwell’s electromagnetic theory requires an accelerated charge to radiate energy continuously. Losing energy at every instant, the electron would spiral steadily inwards until it reached the nucleus. Real atoms survive because that classical rule does not hold inside the atom. Bohr’s first two postulates state that only certain orbits are allowed and that an electron in such a stationary state does not radiate at all. Radiation is emitted only when the electron jumps between allowed states — and since the lowest allowed state has a definite, non-zero radius, there is simply no state below it for the electron to fall into.

Postulate 3. The frequency of radiation absorbed or emitted when a transition occurs between two stationary states that differ in energy by \(\Delta E\) is given by the following expression

\[\nu = \frac{\Delta E}{h} = \frac{E_2 - E_1}{h} \qquad \text{......(4.4)}\]

Where \(E_1\) and \(E_2\) are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.

Notice that this is Planck’s relation \(E = h\nu\) turned round to serve the atom. The energy of the photon is fixed by the gap between two allowed levels, and the frequency follows from that gap.

NotePrinciple / Law

(Bohr’s frequency rule) When an electron makes a transition between two stationary states differing in energy by \(\Delta E\), the frequency of the radiation absorbed or emitted is \(\nu = \Delta E / h\).

Postulate 4. The angular momentum of an electron in a given stationary state can be expressed as

\[m\mathrm{v}r = n \times \frac{h}{2\pi} \qquad \text{where } n = 1,\ 2,\ 3 \qquad \text{......(4.5)}\]

Thus, an electron can move only in those orbits for which its angular momentum is an integral multiple of \(h/2\pi\). Thus only certain fixed orbits are allowed here.

NotePrinciple / Law

(Quantization of angular momentum) An electron may occupy only those orbits in which its angular momentum \(m\mathrm{v}r\) is a whole-number multiple of \(h/2\pi\).

This postulate is what makes the first one work. Postulate 1 says only some orbits are allowed but does not say which; Postulate 4 supplies the selection rule, and out of it come the fixed radii and energies listed in the next subsection.

Understanding Angular Momentum

Angular momentum is a product of moment of inertia (\(I\)) and angular velocity (\(\omega\)), omega. The quantity is the rotational counterpart of ordinary momentum: moment of inertia plays the part of mass, and angular velocity the part of speed. The set-up is drawn in Fig. 4.6a.

Diagram

Fig. 4.6a – A particle of mass m moving along a circular path of radius r with velocity v tangential to the circle, angular velocity ω marked at the centre; label m, r, v and ω.

Angular momentum \(= I \times \omega\) ; but

\[I = m r^{2} \quad \text{and} \quad \omega = \mathrm{v}/r\]

\(\therefore\) Angular momentum \(= mr^{2} \times \mathrm{v}/r = m\mathrm{v}r\)

The two \(r\) terms partly cancel, leaving the compact form \(m\mathrm{v}r\) used in Eq. (4.5), where \(m\) is the mass of the electron in kilogram, \(\mathrm{v}\) its speed in \(\text{m s}^{-1}\) and \(r\) the radius of the orbit in metre.

Niels Bohr came to Manchester as a young Danish physicist to work in Rutherford’s laboratory, arriving just as the nuclear model was raising more questions than it settled. His answer, published in 1913, was bold to the point of being uncomfortable: he kept Rutherford’s nucleus, kept circular orbits, and then simply forbade the electron to radiate while in them. Many senior physicists found the mixture of classical and quantum rules hard to accept. What silenced most of the objections was that the model reproduced the hydrogen spectrum, including the value of the Rydberg constant, from first principles. Bohr received the Nobel Prize in Physics in 1922.

4.6.2 Results of Bohr’s Theory

Bohr’s theory is used to derive the energies of orbits, that is, the stationary states, in the hydrogen atom. The results of Bohr’s theory for the hydrogen atom are summarized here.

a. The stationary states for the electron are numbered \(n = 1, 2, 3.......\) These integers are known as principal quantum numbers.

b. The radii of the stationary states are

\[r_n = n^{2}a_0\]

where \(a_0 = 52.9\ \text{pm}\) (picometre). Thus, the radius of the first stationary state, called the Bohr radius, is 52.9 pm.

Because the radius goes as \(n^{2}\), the orbits are not evenly spaced. The second orbit lies four times as far out as the first, the third nine times as far, and so on — the atom’s allowed shells get rapidly more widely separated.

c. The most important property associated with the electron is the energy of its stationary state. It is given by the expression.

\[E = -R_H\left(\frac{1}{n^{2}}\right), \text{ where } n = 1, 2, 3, \qquad \text{......(4.6)}\]

\(R_H\) is the Rydberg constant for hydrogen and its value in joules is \(2.18\times10^{-18}\ \text{J}\).

The lowest energy state is called the ground state. Energy of the ground \(1/1^{2}\) state is

\[E_1 = -2.18\times10^{-18}\ 1/1^{2} = -2.18\times10^{-18}\ \text{J}\]

Energy of the stationary state corresponding to \(n = 2\) is

\[E_2 = -2.18\times10^{-18}\ (1/(2)^{2}) = -0.545\times10^{-18}\ \text{J}\]

Q. What does the negative sign of electron energy convey?

A. A free electron at rest is an electron that is at infinity from the nucleus and does not experience any electrostatic force of attraction towards the nucleus, and therefore it is assigned the energy value of zero. The negative sign means that the energy of the electron in the atom is lower than the energy of a free electron at rest. Mathematically this corresponds to setting ‘\(n\)’ equal to infinity in the equation so that \(E_\infty = 0\). As the electron gets close to the nucleus, ‘\(n\)’ decreases, \(E_n\) becomes large in absolute value and more and more negative. Thus stationary states with smaller values of ‘\(n\)’ have large and negative energy. The negative sign corresponds to the attractive force between the electron and the nucleus.

energy of an electron in a stationary state (SI unit: joule, J. Sign convention — the zero of energy is taken as the free electron at rest, infinitely far from the nucleus. Every bound state therefore has a negative energy. A more negative value means a more tightly bound, more stable state; \(n = 1\) is the most negative and hence the most stable. When comparing two levels, remember that \(-2.18\times10^{-18}\ \text{J}\) is lower than \(-0.545\times10^{-18}\ \text{J}\).)

Using the energies of the \(n = 1\) and \(n = 2\) states of the hydrogen atom given above, calculate the energy that must be absorbed to excite the electron from the ground state to the second stationary state.

Solution: Energy must be supplied to lift the electron from the lower state to the higher one, and the amount needed is exactly the gap between them.

\[\Delta E = E_2 - E_1\]

\[\Delta E = (-0.545\times10^{-18}\ \text{J}) - (-2.18\times10^{-18}\ \text{J})\]

Subtracting a negative number is the same as adding its magnitude:

\[\Delta E = (-0.545 + 2.18)\times10^{-18}\ \text{J}\]

\[\Delta E = 1.635\times10^{-18}\ \text{J}\]

The answer is positive, which is the sign we expect for energy absorbed by the atom. Had the electron fallen from \(n = 2\) to \(n = 1\), the same amount of energy would have been given out as a photon.

Calculate the radius of the third stationary state of the hydrogen atom.

Solution: For hydrogen the radii of the stationary states are given by

\[r_n = n^{2}a_0, \qquad a_0 = 52.9\ \text{pm}\]

For the third stationary state, \(n = 3\):

\[r_3 = 3^{2}\times52.9\ \text{pm} = 9\times52.9\ \text{pm}\]

\[r_3 = 476.1\ \text{pm}\]

The third orbit lies nine times as far from the nucleus as the first — a direct consequence of the \(n^{2}\) dependence.

Hydrogen-Like Species

Bohr’s theory is applicable to the hydrogen atom and to hydrogen-like species, which contain only one extranuclear electron. The restriction matters: with two or more electrons present, each electron would feel not only the pull of the nucleus but also the repulsion of the others, and Bohr’s simple picture does not allow for that.

How many electrons are present in \(^{2}_{1}\mathrm{H}\), \(_{2}\mathrm{He}\) and \(\mathrm{He^{\oplus}}\)? Which of these are hydrogen like species?

Solution: Hydrogen-like species contain only one electron.

\(^{2}_{1}\mathrm{H}\) number of protons = 1 = number of electrons

\(_{2}\mathrm{He}\) : number of protons = 2 = number of electrons

\(\mathrm{He^{\oplus}}\) : (number of electron in He) \(-1\) = (number of electron in \(\mathrm{He^{\oplus}}\)) \(= (2 - 1) = 1\)

Thus \(^{2}_{1}\mathrm{H}\) and \(\mathrm{He^{\oplus}}\) are hydrogen-like species.

Note that the test is the electron count alone, not the nuclear charge. \(\mathrm{He^{\oplus}}\) has twice the nuclear charge of hydrogen, yet it qualifies, because only one electron is left to be described.

d. Bohr theory can be applied to hydrogen like species. For example \(\mathrm{He^{\oplus}}\), \(\mathrm{Li^{2\oplus}}\), \(\mathrm{Be^{3\oplus}}\) and so on. Energies of the stationary states associated with these species are given by :

\[E_n = -2.18\times10^{-18}\left(\frac{Z^{2}}{n^{2}}\right)\ \text{J} \qquad \text{.......(4.7)}\]

and radii by the expression

\[r_n = \frac{52.9(n^{2})}{Z}\ \text{pm} \qquad \text{.......(4.8)}\]

where \(Z\) is the atomic number. From the above expression it can be seen that the energy decreases (becomes more negative) and the radius becomes smaller as the value of \(Z\) increases.

Both equations reduce to the hydrogen forms when \(Z = 1\), as they must. The physical reason behind the trend is straightforward: a larger nuclear charge pulls the single electron more strongly, drawing it closer in and binding it more tightly.

e. Velocities of electrons can also be calculated from the Bohr theory. Qualitatively it is found that the magnitude of the velocity of an electron increases with increase of \(Z\) and decreases with increase in the principal quantum number \(n\).

Calculate the radius and energy associated with the first orbit of \(\mathrm{He^{\oplus}}\).

Solution: \(\mathrm{He^{\oplus}}\) is a hydrogen-like species having the nuclear charge \(Z = 2\) and for the first orbit \(n = 1\)

radius of first orbit of \(\mathrm{He^{\oplus}}\)

\[r_1 = \frac{52.9(n^{2})}{Z} = \frac{52.9\times1^{2}}{2}\]

Carrying out this division gives 26.45 pm — exactly half the Bohr radius, as expected for a doubled nuclear charge.

Energy of first orbit of \(\mathrm{He^{\oplus}}\)

\[E_n = -2.18\times10^{-18}\left(\frac{Z^{2}}{n^{2}}\right)\ \text{J}\]

\[= -2.18\times10^{-18}\left(\frac{2^{2}}{1^{2}}\right)\]

\[= -8.72\times10^{-18}\ \text{J}\]

Compare this with \(-2.18\times10^{-18}\ \text{J}\) for hydrogen: because the energy depends on \(Z^{2}\), doubling the nuclear charge makes the electron four times more tightly bound.

(question only — no solution) Using Eq. (4.7) and Eq. (4.8), calculate the radius and the energy of the electron in the first orbit of \(\mathrm{Li^{2\oplus}}\) (\(Z = 3\)). State whether this electron is more or less tightly held than the electron in the first orbit of \(\mathrm{He^{\oplus}}\), and give the reason.

4.6.3 Explanation of the Line Spectrum of Hydrogen Using Bohr Theory

The line emission spectrum (Fig. 4.5) obtained from atomic hydrogen can be explained quantitatively using Bohr theory.

According to the second postulate of Bohr theory, radiation is emitted when the electron moves from an outer orbit of higher principal quantum number (\(n_i\)) to an inner orbit of lower principal quantum number (\(n_f\)). The subscripts stand for initial and final, and in emission the initial orbit is always the outer one.

The energy difference (\(\Delta E\)) between the initial and final orbit of the electronic transition corresponds to the energy of the emitted radiation. From the third postulate of Bohr theory \(\Delta E\) can be expressed as

\[\Delta E = E_i - E_f \qquad \text{.......(4.9)}\]

According to the results derived from Bohr theory the energy \(E\) of an orbit is related to its principal quantum number ‘\(n\)’ by the Eq. (4.6).

\[E = -R_H\left(\frac{1}{n^{2}}\right) \qquad \text{.......(4.6)}\]

Combining these two Eq. (4.9) and Eq. (4.6) we get:

\[\Delta E = \left[-\frac{R_H}{n_i^{2}}\right] - \left[-\frac{R_H}{n_f^{2}}\right] = R_H\left[\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right]\]

The two minus signs in the middle step are easy to mishandle. Removing the brackets turns the second term into \(+R_H/n_f^{2}\), and taking \(R_H\) out as a common factor leaves the bracket in the order shown, with the final-orbit term first.

Substituting the value of \(R_H\) in joules we get

\[\Delta E = 2.18\times10^{-18}\left[\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right]\ \text{J} \qquad \text{.......(4.10)}\]

From Energy Difference to Wavenumber

This expression can be rewritten in terms of the wavenumber of the emitted radiation in the following steps.

We know: \((\Delta E)\ \text{J} = (h)\ \text{J s} \times (\nu)\ \text{Hz}\) … (4.1)

and by definition

\[(\bar{\nu})\ \text{cm}^{-1} = \frac{(\nu)\ \text{Hz}}{(c)\ \text{cm s}^{-1}} \qquad \text{.......(4.11)}\]

combining these Eq. (4.10), Eq. (4.1) and Eq. (4.11) we get

\[\therefore\ (\bar{\nu})\ \text{cm}^{-1} = \frac{(\Delta E)}{(h)\ \text{J s}} \times \frac{1}{(c)\ \text{cm s}^{-1}}\]

\[= \frac{2.18\times10^{-18}}{6.626\times10^{-34}\times3\times10^{10}}\left[\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right]\ \text{cm}^{-1}\]

\[\therefore\ (\bar{\nu})\ \text{cm}^{-1} = 109677\left[\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right]\ \text{cm}^{-1}\]

The speed of light appears here as \(3\times10^{10}\ \text{cm s}^{-1}\) rather than \(3\times10^{8}\ \text{m s}^{-1}\), because the wavenumber is wanted in \(\text{cm}^{-1}\) and every length in the calculation must therefore be in centimetre.

This appears like the Rydberg Eq. (4.3), where \(n_f = n_1\) and \(n_i = n_2\).

Look carefully at what has just happened. The number 109677 was not put into the calculation; it came out of it, assembled from \(R_H\) in joules, Planck’s constant and the speed of light. A constant that Balmer and Rydberg had obtained purely by fitting measurements had been produced from a model of the atom.

In other words, Bohr theory successfully accounts for the empirical Rydberg equation for the line emission spectrum of hydrogen. In the Rydberg equation ‘\(n_1\)’ and ‘\(n_f\)’ are integers. Bohr’s theory assigns physical meaning to them as principal quantum numbers corresponding to the concentric orbits. The integers in the Rydberg equation stand for the final orbit, \(n_f\) of the electronic transition, and \(n_1\) for the initial orbit.

Q3. When hydrogen gas is made to glow in a discharge tube, why does it give only a few sharp coloured lines instead of a smooth rainbow of all colours?

A. Because the electron is allowed only certain fixed energies. The discharge lifts electrons into higher stationary states, and light is emitted only when an electron drops back to a lower state. The energy of the photon emitted equals the gap between two allowed levels, and by Bohr’s frequency rule \(\nu = \Delta E/h\), a fixed gap gives a fixed frequency. Since there are only certain allowed levels, there are only certain possible gaps — and therefore only certain possible frequencies. A continuous rainbow would require the electron to be able to lose any amount of energy at all, which is exactly what quantization forbids.

Which Series Comes from Which Transition

The emission lines comprising the five series thus, are the result of electronic transitions from the excited hydrogen atoms.

The Lyman series is the result of moving of electron excited to higher orbits of \(n_2 = n_i = 2, 3, 4,.....\) etc. to lower orbits of \(n_1 = n_f = 1\); the Balmer series results from electron from \(n_2 = n_i = 3, 4,........\) to the lower orbit of \(n_1 = n_f = 2\), so on and so forth.

In short, a series is named by where the electron lands, not by where it starts. Every transition ending at \(n = 1\) belongs to the Lyman series, every transition ending at \(n = 2\) to the Balmer series, and so on. Since the levels crowd together as \(n\) increases, the biggest energy gaps — and therefore the highest-frequency photons — are those that end at \(n = 1\), which is why the Lyman series lies in the ultraviolet.

The electronic transitions giving rise to the different emission line series of atomic hydrogen are shown in Fig. 4.6.

Diagram

Fig. 4.6 – Electronic transitions in the hydrogen spectrum: horizontal energy levels labelled n = 1, 2, 3, 4, 5, 6 and n = ∞, drawn closer together as n rises; downward arrows grouped into the Lyman series (all ending at n = 1), Balmer series (ending at n = 2), Paschen series (n = 3), Bracket series (n = 4) and Pfund series (n = 5); mark the wavelength axis and label the ionisation enthalpy of hydrogen, 1312 kJ mol⁻¹.

Fig. 4.6, the transition diagram ((a) Why are the energy levels drawn closer and closer together as \(n\) increases? (b) Which of the five series involves the largest energy gaps, and how does the figure show this? (c) The figure marks the ionisation enthalpy of hydrogen as 1312 kJ mol⁻¹ — which transition on the diagram does this quantity correspond to?)

What is the wavelength of the photon emitted during the transition from the orbit of \(n = 5\) to that of \(n = 2\) in hydrogen atom?

Solution: The wavenumber of the transition is given by the Rydberg expression

\[(\bar{\nu}) = 109677\left[\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right]\ \text{cm}^{-1}\]

Here \(n_1 = 2\) and \(n_2 = 5\)

\[\therefore\ (\bar{\nu}) = 109677\left[\frac{1}{2^{2}} - \frac{1}{5^{2}}\right]\]

\[= 109677\left[\frac{1}{4} - \frac{1}{25}\right]\]

\[= 109677 \times \frac{21}{100} = 23032.17\ \text{cm}^{-1}\]

\[\lambda = \frac{1}{23032.17}\]

\[\lambda = \frac{1}{\bar{\nu}} = 4.34\times10^{-5}\ \text{cm}\]

wavelength of photon

\[\lambda = 4.34\times10^{-5}\ \text{cm} = 4.34\times10^{-5}\times10^{7}\ \text{nm} = 434\ \text{nm}\]

Two steps deserve a second look. The bracket was evaluated as \(\frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100}\). And the final conversion uses \(1\ \text{cm} = 10^{7}\ \text{nm}\), which is why the power of ten jumps by seven. A wavelength of 434 nm lies in the visible region — as it should, since the transition ends at \(n = 2\) and therefore belongs to the Balmer series.

Hydrogen’s spectral lines are read far beyond the laboratory. Because every element emits its own fixed set of wavelengths, astronomers can identify the composition of a star from the light it sends us, and hydrogen — the most abundant element in the universe — is the first thing they look for. The same lines also carry information about motion: if the whole pattern of hydrogen lines from a distant galaxy is found shifted towards longer wavelengths, the galaxy is receding from us. An atomic model built to explain a discharge tube full of hydrogen ended up being one of the chief tools for surveying the universe.

(question only — no solution) Calculate the wavenumber (in \(\text{cm}^{-1}\)) and the wavelength (in nm) of the radiation emitted when an electron in a hydrogen atom falls from the orbit \(n = 4\) to the orbit \(n = 2\). To which series does this line belong, and in which region of the spectrum does it lie?

4.6.4 Limitations of Bohr Model

For all its success with hydrogen, Bohr’s model soon ran into difficulties it could not overcome.

  1. Bohr’s atomic model failed to account for the finer details of the hydrogen atom spectrum observed in sophisticated spectroscopic experiments. When instruments improved, several lines that Bohr treated as single lines turned out to be closely spaced groups of lines.

  2. The Bohr model was unable to explain the spectrum of atoms other than hydrogen. Even helium, with just two electrons, lay beyond its reach.

  3. Bohr theory could not explain the splitting of spectral lines in the presence of a magnetic field (the Zeeman effect) or an electric field (the Stark effect).

  4. Bohr theory failed to explain the ability of atoms to form molecules by chemical bonds. For a chemist this is the most serious failure of all — a model of the atom that cannot account for bonding cannot support chemistry.

It was, therefore, thought that a better theory was needed to explain the salient features of atomic structure.

Rutherford model vs Bohr model | Feature | Rutherford model | Bohr model | |—|—|—| | Nucleus | Small, dense, positively charged | Same, retained unchanged | | Electron paths | Orbits of any radius | Only certain allowed orbits (stationary states) | | Energy of the electron | Not specified | Fixed for each orbit, \(E = -R_H(1/n^{2})\) | | Radiation while revolving | Continuous emission predicted | No emission in a stationary state | | Stability of the atom | Predicts collapse | Explains stability | | Line spectrum of hydrogen | Not explained | Explained quantitatively, reproduces \(R_H\) | | Atoms other than hydrogen | Not explained | Still not explained | | Chemical bonding | Not explained | Not explained |

4.6.5 Reasons for Failure of the Bohr Model

With the limitations of the Bohr model for the hydrogen atom becoming transparent, attempts were made to develop a better and more general model for the atom. This was possible because two important developments took place after the Bohr model was postulated.

These developments were:

  1. de Broglie’s proposal of dual behaviour of matter, and
  2. Heisenberg uncertainty principle.

Each of these strikes at a different assumption buried inside Bohr’s picture.

De Broglie’s Dual Behaviour of Matter

In the Bohr model an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus.

In contrast to this de Broglie proposed in 1924 that matter should exhibit a dual behaviour, that is, both particle and wave like properties. This means that an electron should have momentum, \(p\), a property of a particle, as well as wavelength, \(\lambda\), a property of a wave.

The step being taken here is worth appreciating. Radiation had already been shown to be both wave and particle; de Broglie simply asked whether the same double life might be led by matter.

He gave the following relation between \(\lambda\) and \(p\) of a material particle.

\[\lambda = \frac{h}{m\mathrm{v}} = \frac{h}{p}\]

Here \(h\) is Planck’s constant in J s, \(m\) is the mass of the particle in kg and \(\mathrm{v}\) its velocity in \(\text{m s}^{-1}\); the wavelength \(\lambda\) comes out in metre.

NotePrinciple / Law

(de Broglie relation) Every moving material particle has an associated wavelength given by \(\lambda = h/m\mathrm{v}\), which is inversely proportional to its momentum.

De Broglie’s prediction was confirmed by diffraction experiments (a wave property). Diffraction is something only waves do, so obtaining a diffraction pattern from a beam of electrons settled the matter.

Q. If all moving matter has a wavelength, why do we never see a moving cricket ball behave like a wave?

A. Because \(\lambda = h/m\mathrm{v}\) and \(h\) is extraordinarily small. For an object of ordinary mass the wavelength works out to something far smaller than any object it could possibly bend around, so no wave behaviour is ever observed. Only for particles as light as an electron does the wavelength become comparable to atomic dimensions, and only then does the wave nature show itself.

Calculate the wavelength associated with an electron moving with a velocity of \(1.0\times10^{6}\ \text{m s}^{-1}\). (Mass of electron \(= 9.10938\times10^{-31}\ \text{kg}\), \(h = 6.626\times10^{-34}\ \text{J s}\))

Solution: By the de Broglie relation,

\[\lambda = \frac{h}{m\mathrm{v}}\]

Substituting the values:

\[\lambda = \frac{6.626\times10^{-34}\ \text{J s}}{(9.10938\times10^{-31}\ \text{kg})\times(1.0\times10^{6}\ \text{m s}^{-1})}\]

\[\lambda = \frac{6.626\times10^{-34}}{9.10938\times10^{-25}}\ \text{m}\]

\[\lambda = 7.27\times10^{-10}\ \text{m}\]

This is about 727 pm — of the same order as the size of an atom. That is precisely why the wave nature of the electron cannot be ignored when describing an atom, while it can be ignored completely for any everyday object.

The wave nature of the electron is the working principle of the electron microscope. An ordinary optical microscope cannot resolve detail much finer than the wavelength of visible light, a few hundred nanometres. A beam of fast-moving electrons has a de Broglie wavelength thousands of times shorter, so an instrument that focuses electrons instead of light can resolve far finer detail — down to the arrangement of particles in a catalyst or the structure of a virus. Chemists use such instruments routinely to examine surfaces, nanomaterials and crystal defects. As a small exercise, before reading on, predict what happens to the resolving power of such a microscope if the electrons are accelerated to a higher velocity.

(question only — no solution) Calculate the de Broglie wavelength associated with (a) an electron of mass \(9.10938\times10^{-31}\ \text{kg}\) moving at \(2.0\times10^{6}\ \text{m s}^{-1}\), and (b) a ball of mass 0.10 kg moving at \(20\ \text{m s}^{-1}\). Comment on why the wave nature is significant in only one of the two cases.

De Broglie’s proposal was made in his doctoral thesis, and it was so unusual that his examiners were unsure what to make of it. The idea was taken seriously partly because Einstein, on being shown the work, judged it sound. Within a few years, experiments in which beams of electrons were diffracted by crystals showed the predicted wave behaviour directly, and de Broglie was awarded the Nobel Prize in Physics in 1929 — an unusually short interval between a thesis and a prize.

Heisenberg Uncertainty Principle

In the year 1927 Werner Heisenberg stated the uncertainty principle: “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.”

In other words the position and momentum of an electron cannot be determined with the same certainty. If the certainty of determination of one property of the two is high, it means that the uncertainty of its determination is low. In that case the uncertainty of determination of the other property is very high.

NotePrinciple / Law

(Heisenberg uncertainty principle) The position and the momentum of an electron cannot both be determined exactly at the same instant; the more precisely one of them is known, the less precisely the other can be known.

Mathematically the Heisenberg uncertainty principle is expressed as:

\[\Delta x \times \Delta p_x \geq \frac{h}{4\pi}\]

\[\Delta x \times \Delta(m\mathrm{v}_x) \geq \frac{h}{4\pi}\]

Since momentum is mass times velocity and the mass of the electron is a fixed quantity, the mass may be taken out of the uncertainty and carried to the other side, giving the form in which the velocity appears by itself:

\[\Delta x \times \Delta \mathrm{v}_x \geq \frac{h}{4\pi m}\]

Here \(\Delta x\) is the uncertainty in position and \(\Delta p_x\) (or \(\Delta \mathrm{v}_x\)) is the uncertainty in momentum. Both are uncertainties, not values — the product of the two errors cannot be pushed below a certain floor, no matter how good the apparatus.

A further implication of the uncertainty principle is that for an electron having certain energy one can only determine its probability at a particular point \(x\) around the nucleus. This single sentence marks the end of the orbit and the beginning of the modern picture: we may ask how likely the electron is to be found somewhere, but not where it is at a given moment.

Bohr’s model describes concentric orbits as well defined paths of the electron rotating about the nucleus and calculates the energy of the electron occupying these orbits. The Bohr model assumes that both the position and the momentum of the electron in the hydrogen atom are known exactly at the same time, which is ruled out by the Heisenberg uncertainty principle.

Hence no attempt was made to extend the Bohr model to other atoms. A different approach to the atomic model, which would account for the particle duality of matter and would be consistent with the Heisenberg uncertainty principle, was required. This became possible with the development of quantum mechanics.

Flowchart

FC 4.4 – Why the Bohr model had to be replaced: Rutherford model → predicts collapsing atom and gives no electron energies → Bohr postulates (fixed orbits, no radiation in a stationary state, ν = ΔE/h, mvr = nh/2π) → explains hydrogen spectrum and derives R_H → but fails for multi-electron atoms, fine structure, Zeeman and Stark effects, chemical bonding → two new ideas arrive: de Broglie’s wave nature of matter (1924) and Heisenberg’s uncertainty principle (1927) → a definite orbit becomes meaningless → quantum mechanical model of the atom

(question only — no solution) Explain, in two or three sentences each, why the Heisenberg uncertainty principle makes the idea of a definite circular orbit unacceptable, and why this objection does not arise for a planet going round the sun.

4.7 Quantum Mechanical Model of Atom

A new branch of science, called quantum mechanics, was developed in 1926 by Werner Heisenberg and Erwin Schrodinger based on the uncertainty principle and wave motion, respectively. Quantum mechanics based on the ideas of wave motion will be discussed here.

Schrodinger developed the fundamental equation of quantum mechanics which incorporates the wave particle duality of matter. Since de Broglie had shown that a moving electron has a wavelength, the sensible next step was to write down an equation of the kind used for waves and apply it to the electron in an atom.

The Schrodinger equation or wave equation is written as

\[\hat{H}\psi = E\psi \qquad \text{.......(4.12)}\]

Here \(\hat{H}\) is a mathematical operator called the Hamiltonian, \(\psi\) (psi) is the wave function and \(E\) the total energy of the system. An operator is simply an instruction to perform certain mathematical operations on whatever follows it; the Hamiltonian is the operator that corresponds to the total energy of the system.

Solving the Schrodinger equation is beyond the scope of this book. It may, however, be noted that the solution of the Schrodinger equation gives \(E\) and \(\psi\) — that is, it delivers the allowed energies of the electron and the mathematical functions that describe it, without either of them having to be assumed in advance. This is the deep difference from Bohr’s approach, where the allowed states had to be postulated.

4.7.1 Schrodinger Equation

When the Schrodinger equation is solved for the hydrogen atom, the possible values of energy (\(E\)) that the electron may have, along with the corresponding wave function (\(\psi\)), are obtained.

As a natural consequence of solving this equation, a set of three quantum numbers characteristic of the quantized energy levels and the corresponding wave functions are obtained. These are: principal quantum number (\(n\)), azimuthal quantum number (\(l\)) and magnetic quantum number (\(m_l\)).

The phrase natural consequence is important. Bohr had to insert his quantum number by hand through the angular-momentum postulate; here the whole numbers emerge from the mathematics of the equation itself, in the same way that only certain whole numbers of half-wavelengths can fit on a stretched string.

The solution of the Schrodinger wave equation led to three quantum numbers and successfully predicted the features of the hydrogen atom emission spectrum. Splitting of spectral lines in multi-electron atomic emission spectra could not be explained through such a model. These were explained by George Uhlenbeck and Samuel Goudsmit (1925) who proposed the presence of the fourth quantum number called the electron spin quantum number, \(m_s\).

What the Wave Function Means

The wave function, \(\psi\), as such does not have any physical meaning. It is not a picture of the electron and it cannot be measured. What can be measured is its square.

The probability of finding an electron at a point within an atom is proportional to \(\psi^{2}\) in the neighbourhood of that point (within a tiny volume element) around it.

NoteDefinition

The wave function \(\psi\) is the mathematical function obtained by solving the Schrodinger equation; its square, \(\psi^{2}\), at any point gives the probability of finding the electron in a tiny volume around that point.

Q. Why work with \(\psi^{2}\) rather than with \(\psi\) itself?

A. Because \(\psi\) can be positive or negative, and a negative probability is meaningless. Squaring removes the sign while keeping the size of the function, and it is the squared quantity that experiments actually measure. The sign of \(\psi\) is not wasted information, though — you will meet it again in Chapter 5, where the signs of overlapping wave functions decide whether a chemical bond forms.

Schrodinger published his wave equation in 1926, and within months it was shown to give exactly the same predictions as Heisenberg’s rather different-looking matrix approach — two independent routes to the same quantum mechanics. What the equation did not come with was an interpretation. Schrodinger himself initially thought of \(\psi\) as describing a smeared-out electron. It was Max Born, later the same year, who proposed that \(\psi^{2}\) should be read as a probability density, and this is the interpretation that survived. Schrodinger received the Nobel Prize in Physics in 1933, Born in 1954 — a long wait for an idea that had become standard textbook material decades earlier.

4.7.2 Atomic Orbitals and Quantum Numbers

Many wave functions are possible for an electron, and therefore, many atomic orbitals are present in an atom. Thus the wave functions or atomic orbitals form the basis of the quantum mechanical electronic structure of an atom.

Various orbitals in an atom differ in size, shape and orientation with respect to the nucleus depending upon the value of \(\psi^{2}\).

NoteDefinition

An atomic orbital is a wave function of an electron in an atom; the region around the nucleus where the probability of finding that electron is high is described by the square of this wave function.

Each orbital is designated by three quantum numbers labelled as \(n\), \(l\) and \(m_l\), and each electron is assigned four quantum numbers, viz, \(n\), \(l\), \(m_l\) and \(m_s\). Three numbers to describe the room, and a fourth to describe the electron sitting in it — that is the arrangement to keep in mind.

orbit vs orbital | Feature | Orbit (Bohr) | Orbital (quantum mechanical) | |—|—|—| | Meaning | A definite circular path of fixed radius | A region of space where the probability of finding the electron is high | | Basis | Bohr’s postulates | Solution of the Schrodinger equation | | Position of the electron | Assumed known exactly at every instant | Only a probability can be stated | | Consistent with the uncertainty principle? | No | Yes | | Shape | Always circular | Spherical, dumb-bell, double dumb-bell and others | | Maximum number of electrons | \(2n^{2}\) in the \(n\)th orbit | 2 in any one orbital |

Principal Quantum Number (\(n\))

The principal quantum number ‘\(n\)’ is a positive integer with values of \(n\) being 1, 2, 3, 4, ………….. It identifies the shell.

Atomic orbitals having the same value of ‘\(n\)’ belong to the same shell. With increase of ‘\(n\)’, the number of allowed orbitals in that shell increases and is given by ‘\(n^{2}\)’. A set of orbitals with a given value of ‘\(n\)’ constitutes a single shell. Shells are represented by symbols K, L, M, N,……… and so on, as shown in Table 4.6.

Table 4.6 : Allowed orbitals in the first four shells

Principal quantum number \(n\) Shell symbol Allowed number of orbitals \(n^{2}\) Size of shell
1 K 1 increases ↓
2 L 4
3 M 9
4 N 16

With an increase of ‘\(n\)’, the distance from the nucleus and the size of the shell increases, and also the energy increases (becoming lesser and lesser negative). Recall from Section 4.6.2 that bound states have negative energies, so “less negative” means “higher up”.

In hydrogen-like species the energy of the orbital depends only on the value of ‘\(n\)’. In the case of multi-electron atoms the energy of the orbital depends on two quantum numbers, ‘\(n\)’ and ‘\(l\)’ as well.

NoteDefinition

The principal quantum number \(n\) identifies the shell of an orbital, and mainly determines the size of the orbital and the energy of the electron occupying it.

Azimuthal Quantum Number (\(l\))

The azimuthal quantum number, \(l\), is also called the subsidiary quantum number.

Atomic orbitals with the same value of ‘\(n\)’ but different values of ‘\(l\)’ constitute a subshell belonging to the shell for the given ‘\(n\)’. The number of subshells in a shell is equal to ‘\(n\)’. Thus, the third shell contains three subshells (with three different values of ‘\(l\)’), the second shell contains two subshells and the first shell contains only one subshell.

The values of ‘\(l\)’ range from 0 to (\(n - 1\)). Thus, the K shell (with \(n = 1\)) contains only one subshell having \(l = 0\).

The subshells or sub-levels having ‘\(l\)’ equal to 0, 1, 2, 3, ….. are represented by the symbols s, p, d, f,…………., respectively.

NoteDefinition

The azimuthal quantum number \(l\) identifies the subshell to which an orbital belongs and decides the shape of the orbital; for a given \(n\) it may take the values \(0, 1, 2, \ldots (n-1)\).

The letters s, p, d and f look arbitrary, but they are the initials of words used by spectroscopists long before quantum numbers existed. Studying the emission lines of the alkali metals, they had sorted the observed series of lines into families they described as sharp, principal, diffuse and fundamental, according to the appearance of the lines. When the subshells were later identified, the old labels were simply carried over, in that order, and the alphabet was continued from f onwards for any further subshells. A student today is therefore using the shorthand of nineteenth-century spectroscopy every time an electronic configuration is written down.

Magnetic Orbital Quantum Number (\(m_l\))

The magnetic orbital quantum number, \(m_l\), gives information about the relative spatial orientation of the orbitals in a given subshell.

For any subshell (defined by the ‘\(l\)’ value) \((2l + 1)\) values of \(m_l\) are possible which range through :

\[m_l = -l,\ -(l-1),\ -(l-2),\ ..........,0,.....,(l-2),\ (l-1),\ l\]

Thus for the subshell ‘s’ with \(l = 0\), the only allowed value of \(m_l = 0\). In other words, the ‘s’ subshell has only one orbital in it. For the subshell ‘p’ having \(l = 1\), the allowed values of \(m_l\) are \(-1, 0, +1\). Thus the ‘p’ subshell contains three orbitals having distinct orientations, and so on.

Notice that \(m_l\) does not change the size or the shape of an orbital — only the direction in which it points. The three p orbitals are identical in every respect except that they lie along three mutually perpendicular directions.

NoteDefinition

The magnetic orbital quantum number \(m_l\) gives the orientation in space of an orbital within its subshell, and takes \((2l+1)\) values from \(-l\) to \(+l\) including zero.

The sum of orbitals in the constituent subshells gives the total number of orbitals in the concerned shell and is given by \(n^{2}\), as worked out in Table 4.7.

Table 4.7 : Distribution of orbitals in shells and subshells

Shell Principal quantum number \(n\) Total orbitals in the shell \(n^{2}\) Total number of subshells in a shell \(n\) Azimuthal quantum number \(l\) Number of orbitals in the subshell \(2l+1\) Sum of orbitals in all the subshells
K \(n = 1\) \(1^{2} = 1\) 1 \(l = 0\) \(2\times0+1 = 1\) 1
L \(n = 2\) \(2^{2} = 4\) 2 \(l = 0\)
\(l = 1\)
\(2\times0+1 = 1\)
\(2\times1+1 = 3\)
\(1 + 3 = 4\)
M \(n = 3\) \(3^{2} = 9\) 3 \(l = 0\)
\(l = 1\)
\(l = 2\)
\(2\times0+1 = 1\)
\(2\times1+1 = 3\)
\(2\times2+1 = 5\)
\(1 + 3 + 5 = 9\)
Electron Spin Quantum Number (\(m_s\))

The electron spin quantum number, \(m_s\), specifies the spin state of the electron in an orbital. An electron spins around its axis. This imparts spin angular momentum to the electron.

The two orientations which the spin angular momentum of an electron can take up give rise to the spin states which can be distinguished from each other by the spin quantum number, \(m_s\), which can be either \(+1/2\) or \(-1/2\). The two spin states are represented by two arrows, ↑ (pointing up) and ↓ (pointing down) and thus have opposite spins.

“An orbital can accommodate maximum two electrons and they must have opposite spins.”

This is known as the Pauli exclusion principle which will be dealt with in Section 4.7.5.

NoteDefinition

The electron spin quantum number \(m_s\) describes the spin state of an electron and can have only two values, \(+1/2\) and \(-1/2\).

the four quantum numbers | Quantum number | Symbol | Allowed values | What it tells us | Belongs to | |—|—|—|—|—| | Principal | \(n\) | 1, 2, 3, 4 … | Shell; size of the orbital and mainly its energy | the orbital | | Azimuthal (subsidiary) | \(l\) | 0 to \((n-1)\) | Subshell (s, p, d, f); shape of the orbital | the orbital | | Magnetic orbital | \(m_l\) | \(-l\) to \(+l\), that is \((2l+1)\) values | Orientation of the orbital in space | the orbital | | Electron spin | \(m_s\) | \(+1/2\) or \(-1/2\) | Spin state of the electron | the electron |

The Table 4.8 shows orbitals in the first four shells with the three quantum numbers for each orbital.

Table 4.8 : Orbital distribution in the first four shells

Symbol of Shell Value of Principal quantum number (\(n\)) Number of subshells Value of Azimuthal Quantum number \(l\) Symbol of subshell Total Number of orbitals in the subshell \(= 2l+1\) Values of the magnetic quantum number \(m_l\) for the subshell
K \(n = 1\) 1 \(l = 0\) 1s \(2\times0+1 = 1\) \(m_l = 0\)
L \(n = 2\) 2 \(l = 0\) 2s \(2\times0+1 = 1\) \(m_l = 0\)
\(l = 1\) 2p \(2\times1+1 = 3\) \(m_l = -1\), \(m_l = 0\), \(m_l = +1\)
M \(n = 3\) 3 \(l = 0\) 3s \(2\times0+1 = 1\) \(m_l = 0\)
\(l = 1\) 3p \(2\times1+1 = 3\) \(m_l = -1\), \(m_l = 0\), \(m_l = +1\)
\(l = 2\) 3d \(2\times2+1 = 5\) \(m_l = -2\), \(m_l = -1\), \(m_l = 0\), \(m_l = +1\), \(m_l = +2\)
N \(n = 4\) 4 \(l = 0\) 4s \(2\times0+1 = 1\) \(m_l = 0\)
\(l = 1\) 4p \(2\times1+1 = 3\) \(m_l = -1\), \(m_l = 0\), \(m_l = +1\)
\(l = 2\) 4d \(2\times2+1 = 5\) \(m_l = -2\), \(m_l = -1\), \(m_l = 0\), \(m_l = +1\), \(m_l = +2\)
\(l = 3\) 4f \(2\times3+1 = 7\) \(m_l = -3\), \(m_l = -2\), \(m_l = -1\), \(m_l = 0\), \(m_l = +1\), \(m_l = +2\), \(m_l = +3\)

How many orbitals make the N shell? What is the subshell wise distribution of orbitals in the N shell?

Solution: For the N shell the principal quantum number \(n = 4\) \(\therefore\) Total number of orbitals in N shell \(= n^{2} = 4^{2} = 16\). The total number of subshells in the N shell \(= n = 4\). The four subshells with their azimuthal quantum numbers and the constituent orbital number are as shown below

Azimuthal quantum number \(l\) | Symbol of subshell | number of orbitals \(2l + 1\) |

|—|—|—|

\(l = 0\) | s | \((2\times0)+1 = 1\) |
\(l = 1\) | p | \((2\times1)+1 = 3\) |
\(l = 2\) | d | \((2\times2)+1 = 5\) |
\(l = 3\) | f | \((2\times3)+1 = 7\) |

As a check, the four subshells contain \(1 + 3 + 5 + 7 = 16\) orbitals, which agrees with \(n^{2} = 16\).

For \(n = 3\), list all the permitted values of \(l\) and, for each of them, all the permitted values of \(m_l\). Hence state the total number of orbitals in this shell.

Solution: The azimuthal quantum number runs from 0 to \((n-1)\). For \(n = 3\), therefore, \(l = 0, 1, 2\).

For \(l = 0\) (3s subshell): \(m_l = 0\) — that is 1 orbital.

For \(l = 1\) (3p subshell): \(m_l = -1, 0, +1\) — that is 3 orbitals.

For \(l = 2\) (3d subshell): \(m_l = -2, -1, 0, +1, +2\) — that is 5 orbitals.

Total number of orbitals \(= 1 + 3 + 5 = 9\), which is \(n^{2} = 3^{2} = 9\), as expected.

(question only — no solution) State, with a reason, whether each of the following sets of quantum numbers (\(n\), \(l\), \(m_l\), \(m_s\)) is permitted: (a) \((2, 1, -1, +1/2)\), (b) \((3, 3, 0, -1/2)\), (c) \((4, 2, +3, +1/2)\), (d) \((1, 0, 0, +1)\).

4.7.3 Shapes of Atomic Orbitals

The probability of finding an electron at a given point in an atom is proportional to the square of the wave function \(\psi^{2}\) at that point. According to Max Born, \(\psi^{2}\) at a point in an atom is the probability density of the electron at that point.

Figure 4.7 (a) shows the probability density diagrams of 1s and 2s atomic orbitals. These diagrams appear like a cloud, denser where the electron is more likely to be found and thinner where it is less likely.

Diagram

Fig. 4.7 (a) – Probability density plots of the 1s and 2s orbitals drawn as clouds of dots, densest near the nucleus; the 2s cloud shown larger and with a clear blank ring within it. Label 1s, 2s and the node in 2s.

The electron cloud of the 2s orbital shows one node, which is a region with nearly zero probability density and displays the change of sign for its corresponding wavefunction.

NoteDefinition

A node is a region within an orbital where the probability density \(\psi^{2}\) is almost zero, and across which the wave function changes sign.

Boundary Surface Diagrams

Figure 4.7 (b) shows the boundary surface diagram of atomic orbitals 1s and 2s, which are spherical in shape.

Here, a boundary surface is drawn in space for an orbital such that the value of the probability density \(\psi^{2}\) is constant and encloses a region where the probability of finding the electron is typically more than 90%. Such a boundary surface diagram is a good representation of the shape of an orbital, and it is the representation used throughout the rest of this book.

Diagram

Fig. 4.7 (b) – Boundary surface diagrams of the 1s and 2s orbitals: two spheres, the 2s sphere noticeably larger than the 1s sphere. Label 1s and 2s.

Q. Why is the boundary surface drawn to enclose about 90% of the probability rather than all of it?

A. Because the value of \(\psi^{2}\) at any finite distance from the nucleus is never zero. Therefore a boundary surface enclosing 100% probability density — which occurs only at infinity — cannot be drawn. A surface enclosing about 90% is a practical compromise: it gives a definite shape and size while acknowledging that the electron cloud has no hard edge.

Q1. If an atom is mostly empty space, why does a gold ring feel completely solid and not hollow?

A. “Empty” refers only to the nucleus taking up a vanishingly small part of the atom’s volume. The rest of that volume is not truly empty — it is filled by the electron cloud, the region described by \(\psi^{2}\), and this cloud extends right out to the edge of the atom with no hard boundary. When your finger touches the ring, the electron clouds of the atoms in your skin meet the electron clouds of the gold atoms. Both are negatively charged, so they repel each other strongly and neither can pass through the other. What we call solidity is this electrical repulsion between electron clouds, not a collision between lumps of matter.

s Orbitals

The s orbitals are spherical in shape. Their size increases with increase of \(n\). It means that the electron is located farther away from the nucleus as the principal quantum number \(n\) increases.

Because a sphere looks the same from every direction, an s orbital has no preferred direction in space — consistent with the fact that \(l = 0\) allows only the single value \(m_l = 0\).

p Orbitals

Figure 4.8 displays the boundary surface diagram with the nucleus being at the origin, for the three 2p orbitals (\(n = 2\) and \(l = 1\)). It can be seen that each p orbital has two lobes on the two sides of a nodal plane passing through the nucleus. (A nodal plane has \(\psi^{2}\) very close to zero.)

Diagram

Fig. 4.8 – Shapes of the three 2p orbitals: three dumb-bell shaped orbitals drawn with x, y and z axes and the nucleus at the origin, the lobes lying along the x, y and z axes respectively; label 2pₓ, 2p_y, 2p_z and mark the nodal plane through the nucleus in each.

NoteDefinition

A nodal plane is a plane passing through the nucleus on which the probability density \(\psi^{2}\) is very close to zero, so the electron is essentially never found there.

The size and energy of the dumbell shaped three 2p orbitals are the same. Their orientations in space are, however, different. The lobes of the three 2p orbitals are along the \(x\), \(y\) and \(z\) axes. Accordingly the corresponding orbitals are designated as \(2p_x\), \(2p_y\) and \(2p_z\).

The size and energy of the orbitals in the p subshell increase with the increase of the principal quantum number. A 3p orbital therefore has the same shape as a 2p orbital but is larger and of higher energy.

d Orbitals

There are five orbitals associated with the d subshell. They are designated by \(d_{xy}\), \(d_{yz}\), \(d_{xz}\), \(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\). The shapes of the five 3d orbitals are shown in Fig. 4.9.

Diagram

Fig. 4.9 – Shapes of the five 3d orbitals with x, y and z axes marked: (a) d_xy, (b) d_xz, (c) d_yz — each with four lobes lying between the axes or in the stated planes; (d) d_x²−y² with four lobes along the x and y axes; (e) d_z² with two lobes along the z axis and a ring around the middle.

In spite of the difference in their shapes, the five d orbitals are equivalent in energy. The shapes of 4d, 5d, 6d…… orbitals are similar to those of 3d orbitals, but their respective size and energies are large, or they are said to be more diffused.

The shapes of the d orbitals are not a mathematical curiosity — they are the reason transition metals behave as they do. Because the five d orbitals point in different directions, the groups attached to a metal ion in a compound affect them unequally, splitting them into sets of slightly different energy. Electrons moving between these sets absorb visible light, which is why compounds of copper, chromium and cobalt are strongly coloured while compounds of sodium and calcium are not. The same partly filled d orbitals allow transition metals to grip and release small molecules, which is what makes iron, nickel and platinum such effective catalysts in industry. You will build on this in later chapters.

An atom has two electrons in its 4s orbital. Write the values of the four quantum numbers for each of them.

Solution: For the 4s orbital, 4 stands for the principal quantum number \(n\); s stands for the subshell s having the value of azimuthal quantum number \(l = 0\). In the ‘s’ subshell there is only one orbital and it has magnetic quantum number \(m_l = 0\). The two electrons in this orbital have opposite spins. Thus the four quantum numbers of the two electrons in the 4s orbital are :

| \(n\) | \(l\) | \(m_l\) | \(m_s\) |

|—|—|—|—|—|

electron 1 | 4 | 0 | 0 | \(+1/2\) |
electron 2 | 4 | 0 | 0 | \(-1/2\) |

The first three quantum numbers are identical because both electrons occupy the same orbital; only the fourth can distinguish them, and it does.

4.7.4 Energies of Orbitals

The energy of an electron in the hydrogen atom or hydrogen-like species is determined by the principal quantum number alone. This is because the only interaction in these species is the attraction between the electron and the nucleus. With a single electron present, there is no other electron to repel it, so nothing else can influence its energy.

An increasing order of energies of orbitals in the hydrogen atom is given by

\[1\mathrm{s} < 2\mathrm{s} = 2\mathrm{p} < 3\mathrm{s} = 3\mathrm{p} = 3\mathrm{d} < 4\mathrm{s} = 4\mathrm{p} = 4\mathrm{d} = 4\mathrm{f} < ..........\]

Of course the shapes of the concerned s, p, d, f orbitals are different, as described earlier. The orbitals with the same energy, the corresponding wave functions being different, are called degenerate orbitals.

NoteDefinition

Degenerate orbitals are orbitals that have different wave functions but exactly the same energy.

Thus, in the hydrogen atom 2s and 2p are degenerate orbitals.

Why Multi-Electron Atoms Are Different

In multi-electron atoms there is mutual repulsion among the electrons. An electron in such an atom is therefore not simply pulled by the nucleus; it is also pushed by every other electron, and how strongly it feels the nucleus depends on how well the other electrons come between it and the nucleus.

The energy of an electron in a multi-electron atom, therefore, depends both on the principal quantum number, \(n\), and the azimuthal quantum number, \(l\).

The lower the sum \((n + l)\) for an orbital, the lower is its energy. If two orbitals have the same \((n + l)\) values then the orbital with the lower value of \(n\) is of lower energy. This is called the \((n + l)\) rule.

NotePrinciple / Law

(The \((n+l)\) rule) In a multi-electron atom, the orbital with the lower value of \((n + l)\) has the lower energy; if two orbitals have the same \((n + l)\) value, the one with the lower \(n\) has the lower energy.

From the \((n + l)\) rule the increasing order of energy of orbitals in multi-electron atoms can be written as:

\[1\mathrm{s} < 2\mathrm{s} < 2\mathrm{p} < 3\mathrm{s} < 3\mathrm{p} < 4\mathrm{s} < 3\mathrm{d} < 4\mathrm{p} < 5\mathrm{s} < 4\mathrm{d} < 5\mathrm{p} < 6\mathrm{s} < 4\mathrm{f} < 5\mathrm{d} < 6\mathrm{p} < 7\mathrm{s}............\]

The worked \((n + l)\) values are set out in Table 4.9.

Table 4.9 : Dependance of orbital energy on \((n + l)\) value

Orbital Energy Principal quantum number \(n\) Azimuthal quantum number \(l\) \((n + l)\)
1s \(n = 1\) \(l = 0\) \(1 + 0 = 1\)
2s \(n = 2\) \(l = 0\) \(2 + 0 = 2\)
2p \(n = 2\) \(l = 1\) \(2 + 1 = 3\); \(n = 2\) (lower)
3s \(n = 3\) \(l = 0\) \(3 + 0 = 3\); \(n = 3\) (higher)
3p \(n = 3\) \(l = 1\) \(3 + 1 = 4\); \(n = 3\) (lower)
4s \(n = 4\) \(l = 0\) \(4 + 0 = 4\); \(n = 4\) (higher)
3d \(n = 3\) \(l = 2\) \(3 + 2 = 5\); \(n = 3\) (lower)
4p \(n = 4\) \(l = 1\) \(4 + 1 = 5\); \(n = 4\) (higher)

(Orbital energy increases down the table.)

Q. Why does the 4s orbital fill before the 3d orbital, when 3 is smaller than 4?

A. Because in a multi-electron atom it is \((n + l)\), not \(n\), that decides the order. For 4s, \((n + l) = 4 + 0 = 4\); for 3d, \((n + l) = 3 + 2 = 5\). The smaller sum wins, so 4s lies lower in energy and is filled first. Note that the rule applies to multi-electron atoms; in hydrogen, where there is no electron–electron repulsion, 3d actually lies below 4s.

In a multi-electron atom, electrons occupy different orbitals. The lowest total electronic energy corresponds to the most stable, that is, the ground state of an atom. The orbital wise distribution of electrons in the ground state can be understood from what is called the aufbau principle.

(question only — no solution) Using the \((n + l)\) rule, arrange the following orbitals in increasing order of energy in a multi-electron atom, showing the \((n + l)\) value for each: 5s, 4d, 4f, 6s, 5p.

4.7.5 Aufbau Principle

‘Aufbau’ is a German word meaning ‘building up’. The building up of an orbital means filling up of orbitals with electrons in the ground state of an atom.

The aufbau principle is based on: (i) increasing order of energies of orbitals, (ii) Pauli’s exclusion principle, and (iii) Hund’s rule of maximum multiplicity.

NotePrinciple / Law

(Aufbau principle) In the ground state of an atom, electrons are filled into the available orbitals one by one in order of increasing energy, subject to the Pauli exclusion principle and Hund’s rule of maximum multiplicity.

These three ingredients answer three different questions: which orbital comes next, how many electrons it can hold, and how electrons distribute themselves among orbitals of equal energy. We take them in turn.

The Increasing Order of Energies of Orbitals

As seen in Section 4.7.4, the increasing order of energies of orbitals is decided by the \((n + l)\) value. Electrons in the ground state atom are filled in the orbitals in an increasing order of energy.

Fig. 4.10 shows a useful method to remember this increasing order of orbital energy. The subshells are written out in rows and read along a set of parallel diagonal arrows; following the arrows from the top gives exactly the sequence obtained from the \((n + l)\) rule, without having to compute the sums each time.

Diagram

Fig. 4.10 – Increasing order of orbital energy: rows listing 1s / 2s 2p / 3s 3p 3d / 4s 4p 4d 4f / 5s 5p 5d 5f / 6s 6p 6d / 7s 7p, with parallel diagonal arrows drawn through them from upper right to lower left; the reading order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p … indicated.

Fig. 4.10, the diagonal filling chart ((a) Follow the arrows and write down the first twelve subshells in the order in which they fill. (b) Which subshell does the chart place immediately after 4p, and does this agree with the \((n + l)\) rule? Show the calculation. (c) Explain why the chart places 4f after 6s even though 4 is smaller than 6.)

Pauli Exclusion Principle

The capacity of an orbital to accommodate electrons is decided by the Pauli exclusion principle. Wolfgang Pauli (1926) recognized that “No two electrons in an atom can have the same set of four quantum numbers.”

Another way to state this Pauli exclusion principle is: “Only two electrons can occupy the same orbital and they must have opposite spins.”

NotePrinciple / Law

(Pauli exclusion principle) No two electrons in an atom can have the same set of all four quantum numbers; equivalently, an orbital can hold at most two electrons, and these must have opposite spins.

The Pauli exclusion principle implies that for electrons belonging to the same orbital, the spin quantum number ‘\(m_s\)’ must be different since the other three quantum numbers \(n\), \(l\) and \(m_l\) are the same. There are only two values that ‘\(m_s\)’ can take, which are \(+1/2\) and \(-1/2\).

An orbital thus can accommodate only two electrons with opposite spins, so that the fourth quantum number is different for the two occupying electrons. These two electrons with opposite spins occupying the same orbital are called an electron pair.

This principle is illustrated with the helium atom, He (\(Z = 2\)). Its electronic configuration is \(1\mathrm{s}^{2}\), drawn as ↑↓ in a single box. Both electrons are in the 1s orbital. The two non-identical combinations of the four quantum numbers of both electrons in helium are given in Table 4.10.

Table 4.10

Electron \(n\) \(l\) \(m_l\) \(m_s\) Set of values of four quantum numbers
1st electron 1 0 0 \(+1/2\) \((1, 0, 0, +1/2)\)
2nd electron 1 0 0 \(-1/2\) \((1, 0, 0, -1/2)\)

It implies that two electrons in the same atom always have a different set of quantum numbers; that means, the set of (\(n\), \(l\), \(m_l\)) is the same and the \(m_s\) is different.

Since an orbital can accommodate up to two electrons only, the capacity of a shell with the principal quantum number \(n\), to accommodate electrons is given by \(2n^{2}\). The reasoning is short: a shell has \(n^{2}\) orbitals, and each orbital holds two electrons.

Hund’s Rule of Maximum Multiplicity

Filling of electrons in the orbitals belonging to the same subshell (orbitals of equal energy or degenerate orbitals) follows Hund’s rule of maximum multiplicity. As per this rule “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”

NotePrinciple / Law

(Hund’s rule of maximum multiplicity) Electrons entering a set of degenerate orbitals occupy them singly, with parallel spins, before any orbital receives a second electron.

The physical reason is repulsion. Two electrons crowded into one orbital repel each other more strongly than two electrons kept in separate orbitals of the same energy, so the atom keeps them apart for as long as it can.

Consider, for example, the filling of the p subshell. The p subshell has three degenerate orbitals. Here pairing of electrons starts when the fourth electron enters the p subshell. The electronic configuration of four electrons occupying p-orbitals will then be ↑↓ | ↑ | ↑ and not ↑↓ | ↑↓ | (empty).

It is observed that a half-filled and fully filled set of degenerate orbitals has extra stability. Keep this statement in mind — it is the key to the two anomalous configurations discussed shortly.

Hund’s rule has a directly observable consequence: unpaired electrons make a substance magnetic. A species with one or more unpaired electrons is drawn into a magnetic field and is called paramagnetic, while one in which all electrons are paired is weakly pushed out and is called diamagnetic. Liquid oxygen is the classic demonstration — it clings between the poles of a magnet, because the oxygen molecule carries unpaired electrons. Chemists use exactly this behaviour in reverse, measuring how strongly a compound is attracted to a magnetic field in order to count the unpaired electrons in a metal ion whose structure they are trying to establish.

Flowchart

FC 4.5 – Filling electrons into orbitals: total electrons known → pick the next orbital by the (n + l) rule (use the Fig. 4.10 diagonal chart) → is the subshell a set of degenerate orbitals? if yes → place one electron in each orbital with parallel spins (Hund’s rule) before pairing → each orbital takes a maximum of two electrons with opposite spins (Pauli) → subshell full? if yes → move to the next orbital in the energy order; if electrons remain, repeat → all electrons placed → ground-state configuration obtained

4.7.6 Electronic Configuration of Atoms and Its Representation

The electronic configuration of an atom is the distribution of its electrons in orbitals. The electronic configuration can be written by applying the aufbau principle.

NoteDefinition

The electronic configuration of an atom is the arrangement of all its electrons among the available orbitals in the ground state.

There are two methods of representing electronic configuration:

  1. Orbital notation: \(n\mathrm{s}^{a}\,n\mathrm{p}^{b}\,n\mathrm{d}^{c}..........\)

  2. Orbital diagram: a row of boxes, one box per orbital

In the orbital notation method, a shell is represented by the principal quantum number followed by the respective symbol of the subshell, and the number of electrons occupying that subshell is written as a superscript on the right side of the symbol.

In the orbital diagram method, each orbital in a subshell is represented by a box and the electron is represented by an arrow (↑ for up spin and ↓ for low spin) placed in the respective boxes. In this second method all the four quantum numbers of the electron are accounted for — the box tells us \(n\), \(l\) and \(m_l\), and the direction of the arrow tells us \(m_s\).

Electronic configuration of a few elements is illustrated in Table 4.11.

Table 4.11 : Representation of electronic configuration

Element symbol Orbital notation Orbital diagram
\(_{1}\mathrm{H}\) \(1\mathrm{s}^{1}\) 1s: ↑
\(_{2}\mathrm{He}\) \(1\mathrm{s}^{2}\) 1s: ↑↓
\(_{3}\mathrm{Li}\) \(1\mathrm{s}^{2}2\mathrm{s}^{1}\) 1s: ↑↓   2s: ↑
\(_{4}\mathrm{Be}\) \(1\mathrm{s}^{2}2\mathrm{s}^{2}\) 1s: ↑↓   2s: ↑↓
\(_{9}\mathrm{F}\) \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{5}\) 1s: ↑↓   2s: ↑↓   2p: ↑↓ | ↑↓ | ↑

Notice fluorine’s 2p arrangement. Two of the three 2p orbitals are paired and one holds a single electron, exactly as Hund’s rule requires once the fourth, fifth and sixth p electrons have entered.

Write the electronic configuration of \(_{18}\mathrm{Ar}\) and \(_{19}\mathrm{K}\) using the orbital notation and orbital diagram method.

Solution: From the atomic numbers it is understood that 18 electrons are to be filled in the Ar atom and 19 electrons are to be filled in the K atom. These are to be filled in the orbitals according to the aufbau principle. The electronic configuration of these atoms can be represented as:

| Orbital notation | Orbital diagram |

|—|—|—|

\(_{18}\mathrm{Ar}\) | \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}\,3\mathrm{s}^{2}3\mathrm{p}^{6}\) | 1s: ↑↓   2s: ↑↓   2p: ↑↓ | ↑↓ | ↑↓   3s: ↑↓   3p: ↑↓ | ↑↓ | ↑↓ |
\(_{19}\mathrm{K}\) | \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}\,3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{1}\) | 1s: ↑↓   2s: ↑↓   2p: ↑↓ | ↑↓ | ↑↓   3s: ↑↓   3p: ↑↓ | ↑↓ | ↑↓   4s: ↑ |

The nineteenth electron of potassium goes into 4s and not into 3d, because \((n + l)\) is 4 for 4s and 5 for 3d.

Condensed Orbital Notation

The orbital notation of the electronic configuration of an element with a high atomic number comprises a long train of symbols of orbitals with an increasing order of energy. It can be condensed by dividing it into two parts.

The electronic configuration of the preceding inert gas is a part of the electronic configuration of any element. In the condensed orbital notation it is implied by writing the symbol of that inert gas in a square bracket. It is the core part of the electronic configuration of that element. The outer configuration is specific to a particular element and written immediately after the bracket.

For example, the orbital notation of potassium, K (\(Z = 19\)) is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{1}\). Its core part is the electronic configuration of the preceding inert gas argon, ‘Ar : \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}\)’, while ‘\(4\mathrm{s}^{1}\)’ is an outer part.

Therefore the condensed orbital notation of the electronic configuration of potassium is ‘K : \([\mathrm{Ar}]\,4\mathrm{s}^{1}\).’

This shorthand is not merely a saving of ink. It puts the chemically active outer electrons in plain view, and it is those electrons that determine how an element reacts.

Table 4.12 displays the detailed and condensed orbital notations of the electronic configuration of various elements with atomic numbers from 1 to 30.

Table 4.12 Electronic configuration of the first thirty elements

Atomic Number Element K (\(n = 1\)) L (\(n = 2\)) M (\(n = 3\)) N (\(n = 4\)) Condensed notation of electronic configuration
1 Hydrogen \(1\mathrm{s}^{1}\) \(1\mathrm{s}^{1}\)
2 Helium \(1\mathrm{s}^{2}\) \(1\mathrm{s}^{2}\)
3 Lithium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{1}\) \([\mathrm{He}]\,2\mathrm{s}^{1}\)
4 Beryllium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}\) \([\mathrm{He}]\,2\mathrm{s}^{2}\)
5 Boron \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{1}\) \([\mathrm{He}]\,2\mathrm{s}^{2}2\mathrm{p}^{1}\)
6 Carbon \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{2}\) \([\mathrm{He}]\,2\mathrm{s}^{2}2\mathrm{p}^{2}\)
7 Nitrogen \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{3}\) \([\mathrm{He}]\,2\mathrm{s}^{2}2\mathrm{p}^{3}\)
8 Oxygen \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{4}\) \([\mathrm{He}]\,2\mathrm{s}^{2}2\mathrm{p}^{4}\)
9 Fluorine \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{5}\) \([\mathrm{He}]\,2\mathrm{s}^{2}2\mathrm{p}^{5}\)
10 Neon \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \([\mathrm{He}]\,2\mathrm{s}^{2}2\mathrm{p}^{6}\)
11 Sodium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{1}\) \([\mathrm{Ne}]\,3\mathrm{s}^{1}\)
12 Magnesium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}\)
13 Aluminium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{1}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}3\mathrm{p}^{1}\)
14 Silicon \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{2}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}3\mathrm{p}^{2}\)
15 Phosphorous \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{3}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}3\mathrm{p}^{3}\)
16 Sulfur \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{4}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}3\mathrm{p}^{4}\)
17 Chlorine \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{5}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}3\mathrm{p}^{5}\)
18 Argon \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}\) \([\mathrm{Ne}]\,3\mathrm{s}^{2}3\mathrm{p}^{6}\)
19 Potassium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}\) \(4\mathrm{s}^{1}\) \([\mathrm{Ar}]\,4\mathrm{s}^{1}\)
20 Calcium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}\)
21 Scandium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{1}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{1}\)
22 Titanium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{2}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{2}\)
23 Vanadium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{3}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{3}\)
24 Chromium \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{5}\) \(4\mathrm{s}^{1}\) \([\mathrm{Ar}]\,4\mathrm{s}^{1}3\mathrm{d}^{5}\)
25 Manganese \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{5}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{5}\)
26 Iron \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{6}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{6}\)
27 Cobalt \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{7}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{7}\)
28 Nickel \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{8}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{8}\)
29 Copper \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{10}\) \(4\mathrm{s}^{1}\) \([\mathrm{Ar}]\,4\mathrm{s}^{1}3\mathrm{d}^{10}\)
30 Zinc \(1\mathrm{s}^{2}\) \(2\mathrm{s}^{2}2\mathrm{p}^{6}\) \(3\mathrm{s}^{2}\,3\mathrm{p}^{6}3\mathrm{d}^{10}\) \(4\mathrm{s}^{2}\) \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{10}\)

Flowchart

FC 4.6 – Writing an electronic configuration: read Z from the periodic table → Z = number of electrons in the neutral atom → fill orbitals in the (n + l) order using Fig. 4.10 → apply Pauli (max 2 per orbital, opposite spins) and Hund (singly first in degenerate orbitals) → check for the half-filled/fully filled exception at Cr and Cu → write the full orbital notation → replace the preceding inert-gas core by its symbol in square brackets to get the condensed notation. Branch: for an ion, remove electrons from the outermost shell (highest n) first, or add electrons following the same filling order.

Electronic Configurations of Cu and Cr

Chromium : The atomic number of chromium is 24. The expected electronic configuration is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{2}3\mathrm{d}^{4}\); in that case 3d is not half-filled. Hence, it has less stability.

Interelectronic repulsion makes one 4s electron enter into one of the empty 3d orbitals, thereby both 4s and 3d orbitals become half-filled so that the chromium atom acquires extra stability. Its electronic configuration is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{1}3\mathrm{d}^{5}\).

Copper : The atomic number of copper is 29. The expected electronic configuration is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{2}3\mathrm{d}^{9}\). Here the 3d orbital is neither half-filled nor fully filled. Due to interelectronic repulsions, one 4s electron is forced to enter into 3d, which makes it completely filled with 4s being half-filled. Hence the copper atom acquires extra stability. Now the electronic configuration of Cu is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{1}3\mathrm{d}^{10}\).

The two cases share one logic. The 4s and 3d subshells lie very close in energy, so the small cost of promoting one 4s electron is more than repaid by the extra stability of a half-filled or fully filled d subshell.

Q5. Electrons fill orbitals in a neat, predictable order for element after element — so why do chromium and copper break that order?

A. Because the aufbau order is a guide to the lowest total energy, not a rule that outranks it. In chromium and copper the 4s and 3d subshells are extremely close in energy, so shifting one electron from 4s to 3d costs very little. In return, chromium gains a half-filled 3d⁵ set alongside a half-filled 4s¹, and copper gains a completely filled 3d¹⁰ alongside a half-filled 4s¹ — and half-filled and fully filled sets of degenerate orbitals possess extra stability. The atom therefore settles into \(4\mathrm{s}^{1}3\mathrm{d}^{5}\) and \(4\mathrm{s}^{1}3\mathrm{d}^{10}\) rather than the expected \(4\mathrm{s}^{2}3\mathrm{d}^{4}\) and \(4\mathrm{s}^{2}3\mathrm{d}^{9}\). The exception, in other words, obeys the deeper rule.

Write the electronic configuration of manganese (\(Z = 25\)) in the condensed orbital notation, and state the number of unpaired electrons it contains.

Solution: Manganese has 25 electrons. Filling in the \((n + l)\) order, the first 18 electrons complete the argon core, the next two enter 4s and the remaining five enter 3d:

\[\mathrm{Mn} : [\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{5}\]

For the count of unpaired electrons, the 4s orbital holds a pair, so it contributes none. The 3d subshell has five degenerate orbitals and exactly five electrons, so by Hund’s rule each orbital takes one electron with parallel spin and no pairing occurs at all:

3d: ↑ | ↑ | ↑ | ↑ | ↑

Number of unpaired electrons = 5. Note that manganese is not an exception to the aufbau order; it simply happens to reach a half-filled 3d subshell in the normal course of filling.

Isoelectronic Species

Atoms and ions having the same number of electrons are isoelectronic. The electronic configuration of isoelectronic species is the same.

NoteDefinition

Isoelectronic species are atoms or ions that contain the same number of electrons and therefore have identical electronic configurations.

Consider \(\mathrm{K^{\oplus}}\) formed by removal of one electron from a K atom, which has 19 electrons (\(Z = 19\)). Therefore \(\mathrm{K^{\oplus}}\) has 18 electrons.

\[\mathrm{K}\ (Z = 19) \longrightarrow \mathrm{K^{\oplus}} + \mathrm{e^{\ominus}}\]

Number of electrons : 19 → 18

Species such as Ar and \(\mathrm{Ca^{2\oplus}}\) containing 18 electrons are isoelectronic with \(\mathrm{K^{\oplus}}\). The electronic configuration of all these species with 18 electrons is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}\).

Take care with the counting. Being isoelectronic is about electrons only — these three species have 18 electrons each but 18, 19 and 20 protons respectively, so they are certainly not identical in every respect.

Find out one dinegative anion and one unipositive cation which are isoelectronic with the Ne atom. Write their electronic configuration using orbital notations and the orbital diagram method.

Solution: Ne has \(Z = 10\). Therefore Ne and its isoelectronic species contain 10 electrons each.

The dinegative anionic species isoelectronic with Ne is obtained by adding two electrons to the atom with \(Z = 8\). This is \(\mathrm{O^{2\ominus}}\). The unipositive cationic species isoelectronic with Ne is obtained by removing one electron from an atom of \(Z = 11\). It is \(\mathrm{Na^{\oplus}}\). These species and their electronic configuration are shown below :

| Number of electrons | Orbital Notation | Orbital diagram |

|—|—|—|—|

Ne | 10 | \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}\) | 1s: ↑↓   2s: ↑↓   2p: ↑↓ | ↑↓ | ↑↓ |
\(\mathrm{O} + 2\mathrm{e^{\ominus}} \longrightarrow \mathrm{O^{2\ominus}}\) | 10 | \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}\) | 1s: ↑↓   2s: ↑↓   2p: ↑↓ | ↑↓ | ↑↓ |
\(\mathrm{Na} - \mathrm{e^{\ominus}} \longrightarrow \mathrm{Na^{\oplus}}\) | 10 | \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}\) | 1s: ↑↓   2s: ↑↓   2p: ↑↓ | ↑↓ | ↑↓ |

Working backwards from the charge is the key step here. A dinegative ion has two electrons more than its neutral atom, so the parent atom must have \(Z = 10 - 2 = 8\); a unipositive ion has one electron fewer, so its parent must have \(Z = 10 + 1 = 11\).

The configuration \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}\) shared by Ar, \(\mathrm{K^{\oplus}}\) and \(\mathrm{Ca^{2\oplus}}\) is a completely filled outer shell, and such an arrangement is remarkably unreactive. This is why argon is pumped into the space around a weld to shield the hot metal from oxygen, why it fills some electric lamps to stop the filament burning away, and why it is used as an inert atmosphere for handling reactive substances in the laboratory. An idea that began as a way of bookkeeping electrons turns out to explain a choice made on factory floors.

(question only — no solution) Write the condensed electronic configuration of (a) \(\mathrm{Fe}\) (\(Z = 26\)), (b) \(\mathrm{Fe^{2+}}\) and (c) \(\mathrm{Fe^{3+}}\), and state the number of unpaired electrons in each. Which of the three would you expect to be most strongly attracted to a magnet?

(question only — no solution) Identify the ion of a Group 16 element and the ion of a Group 1 element that are isoelectronic with \(\mathrm{Ar}\), and write the electronic configuration common to them and to argon.

4.8 Summary

  1. Dalton’s picture of the atom as a hard, indivisible sphere explained the laws of chemical combination but could not explain effects such as the electricity generated when glass or ebonite is rubbed. The discovery of subatomic particles showed the atom to be divisible.

  2. Three subatomic particles concern chemistry. The electron carries a charge of \(-1.6022\times10^{-19}\ \text{C}\) and has a mass of \(9.10938\times10^{-31}\ \text{kg}\); the proton carries an equal positive charge, \(+1.6022\times10^{-19}\ \text{C}\), with a mass of \(1.6726\times10^{-27}\ \text{kg}\); the neutron is uncharged, with a mass of \(1.67493\times10^{-27}\ \text{kg}\). Protons and neutrons occupy the nucleus and are together called nucleons; electrons occupy the extranuclear part.

  3. J. J. Thomson (1897) showed from the cathode ray tube experiment that cathode rays are a stream of negatively charged particles 1837 times lighter than a hydrogen atom and present in all atoms — later named electrons.

  4. Ernest Rutherford (1911) found that a few α-particles bounce back from thin gold foil, which established a massive, positively charged nucleus. In 1919 he transmuted nitrogen into oxygen with liberation of hydrogen, \(^{14}_{7}\mathrm{N} + {}^{4}_{2}\alpha \rightarrow {}^{17}_{8}\mathrm{O} + {}^{1}_{1}\mathrm{H}\), and since hydrogen was always emitted, the hydrogen nucleus was renamed the proton.

  5. James Chadwick (1932) identified the neutron, predicted by Rutherford in 1920, from the protons knocked out of paraffin by the neutral radiation from beryllium.

  6. The atomic number \(Z\) is the number of protons, and equals the number of electrons in a neutral atom; it fixes the chemical identity of the element. The mass number \(A\) is the total number of nucleons, so \(A = Z + N\) and \(N = A - Z\). A composition written as \(^{A}_{Z}\mathrm{X}\) is called a nuclide.

  7. Isotopes have the same \(Z\) but different \(A\); isobars have the same \(A\) but different \(Z\); isotones have the same \(N\). Isotopes of an element occupy the same position in the periodic table and have similar chemical properties, and their fixed natural abundances give the element a weighted average atomic mass — as for chlorine, \((3\times35 + 1\times37)/4 = 35.5\).

  8. Rutherford’s model has two drawbacks. An orbiting electron is in accelerated motion and, by Maxwell’s theory, must radiate energy, spiral inwards and collapse into the nucleus, whereas real atoms are stable. The model also says nothing about how electrons are distributed or what energies they have.

  9. Electromagnetic radiation shows wave–particle duality: it behaves as a wave while propagating (diffraction, interference) and as a stream of photons when it interacts with matter (black-body radiation, photoelectric effect).

  10. A wave is described by wavelength \(\lambda\) (SI unit m), frequency \(\nu\) (Hz), wavenumber \(\bar{\nu} = 1/\lambda\) (commonly \(\text{cm}^{-1}\)) and amplitude, whose square gives the intensity. In vacuum all electromagnetic radiation travels at \(c = 3.0\times10^{8}\ \text{m s}^{-1}\), and \(c = \nu\lambda\).

  11. Max Planck (1900) proposed that radiation is emitted or absorbed only in packets called quanta, of energy \(E = h\nu\), with \(h = 6.626\times10^{-34}\ \text{J s}\). Einstein (1905) used this to explain the photoelectric effect, treating radiation as photons of energy \(h\nu\) having zero rest mass.

  12. Excited atoms emit radiation, and the record of this radiation is an emission spectrum. Hydrogen gives the simplest emission spectrum, consisting of sharp lines rather than a continuous band. Balmer (1885) fitted the visible lines, and Rydberg generalised all five series as \(\bar{\nu} = 109677\left[\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right]\ \text{cm}^{-1}\), with \(R_H = 109677\ \text{cm}^{-1}\). The five series are Lyman (\(n_1 = 1\), ultraviolet), Balmer (\(n_1 = 2\), visible), Paschen (\(n_1 = 3\)), Bracket (\(n_1 = 4\)) and Pfund (\(n_1 = 5\)), the last three in the infrared.

  13. Bohr (1913) proposed that the electron moves only in certain allowed circular orbits called stationary states, that it neither gains nor loses energy while in such a state, that the frequency of radiation for a transition is \(\nu = \Delta E/h\) (Bohr’s frequency rule), and that the angular momentum is quantized as \(m\mathrm{v}r = n\times\dfrac{h}{2\pi}\).

  14. Bohr’s results for hydrogen: the states are numbered by the principal quantum number \(n\); the radii are \(r_n = n^{2}a_0\) with \(a_0 = 52.9\ \text{pm}\); and the energies are \(E = -R_H\left(\dfrac{1}{n^{2}}\right)\) with \(R_H = 2.18\times10^{-18}\ \text{J}\), so that \(E_1 = -2.18\times10^{-18}\ \text{J}\) and \(E_2 = -0.545\times10^{-18}\ \text{J}\). The negative sign shows that a bound electron has less energy than a free electron at rest, for which \(E_\infty = 0\).

  15. For hydrogen-like species, which have only one extranuclear electron (such as \(\mathrm{He^{\oplus}}\), \(\mathrm{Li^{2\oplus}}\), \(\mathrm{Be^{3\oplus}}\)), \(E_n = -2.18\times10^{-18}\left(\dfrac{Z^{2}}{n^{2}}\right)\ \text{J}\) and \(r_n = \dfrac{52.9(n^{2})}{Z}\ \text{pm}\). Energy becomes more negative and radius smaller as \(Z\) increases; electron velocity increases with \(Z\) and decreases with \(n\).

  16. Combining \(\Delta E = E_i - E_f\) with the Bohr energy expression gives \(\Delta E = 2.18\times10^{-18}\left[\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right]\ \text{J}\), which converts to \(\bar{\nu} = 109677\left[\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right]\ \text{cm}^{-1}\). Bohr’s theory thus derived the empirical Rydberg equation and gave physical meaning to its integers as principal quantum numbers. Each series arises from transitions ending at a particular lower orbit.

  17. Bohr’s model failed to account for the finer details of the hydrogen spectrum, for the spectra of atoms other than hydrogen, for the splitting of lines in a magnetic field (Zeeman effect) or an electric field (Stark effect), and for the formation of chemical bonds.

  18. Two later developments explained the failure. De Broglie (1924) proposed that matter also has dual behaviour, with \(\lambda = \dfrac{h}{m\mathrm{v}} = \dfrac{h}{p}\), confirmed by diffraction experiments. Heisenberg (1927) stated that the exact position and exact momentum of an electron cannot be determined simultaneously, \(\Delta x \times \Delta p_x \geq \dfrac{h}{4\pi}\), so only the probability of finding an electron at a point can be stated — which makes a definite orbit meaningless.

  19. Quantum mechanics was developed in 1926 by Heisenberg and Schrodinger. The Schrodinger wave equation is \(\hat{H}\psi = E\psi\), where \(\hat{H}\) is the Hamiltonian operator, \(\psi\) the wave function and \(E\) the total energy. Its solution yields the allowed energies together with three quantum numbers \(n\), \(l\) and \(m_l\). The wave function itself has no physical meaning; according to Max Born, \(\psi^{2}\) gives the probability density of the electron at a point.

  20. Uhlenbeck and Goudsmit (1925) added the fourth quantum number, the electron spin quantum number \(m_s\), which takes the values \(+1/2\) and \(-1/2\). An orbital is designated by \(n\), \(l\) and \(m_l\), while an electron is described by all four.

  21. The principal quantum number \(n\) (values 1, 2, 3 …) identifies the shell K, L, M, N …, which contains \(n^{2}\) orbitals and can hold \(2n^{2}\) electrons. The azimuthal quantum number \(l\) (values 0 to \(n-1\)) identifies the subshell s, p, d, f and decides the shape; a shell has \(n\) subshells. The magnetic quantum number \(m_l\) takes \((2l+1)\) values from \(-l\) to \(+l\) and gives the orientation of an orbital.

  22. s orbitals are spherical and grow larger with increasing \(n\); the 2s orbital has one node, a region of nearly zero probability density. The three 2p orbitals are dumb-bell shaped with two lobes separated by a nodal plane through the nucleus, are equal in size and energy, and lie along the \(x\), \(y\) and \(z\) axes. The five d orbitals — \(d_{xy}\), \(d_{yz}\), \(d_{xz}\), \(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\) — differ in shape but are equal in energy. A boundary surface diagram encloses about 90% of the probability, since a surface enclosing 100% cannot be drawn.

  23. In hydrogen and hydrogen-like species, orbital energy depends on \(n\) alone, so 2s and 2p are degenerate. In multi-electron atoms, electron–electron repulsion makes the energy depend on both \(n\) and \(l\): the lower the \((n + l)\) value the lower the energy, and for equal \((n + l)\) the orbital with lower \(n\) lies lower. This gives the order \(1\mathrm{s} < 2\mathrm{s} < 2\mathrm{p} < 3\mathrm{s} < 3\mathrm{p} < 4\mathrm{s} < 3\mathrm{d} < 4\mathrm{p} < 5\mathrm{s} < 4\mathrm{d} < 5\mathrm{p} < 6\mathrm{s} < 4\mathrm{f} < 5\mathrm{d} < 6\mathrm{p} < 7\mathrm{s}\).

  24. The aufbau principle fills orbitals in increasing order of energy, subject to the Pauli exclusion principle — no two electrons in an atom may have the same set of four quantum numbers, so an orbital holds at most two electrons of opposite spin — and Hund’s rule of maximum multiplicity, by which degenerate orbitals are singly occupied before any pairing begins.

  25. Electronic configurations are written either in orbital notation (\(n\mathrm{s}^{a}n\mathrm{p}^{b}n\mathrm{d}^{c}\)) or as orbital diagrams of boxes and arrows, the latter accounting for all four quantum numbers. The condensed notation replaces the preceding inert-gas core by its symbol in square brackets, as in \(\mathrm{K}:[\mathrm{Ar}]4\mathrm{s}^{1}\).

  26. Chromium and copper are exceptions, because half-filled and fully filled sets of degenerate orbitals have extra stability. Chromium is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{1}3\mathrm{d}^{5}\) and copper is \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}4\mathrm{s}^{1}3\mathrm{d}^{10}\).

  27. Atoms and ions with the same number of electrons are isoelectronic and share the same electronic configuration. Thus \(\mathrm{K^{\oplus}}\), Ar and \(\mathrm{Ca^{2\oplus}}\) all contain 18 electrons and are \(1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{6}3\mathrm{s}^{2}3\mathrm{p}^{6}\).

Flowchart

FC 4.7 – How the model of the atom developed: Dalton’s indivisible sphere → discovery of the electron (Thomson, 1897) → discovery of the nucleus and the proton (Rutherford, 1911, 1919) → discovery of the neutron (Chadwick, 1932) → Rutherford’s nuclear model fails on stability and electron energies → clues from radiation: Planck’s quantum theory (1900) and the line spectrum of hydrogen (Balmer 1885, Rydberg) → Bohr’s model (1913) explains hydrogen and derives R_H → fails for multi-electron atoms, fine structure, Zeeman/Stark effects and bonding → de Broglie’s matter waves (1924) and Heisenberg’s uncertainty principle (1927) → Schrodinger equation (1926) → quantum mechanical model: orbitals, four quantum numbers, aufbau + Pauli + Hund → electronic configuration of atoms

4.9 Check Your Concepts

  1. Dalton described the atom as an indivisible hard sphere. State one everyday observation that this description cannot explain, and say which discovery removed the difficulty.

  2. Thomson used cathode rays produced from electrodes of different metals and obtained the same particles every time. What conclusion about the electron follows from this observation, and why would the conclusion be weaker if only one metal had been tested?

  3. In Rutherford’s scattering experiment, nearly all α-particles passed straight through the gold foil while a very small number bounced back. Explain what each of these two observations tells us about the structure of the atom.

  4. Rutherford found that hydrogen was always emitted whenever an element was transmuted by α-particles. Explain how this single fact led to the naming of the proton.

  5. Chadwick directed the unknown radiation from beryllium onto paraffin rather than onto a metal sheet. Explain why paraffin was a suitable choice of target.

  6. An ion is formed when an atom loses two electrons. State, with reasons, which of the following change and which remain the same: atomic number, mass number, number of neutrons, chemical identity of the element.

  7. Explain why the atomic mass of an element quoted in the periodic table is often a fractional number, although the mass number of every individual nuclide is a whole number.

  8. All the isotopes of an element occupy the same position in the modern periodic table and show similar chemical properties, whereas isobars occupy different positions and show different chemical properties. Give the reason behind both statements.

  9. Read the two statements below and decide whether each is true and whether the second correctly explains the first. Assertion: An electron revolving in a circular orbit around the nucleus should continuously emit radiation. Reason: An electron in circular motion undergoes accelerated motion.

  10. A planet revolving round the sun does not spiral into it, but Rutherford’s electron is predicted to spiral into the nucleus. Identify the single physical difference responsible for this and explain it.

  11. Which properties of light are explained by treating it as a wave, and which are explained by treating it as a stream of photons? State the conclusion that had to be accepted about the nature of light.

  12. Two beams of light of the same colour differ in brightness. State which of the following differ between the two beams and which do not: wavelength, frequency, amplitude, energy of a single photon, number of photons arriving per second.

  13. Arrange the following in increasing order of the energy of one photon, giving your reasoning: microwave radiation, infrared radiation, ultraviolet radiation, radiofrequency radiation.

  14. Balmer’s formula reproduced the visible hydrogen lines accurately in 1885, yet it is not regarded as a theory of the atom. Explain why, and state what Bohr’s model added to it.

  15. Explain how Bohr’s first two postulates together remove the objection that an atom should collapse, and state clearly which classical conclusion Bohr had to abandon.

  16. The energy of an electron in a stationary state is always given a negative value. Explain what this negative sign conveys and what physical situation is taken as the zero of energy.

  17. Arrange the following in increasing order of the energy of the photon emitted, and name the series to which each transition belongs: \(n = 3 \rightarrow n = 2\), \(n = 2 \rightarrow n = 1\), \(n = 4 \rightarrow n = 3\).

  18. Explain why the Lyman series lies in the ultraviolet region while the Paschen, Bracket and Pfund series lie in the infrared region.

  19. \(\mathrm{He^{\oplus}}\) and \(\mathrm{Li^{2\oplus}}\) are both hydrogen-like species. Predict which of the two has the smaller first-orbit radius and which holds its electron more tightly, giving the reason in each case.

  20. State any three limitations of Bohr’s model, and for each one indicate whether it was later removed by the quantum mechanical model.

  21. Explain how the de Broglie relation and the Heisenberg uncertainty principle each make the idea of a fixed circular orbit unacceptable for an electron.

  22. The wave function \(\psi\) has no physical meaning of its own, yet it is central to the quantum mechanical model. Explain the role of \(\psi\) and the meaning attached to \(\psi^{2}\).

  23. Distinguish between an orbit and an orbital on the basis of definition, shape and consistency with the uncertainty principle.

  24. State which quantum number decides each of the following, and justify your choice: (a) the size of an orbital, (b) the shape of an orbital, (c) the orientation of an orbital in space, (d) the number of electrons an orbital can hold.

  25. Explain why a boundary surface diagram is normally drawn to enclose about 90% of the probability density and never 100%.

  26. Distinguish between a node and a nodal plane, giving one example of an orbital that possesses each.

  27. In the hydrogen atom the 2s and 2p orbitals have the same energy, but in a multi-electron atom the 2p orbital lies higher than the 2s orbital. Give the reason for this difference.

  28. Using the \((n + l)\) rule, explain why the 4s orbital is filled before the 3d orbital, and state the value of \((n + l)\) for each.

  29. Three electrons are to be placed in the 2p subshell. Draw the correct orbital diagram and one incorrect orbital diagram, and state which rule the incorrect one violates.

  30. Explain why the actual electronic configuration of chromium is \([\mathrm{Ar}]\,4\mathrm{s}^{1}3\mathrm{d}^{5}\) rather than the expected \([\mathrm{Ar}]\,4\mathrm{s}^{2}3\mathrm{d}^{4}\), and state the general principle involved.

  31. \(\mathrm{K^{\oplus}}\), Ar and \(\mathrm{Ca^{2\oplus}}\) have identical electronic configurations. State one property in which they nevertheless differ, and explain why.

  32. Complete the following: an orbital in the \(\mathrm{d}\) subshell can hold a maximum of ______ electrons; the \(\mathrm{d}\) subshell as a whole can hold a maximum of ______ electrons; and the N shell can hold a maximum of ______ electrons. Justify each answer using quantum numbers.

  33. If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.

  34. Name the element that shows simplest emission spectrum.

  35. State Heisenberg uncertainty principle.

  36. Give the names of quantum numbers.

  37. Identify from the following the isoelectronic species: Ne, \(\mathrm{O^{2-}}\), \(\mathrm{Na^{+}}\) OR Ar, \(\mathrm{Cl^{2-}}\), \(\mathrm{K^{\oplus}}\)

  38. Differentiate between Isotopes and Isobars.

  39. Define the terms:

    1. Isotones
    2. Isoelectronic species
    3. Electronic configuration
  40. State and explain Pauli’s exclusion principle.

  41. State Hund’s rule of maximum multiplicity with suitable example.

  42. Write the drawbacks of Rutherford’s model of an atom.

  43. Write postulates of Bohr’s Theory of hydrogen atom.

  44. Mention demerits of Bohr’s Atomic model.

  45. State the order of filling atomic orbitals following Aufbau principle.

  46. Explain the anomalous behavior of copper and chromium.

  47. Write orbital notations for electrons in orbitals with the following quantum numbers.

    1. \(n = 2\), \(l = 1\)
    2. \(n = 4\), \(l = 2\)
    3. \(n = 3\), \(l = 2\)
  48. Write electronic configurations of Fe, \(\mathrm{Fe^{2+}}\), \(\mathrm{Fe^{3+}}\)

  49. Write condensed orbital notation of electronic configuration of the following elements:

    1. Lithium (Z=3)
    2. Carbon (Z=6)
    3. Oxygen (Z=8)
    4. Silicon (Z=14)
    5. Chlorine (Z=17)
    6. Calcium (Z=20)
  50. Draw shapes of 2s and 2p orbitals.

  51. Explain in brief, the significance of azimuthal quantum number.

  52. If \(n = 3\), what are the quantum number \(l\) and \(m\)?

  53. The electronic configuration of oxygen is written as \(1\mathrm{s}^{2}\,2\mathrm{s}^{2}\,2\mathrm{p}_x^{2}\,2\mathrm{p}_y^{1}\,2\mathrm{p}_z^{1}\) and not as \(1\mathrm{s}^{2}\,2\mathrm{s}^{2}\,2\mathrm{p}_x^{2},\ 2\mathrm{p}_y^{2}\,2\mathrm{p}_z^{0}\), Explain.

  54. Write note on ‘Principal Quantum number’.

  55. Using concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells and orbitals.

  56. Indicate the number of unpaired electrons in :

    1. Si (Z=14)
    2. Cr (Z=24)
  57. An atom of an element contains 29 electrons and 35 neutrons. Deduce -

    1. the number of protons
    2. the electronic configuration of that element

4.10 Practice With Numericals

  1. Using the data of Table 4.1, calculate the ratio of the mass of a neutron to the mass of an electron.

  2. A neutral atom has 15 protons and 16 neutrons. Calculate its mass number, write its nuclide symbol, and state the number of electrons it contains.

  3. The nuclide \(^{75}_{33}\mathrm{As}\) is used in certain semiconductors. Calculate the number of protons, neutrons and electrons in one neutral atom of this nuclide, and the total nuclear charge in coulomb. (\(e = 1.6022\times10^{-19}\ \text{C}\))

  4. Copper occurs in nature as two isotopes, \(^{63}_{29}\mathrm{Cu}\) and \(^{65}_{29}\mathrm{Cu}\), in the relative abundance 7 : 3. Calculate the average atomic mass of copper.

  5. An element has two natural isotopes of mass numbers 24 and 26. If the average atomic mass of the element is 24.5, calculate the percentage abundance of each isotope.

  6. Radiation used in a microwave oven has a frequency of \(2.45\times10^{9}\ \text{Hz}\). Calculate its wavelength in metre and its wavenumber in \(\text{cm}^{-1}\). (\(c = 3.0\times10^{8}\ \text{m s}^{-1}\))

  7. Calculate the frequency and the energy of one photon of radiation of wavelength 300 nm. (\(h = 6.626\times10^{-34}\ \text{J s}\), \(c = 3.0\times10^{8}\ \text{m s}^{-1}\))

  8. A lamp emits monochromatic light of wavelength 500 nm with a power of \(1.0\times10^{-3}\ \text{J s}^{-1}\). Calculate the energy of one photon of this light and hence the number of photons emitted per second.

  9. Calculate the wavenumber in \(\text{cm}^{-1}\) and the wavelength in nm of the spectral line of hydrogen for which \(n_1 = 2\) and \(n_2 = 4\). Name the series to which it belongs. (\(R_H = 109677\ \text{cm}^{-1}\))

  10. Calculate the wavenumber of the second line of the Lyman series of hydrogen, that is, the line for which \(n_1 = 1\) and \(n_2 = 3\), and state the region of the electromagnetic spectrum in which it appears. (\(R_H = 109677\ \text{cm}^{-1}\))

  11. Calculate the radius of the fourth stationary state of the hydrogen atom, given that the Bohr radius is 52.9 pm. How many times larger is it than the first stationary state?

  12. Using \(E = -R_H\left(\dfrac{1}{n^{2}}\right)\) with \(R_H = 2.18\times10^{-18}\ \text{J}\), calculate the energy of the electron in the third stationary state of the hydrogen atom, and the energy released when the electron falls from this state to the ground state.

  13. For the hydrogen-like species \(\mathrm{Be^{3\oplus}}\) (\(Z = 4\)), calculate the radius and the energy of the first orbit using Eq. (4.7) and Eq. (4.8).

  14. An electron in a hydrogen atom makes a transition for which \(\Delta E = 1.635\times10^{-18}\ \text{J}\). Calculate the frequency of the radiation emitted, using Bohr’s frequency rule. (\(h = 6.626\times10^{-34}\ \text{J s}\))

  15. Calculate the de Broglie wavelength associated with a proton of mass \(1.6726\times10^{-27}\ \text{kg}\) moving with a velocity of \(2.0\times10^{5}\ \text{m s}^{-1}\). (\(h = 6.626\times10^{-34}\ \text{J s}\))

  16. A moving electron has a de Broglie wavelength of \(1.0\times10^{-10}\ \text{m}\). Calculate its velocity. (Mass of electron \(= 9.10938\times10^{-31}\ \text{kg}\), \(h = 6.626\times10^{-34}\ \text{J s}\))

  17. The position of an electron is measured with an uncertainty of \(1.0\times10^{-10}\ \text{m}\). Using \(\Delta x \times \Delta p_x \geq \dfrac{h}{4\pi}\), calculate the minimum uncertainty in its momentum.

  18. For the O shell (\(n = 5\)), calculate the number of subshells, the total number of orbitals and the maximum number of electrons it can accommodate. Show the working using the appropriate quantum-number relations.

  19. Write the complete and the condensed electronic configuration of the element with \(Z = 23\), and state the number of unpaired electrons present in its ground state.

  20. An atom of an element contains 24 electrons and 28 neutrons. Deduce its atomic number, its mass number, its nuclide symbol and its condensed electronic configuration, and state the number of unpaired electrons it possesses.

4.11 Objective Questions

  1. The energy difference between the shells goes on ……….. when moved away from the nucleus.
    1. Increasing
    2. decreasing
    3. equalizing
    4. static
  2. The value of Plank’s constant is -
    1. \(6.626\times10^{-34}\ \text{Js}\)
    2. \(6.023\times10^{-24}\ \text{Js}\)
    3. \(1.667\times10^{-28}\ \text{Js}\)
    4. \(6.626\times10^{-28}\ \text{Js}\)
  3. p-orbitals are ……. in shape.
    1. spherical
    2. dumb bell
    3. double dumbell
    4. diagonal
  4. “No two electrons in the same atoms can have identical set of four quantum numbers”. This statement is known as -
    1. Pauli’s exclusion principle
    2. Hund’s rule
    3. Aufbau rule
    4. Heisenberg uncertainty principle
  5. Principal Quantum number describes -
    1. shape of orbital
    2. size of the orbital
    3. spin of electron
    4. orientation of in the orbital electron cloud
  6. Make the pairs:
‘A’ ‘B’
a. Neutrons i. six electrons
b. p-orbital ii. \(-1.6\times10^{-19}\ \text{C}\)
c. charge on electron iii. Ultraviolet region
d. Lyman series iv. Chadwick
  1. Complete the following information about the isotopes in the chart given below : (Hint: Refer to Periodic Table if required)
Substance Mass Number Number of Protons Number of Neutrons Number of Electrons
Carbon-14
Lead-208
Chlorine-35
Uranium-238
Oxygen-18
Radium-223
  1. Match the following :
Element No. of Neutron
a. \(^{40}_{18}\mathrm{Ar}\) i. 7
b. \(^{14}_{6}\mathrm{C}\) ii. 21
c. \(^{40}_{19}\mathrm{K}\) iii. 8
d. \(^{14}_{7}\mathrm{N}\) iv. 22