Atomic structure

Class 11 Chemistry, unit 2, book pages 14–38

Practise 64 MCQs Take the unit quiz Review 40 cards

All matter is built from atoms, and the way electrons are arranged in an atom decides how the element behaves. This unit follows the development of the atomic model from Bohr's shells to sub-shells and orbitals, describes the electron, proton and neutron, and explains the four quantum numbers and the rules (Aufbau, n + l, Pauli and Hund) for writing electronic configurations of atoms and ions. It then explains the shapes of orbitals, free radicals and the trends of ionization energy, uses successive ionization energies to place an element in the periodic table, shows how a mass spectrometer gives relative atomic masses, and ends with the configuration of silicon as a semiconductor.

Learning outcomes (27)

Describe that each atomic shell and sub-shell are further divided into degenerate orbitals having the same energy.
Describe protons, neutrons and electrons in terms of their relative charge and relative masses.
Recognize that the terms atomic and proton number represent the same concept.
Describe the behaviour of beams of protons, neutrons and electrons moving at the same velocity in an electric field.
Determine the number of protons, neutrons and electrons present in both atoms and ions given atomic or proton number, mass or nucleon number and charge.
Explain the change in atomic and ionic radius across a period and down a group.
Determine the electronic configuration of elements and their ions with proton numbers (some examples include: a. simple configuration e.g. 2,8. b. sub-shells e.g. $1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}$ . c. students should be able to determine both of these from the periodic table and are not required to memorize these. d. students should understand that chemical properties of an atom are governed by valence electrons).
Define terms related to electronic configuration (some examples include: shells, sub-shells, orbitals, principal quantum number ( $n$ ), ground state).
Relate quantum numbers to the electronic distribution of elements.
Describe the number of orbitals making up s, p, d and f sub-shells, and the number of electrons that can fill s, p, d and f sub-shells.
Apply the Aufbau principle, Pauli's exclusion principle and Hund's rule to write the electronic configuration of elements.
Describe the order of increasing energy of the sub-shells (s, p, d and f).
Explain the electronic configurations to include the number of electrons in each shell, sub-shell and orbitals.
Explain the electronic configurations in terms of energy of the electrons and inter-electron repulsion.
Determine the electronic configuration of atoms and ions given the proton or electron number and charge.
Illustrate the importance of electronic configurations and the development of new materials for electronic devices (for example, semiconductors such as silicon have a specific electronic configuration that makes them ideal for their use in electronic devices).
Describe the shapes of s, p and d orbitals.
Describe a free radical as a species with one or more unpaired electrons.
Explain that ionization energies are due to the attraction between the nucleus and the outer electrons.
Explain how ionization energy helps account for the trends across the period and down a group of the periodic table.
Account for the variation in successive ionization energies of an element.
Explain the factors influencing the ionization energies of elements in terms of nuclear charge, atomic or ionic radius, shielding by inner shells and sub-shells, and spin pair repulsion.
Deduce the electronic configurations of elements using successive ionization energy data.
Deduce the position of an element in the periodic table using successive ionization energy data.
Explain how a mass spectrometer can be used to determine the relative atomic mass of an element from its isotopic composition.
Perform calculations involving non-integer relative atomic masses and abundance of isotopes from given data, including mass spectra.
Explain the concept of emission spectra. Use the concept of emission spectra to deduce the electronic configuration of elements.

2.1Brief history of atomic models

The idea of the atom was refined in steps. Discharge tube experiments, which began in the later part of the 19th century, modified Dalton's concept of the atom. Later, Niels Bohr's atomic model made the picture clearer, and Heisenberg's uncertainty principle then led to the modern idea of sub-shells and orbitals.

Bohr's atomic model

Bohr's model was the first approach to quantized energy levels, that is electronic shells with fixed energy. It was successful in calculating the radius of the hydrogen atom and the energy of the electron in any shell of the hydrogen atom. The emission spectrum of hydrogen was explained well by Bohr's model.

Figure 2.1: Bohr's atomic model. Electrons move in shells of fixed energy, K, L, M and N. An electron that jumps to a shell of lower energy gives out the energy difference as a photon.

The shells are named K, L, M and N, counting outwards from the nucleus, and the further a shell is from the nucleus, the higher its energy. An electron can jump between shells. When it jumps from a higher to a lower shell, the extra energy leaves the atom as a photon of light.

Where the Bohr model fell short

When high resolution spectrophotometers were made, the fine spectrum of hydrogen was found to consist of split spectral lines. These lines split further in an electric field (the Stark effect) and in a magnetic field (the Zeeman effect). Bohr's model could not explain either effect, so the atomic structure became more complicated.

These discoveries were followed by Heisenberg's uncertainty principle, which gave rise to the concept of sub-shells and orbitals as the parts that make up an electronic shell.

How the picture of the atom developed
Stage	What it contributed	Limit or follow-up
Discharge tube experiments (late 19th century)	Modified the Daltonian concept of the atom	Led to Bohr's model
Bohr's model	Quantized shells of fixed energy; radius of the hydrogen atom and energy of its electron in any shell; hydrogen emission spectrum	Could not explain the Stark and Zeeman splitting of the fine spectral lines
Heisenberg's uncertainty principle	Gave the concept of sub-shells and orbitals inside a shell	The basis of the quantum numbers of Section 2.2

2.1.1Sub-atomic particlesOften examined

An atom is made of three fundamental particles: the electron, the proton and the neutron. They differ in charge and in mass.

Electron: A negatively charged sub-atomic particle with a charge of $1.6022 \times 1 0^{- 19} C$ . Its relative charge is $- 1$ and its mass is $9.1095 \times 1 0^{- 31} kg$ . It was discovered by J. J. Thomson. Electrons are deflected towards the positive pole of an electric field.
Proton: A positively charged sub-atomic particle with a charge equal to that of the electron, but it is $1836$ times heavier than an electron, with a mass of $1.6727 \times 1 0^{- 27} kg$ . Its relative charge is $+ 1$ . It is deflected towards the negative pole of an electric field.
Neutron: A particle with no charge, because it passes through electric and magnetic fields undeflected. It is more massive than a proton, with a mass of $1.6750 \times 1 0^{- 27} kg$ .

Because electrons and protons are charged, they respond to electric and magnetic fields; the book calls them electromagnetic in nature. The neutron, with no charge, does not.

The structure of atoms: the three fundamental particles
Particle	Mass (kg)	Charge (C)	Relative mass	Relative charge
Proton	$1.6727 \times 1 0^{- 27}$	$+ 1.6022 \times 1 0^{- 19}$	$1$	$+ 1$
Neutron	$1.6750 \times 1 0^{- 27}$	$0$	$1$	$0$
Electron	$9.1095 \times 1 0^{- 31}$	$- 1.6022 \times 1 0^{- 19}$	$\frac{1}{1836}$	$- 1$

2.1.2Behaviour of electron, proton and neutron in electric fieldOften examined

In an electric field, electrons and protons feel opposite forces because their charges are opposite. Neutrons are electrically neutral, so they generally feel negligible force and are not deviated.

Electrons are negatively charged. They move against the direction of the electric field, towards the positive pole.
Protons are positively charged. They move in the direction of the electric field, towards the negative pole.
Neutrons are neutral. They experience negligible force and show no deviation.

When electrons, protons and neutrons share the same velocity, their paths diverge because of the electric forces on the electrons and the protons. Electrons curve against the electric field, protons curve with it, and neutrons, having no net charge, continue along their original path with minimal deflection. How sharply a path curves also depends on the trajectory force or velocity.

Figure 2.2: Path of positively and negatively charged particles through the uniform electric field. The electron is deflected much more than the proton because it has a much smaller mass.

Behaviour of the three particles moving at the same velocity in an electric field
Particle	Charge	Direction of deflection	Size of deflection
Electron	$-$	Towards the positive plate	Large
Proton	$+$	Towards the negative plate	Small
Neutron	$0$	None, goes straight on	None

2.1.3Atomic number and mass numberOften examined

Atomic number (proton number), Z: The number of protons in the nucleus of an atom. Since an atom is neutral, the number of electrons revolving around the nucleus equals the number of protons in the nucleus.
Mass number (nucleon number), A: The total number of protons and neutrons in the nucleus. The mass number is approximately equal to the atomic mass of the atom.

The terms atomic number and proton number mean the same thing, and mass number and nucleon number mean the same thing. The relations between the particles are:

Z + N = A (P + N = A)

Mass number

N is the number of neutrons, P the number of protons

number of protons = number of electrons (neutral atom)

electrons in a cation = Z - magnitude of the charge

electrons in an anion = Z + magnitude of the charge

Concept assessment exercise 2.1 (question 1): electrons in ions

How many electrons are present in the following species:

Mg X^{2 +}

Al X^{3 +}

S X^{2 -}

Find the atomic number of each element from the periodic table. A neutral atom has as many electrons as its atomic number.
$Z (Mg) = 12, Z (Al) = 13, Z (S) = 16$
$Mg X^{2 +}$ is a cation with charge $+ 2$ , so it has lost 2 electrons.
$12 - 2 = 10$
$Al X^{3 +}$ is a cation with charge $+ 3$ , so it has lost 3 electrons.
$13 - 3 = 10$
$S X^{2 -}$ is an anion with charge $- 2$ , so it has gained 2 electrons.
$16 + 2 = 18$

Mg X^{2 +} : 10 electrons, Al X^{3 +} : 10 electrons, S X^{2 -} : 18 electrons

Concept assessment exercise 2.1 (question 2): protons, neutrons and electrons

Find the number of electrons, protons and neutrons of

Mg X^{2 +}

Cl X^{-}

Al X^{3 +}

and

S X^{2 -}

using the atomic number (use the periodic table).

The atomic number gives the number of protons. The number of neutrons needs the mass number, which is taken as the whole number nearest to the atomic mass in the periodic table (for chlorine, the more abundant isotope $^{35} Cl$ ).

Protons $= Z$ for every species, because forming an ion changes only the electrons.
$Z : Mg = 12, Cl = 17, Al = 13, S = 16$
Electrons: subtract the charge for a cation, add it for an anion.
$Mg X^{2 +} : 12 - 2 = 10; Cl X^{-} : 17 + 1 = 18; Al X^{3 +} : 13 - 3 = 10; S X^{2 -} : 16 + 2 = 18$
Neutrons $= A - Z$ , using $A = 24, 35, 27, 32$ .
$Mg : 24 - 12 = 12; Cl : 35 - 17 = 18; Al : 27 - 13 = 14; S : 32 - 16 = 16$

Species	Protons	Electrons	Neutrons
$Mg X^{2 +}$	12	10	12
$Cl X^{-}$	17	18	18
$Al X^{3 +}$	13	10	14
$S X^{2 -}$	16	18	16

Mg X^{2 +} : (12, 10, 12), Cl X^{-} : (17, 18, 18), Al X^{3 +} : (13, 10, 14), S X^{2 -} : (16, 18, 16) as (protons, electrons, neutrons)

2.1.4Atomic and ionic radius

Radius of an atom: The average distance of the nucleus from the outermost electrons. It measures the size of an atom. The boundary of the electron cloud is not well defined, so the word average is used.

Periodic trends of atomic radius

Across a period (left to right in the main groups) the atomic and ionic radius decrease, because the nuclear charge increases. The shielding effect stays constant from left to right. In the transition metal series, however, it increases because the electrons are added to the inner d orbitals.
Down a group the atomic radius increases throughout the periodic table. There is one more shell in each period, and the shielding effect increases because of the intervening electrons.

Atomic radius trends: it decreases across period 2 and increases down group 1. The circles are drawn to show the trend, not to an exact scale.

Ionic radius

Cations are always smaller than their parent atoms, because the effective nuclear charge increases and the electron-electron repulsion in the valence shell decreases when electrons are removed.
Anions are bigger than their parent atoms.

From top to bottom in a group the trend of ionic radius is the same as that of atomic radius: it keeps increasing. This is because the ions throughout the group have the same charge.

The trend is debatable across a period, because the elements of a period do not usually form ions with the same charge. The charge of the ion depends on the valence shell electronic configuration, which changes from left to right. If we compare ions with the same charge across a period, the trend is the same as that of the atomic radius, that is, a decrease.

Summary of the radius trends
Change	Effect on radius	Reason
Left to right across a period	Decreases	Nuclear charge increases, shielding stays the same
Top to bottom down a group	Increases	One more shell each time and more shielding
Atom to cation	Cation is smaller	Effective nuclear charge increases, electron repulsion in the valence shell decreases
Atom to anion	Anion is bigger	The extra electrons repel each other more

Concept assessment exercise 2.2: complete the sentences

Fill in the blanks. (1) Across a period from left to right the atomic radius generally ____. (2) Down a group the atomic radius tends to ____. (3) When an atom loses electrons to become a cation, its ionic radius ____ compared to its atomic radius. (4) From top to bottom in a group the ionic radius of cations tends to ____ while that of anions tends to ____. (5) Non-metals tend to form ____ ions with ____ ionic radii compared to their atomic radii.

Across a period the nuclear charge increases while the shells stay the same.
$(1) decreases$
Down a group a new shell is added at each step.
$(2) increase$
A cation has lost electrons, so the effective nuclear charge per electron rises and repulsion falls.
$(3) is smaller (decreases)$
The ions of a group have the same charge and one more shell at each step, so both kinds of ion get bigger.
$(4) increase, increase$
Non-metals gain electrons to complete the valence shell, so they form negative ions (anions), which are bigger than the atoms.
$(5) negative (anions), larger$

(1) decreases; (2) increase; (3) smaller; (4) increase, increase; (5) negative, larger

2.2Quantum numbers: principal and azimuthalOften examined

Expressing the presence of electrons around the nucleus in a mathematical way gave rise to the concept of quantum numbers. A set of four numbers, the quantum numbers, is assigned to every electron in an atom. However big the electronic cloud of an atom is, this set of four numbers is enough to give the exact location (the address) of an electron in the atom. The energy of an electron might be expected to depend on the first two quantum numbers, the principal and the azimuthal quantum number.

The four quantum numbers (the last two are explained in the next two sections)
Quantum number	Symbol	What it describes	Allowed values
Principal	$n$	The shell (size and energy)	$1, 2, 3, \dots$
Azimuthal	$l$	The sub-shell (shape of the orbital)	$0$ to $n - 1$
Magnetic	$m$	The orientation of the orbital in space	$- l$ to $+ l$
Spin	$s$	The spin of the electron	$+ \frac{1}{2}$ or $- \frac{1}{2}$

1. Principal quantum number

The principal quantum number is the expression of Bohr's shell in quantum form. It is represented by $n$ , and its possible values are

n = 1, 2, 3, \dots

It also gives the period number in the periodic table, because each period starts with a new shell. It represents Bohr's shell.
If $n$ is greater, the energy of the shell is greater and so is its size.
The principal quantum number arose from the emission spectrum of hydrogen, which was first explained by Bohr and Max Planck. Each spectral line of the emission spectrum is explained by the possible jumps between shells of the H atom.

2. Azimuthal quantum number

The azimuthal quantum number is denoted by $l$ . It describes the shape of the orbitals of an electronic shell. An orbital is a space around the nucleus where the probability of finding the electron is maximum. One shell may comprise more than one sub-shell. The azimuthal quantum number determines the angular momentum of the electron and specifies the sub-shell within a given principal energy level $n$ . Its values range from $0$ to $n - 1$ .

l = 0, 1, 2, 3, \dots, (n - 1)

The presence of sub-shells is related to the fine lines of the spectrum.
The number of sub-shells in a shell depends on the shell number: the bigger the shell, the more sub-shells it has and the more electrons it can hold.
The azimuthal quantum number is also related to the position of the element in the periodic table: the s, p, d and f sub-shells represent the blocks of the periodic table.

Letters used for the sub-shells
$l$	Sub-shell letter
$0$	s
$1$	p
$2$	d
$3$	f

How many sub-shells each shell has
Shell	$n$	Possible values of $l$	Sub-shells
K	1	0	1s
L	2	0, 1	2s, 2p
M	3	0, 1, 2	3s, 3p, 3d
N	4	0, 1, 2, 3	4s, 4p, 4d, 4f

Table showing possible quantum numbers. Every orbital holds two electrons, with $s = + \frac{1}{2}$ and $s = - \frac{1}{2}$ .
Shell $n$	$l$	Sub-shell	$m$	Orbitals	Electrons	Total electrons in the shell
1 (K)	0	1s	0	1	2	2
2 (L)	0	2s	0	1	2	8
	1	2p	$- 1, 0, + 1$	3	6
3 (M)	0	3s	0	1	2	18
	1	3p	$- 1, 0, + 1$	3	6
	2	3d	$- 2, - 1, 0, + 1, + 2$	5	10
4 (N)	0	4s	0	1	2	32
	1	4p	$- 1, 0, + 1$	3	6
	2	4d	$- 2, - 1, 0, + 1, + 2$	5	10
	3	4f	$- 3, - 2, - 1, 0, + 1, + 2, + 3$	7	14

2.2Magnetic quantum numberOften examined

The splitting of the fine spectral lines under the influence of a magnetic field can be explained with the magnetic quantum number. It is also called the orientation quantum number and is denoted by $m$ . Its possible values run from $- l$ to $+ l$ , including zero:

m = - l, \dots, - 2, - 1, 0, + 1, + 2, \dots, + l

Each value of $m$ is one orbital of the sub-shell, so the number of values of $m$ is the number of orbitals. Orbitals of one sub-shell that have the same energy are called degenerate orbitals.

Magnetic quantum number values and the orbitals they represent
Sub-shell	$l$	Values of $m$	Number of orbitals	Orbitals
s	0	$0$	1	s
p	1	$- 1, 0, + 1$	3	$p_{x}, p_{y}, p_{z}$
d	2	$- 2, - 1, 0, + 1, + 2$	5	$d_{x y}, d_{y z}, d_{x z}, d_{x^{2} - y^{2}}, d_{z^{2}}$
f	3	$- 3$ to $+ 3$	7	seven f orbitals

s orbital

For an s orbital $l = 0$ , so there is only one value of the magnetic quantum number, which is also zero. Under the influence of a magnetic field the s orbital therefore remains spherically symmetrical.

p orbitals

For a p orbital $l = 1$ , so $m$ has three values, $+ 1$ , $0$ and $- 1$ . They show that the p sub-shell splits into three degenerate orbitals, $p_{x}$ , $p_{y}$ and $p_{z}$ , which are oriented in space along the x, y and z axes respectively.

d orbitals

For the d orbital $l = 2$ , so there are five values of $m$ : $- 2, - 1, 0, + 1, + 2$ . They represent five orbitals oriented at different positions in space around the nucleus. They are named $d_{x y}$ , $d_{y z}$ , $d_{x z}$ , $d_{x^{2} - y^{2}}$ and $d_{z^{2}}$ .

f orbitals

The f orbitals are seven in number, with magnetic quantum number values from $- 3$ to $+ 3$ .

Worked example: allowed values for 3d and 4f

For the 3d and 4f sub-shells give

n

l

, all values of

m

, the number of orbitals and the maximum number of electrons. Why do 2d and 3f not exist?

Read $n$ from the number and $l$ from the letter ( $s = 0, p = 1, d = 2, f = 3$ ). For 3d, $n = 3$ and $l = 2$ . For 4f, $n = 4$ and $l = 3$ .
The magnetic quantum number runs from $- l$ to $+ l$ . For 3d: $- 2, - 1, 0, + 1, + 2$ , which is $5$ values. For 4f: $- 3$ to $+ 3$ , which is $7$ values.
$2 l + 1 = 52 l + 1 = 7$
Each value of $m$ is one orbital, and an orbital holds two electrons of opposite spin.
$5 \times 2 = 107 \times 2 = 14$
For 2d we would need $l = 2$ , but $l$ goes only from $0$ to $n - 1$ . With $n = 2$ the largest $l$ is $1$ , so there is no 2d. In the same way $n = 3$ allows $l$ up to $2$ only, so there is no 3f.
$l_{m a x} = n - 1$

Sub-shell	$n$	$l$	Values of $m$	Orbitals	Maximum electrons
3d	3	2	$- 2, - 1, 0, + 1, + 2$	5	10
4f	4	3	$- 3, - 2, - 1, 0, + 1, + 2, + 3$	7	14

3d: 5 orbitals, 10 electrons; 4f: 7 orbitals, 14 electrons

2.2Spin quantum numberOften examined

A maximum of two electrons can live in one orbital defined by the magnetic quantum number. These two electrons must also be told apart, so they have different identification: the two electrons in one orbital must have opposite spin. If one spins clockwise, the other spins anticlockwise.

The spin quantum number describes the spin of an electron. Goudsmit proved in 1925 that an electron also spins about its own axis. This is called self-rotation and it causes a spin magnetic field around the electron. Two electrons can stay in one orbital only if they cancel the spin magnetic field of each other, and for this purpose they have opposite spins.

A clockwise spin is represented by $+ \frac{1}{2}$ and the symbol $↑$ .
An anticlockwise spin is represented by $- \frac{1}{2}$ and the symbol $↓$ .

Figure 2.5: Spin quantum number. An electron spins about its own axis, clockwise (

s = + \frac{1}{2}

) or anticlockwise (

s = - \frac{1}{2}

2.2.1Quantum numbers and electronic configuration

Electronic configuration: The distribution of electrons around the nucleus in orbits and, in turn, in orbitals.

A complete electronic configuration must clearly represent all four quantum numbers. With the help of the four quantum numbers one should be able to write the complete configuration.

The book illustrates this with an electron of the 2p sub-shell. Its address is read off the four numbers:

Reading the four quantum numbers of an electron (the book's figure)

An electron has n=2, l=1, m=+1 and s=-\tfrac12. Where is it and how is it spinning?

n=2 means the second shell, the L shell.
n=2
l=1 is the p sub-shell, so the electron is in the 2p sub-shell.
l=1\ \Rightarrow\ \text{p sub-shell}
m=+1 picks one of the three 2p orbitals, here the orbital with m=+1.
m=+1
s=-\tfrac12 means an anticlockwise spin, drawn as \downarrow.
s=-\tfrac12\ \Rightarrow\ \downarrow

2p\ \text{orbital with } m=+1,\ \text{spin } -\tfrac12

2.2.2Emission spectrum and electronic configuration of elements

The concept of emission spectra is used to determine the electronic configuration of elements. When atoms absorb energy, for example through heat or electricity, their electrons move to a higher energy level. When these electrons return to their lower energy levels, they release the absorbed energy in the form of light. The emitted light can be analysed to show the unique spectral lines, or pattern, of each element. This process is essential in spectroscopy and is widely used to identify and understand the electronic structure of atoms.

The emission spectra are used to infer the electronic configuration of elements as follows.

1. Line spectrum analysis

Each element has a characteristic line spectrum, consisting of certain wavelengths of light that are emitted when electrons move between energy levels. By analysing the spectral lines of an emission spectrum, one can infer the energy levels and the transitions that the electrons undergo in an atom.

2. Quantized energy levels

The observation of distinct lines supports Bohr's idea that electrons occupy specific, quantized energy levels (shells).

3. Electronic configuration deduction

The number and position of the lines in the spectrum can be used to determine the number of shell electrons and the distribution of the shell electrons. For instance, the number and arrangement of lines in the spectrum indicate the number of major energy levels or shells.

A hydrogen atom is a classic example of an emission spectrum. Its emission spectrum is composed of distinct lines that represent the electron transitions in the hydrogen atom. It helps to determine the energy levels of hydrogen and its electronic configuration.
A sodium atom has a characteristic yellow emission line. Analysing this line helps to determine the electronic configuration of sodium, including the distribution of its electrons in the energy levels.

The hydrogen emission spectrum: electrons falling to a lower level emit light. Each series of lines is made of jumps that end on the same lower level (arrows from n=6 are shown as examples).

Concept assessment exercise 2.3: principal, azimuthal and magnetic quantum numbers

(1) Explain the role of the principal quantum number n in describing the energy levels of electrons within an atom. (2) When n=3, how many different possible values can the azimuthal quantum number l take? (3) Describe the significance of the magnetic quantum number m in relation to the arrangement of electrons within an atom.

Question 1. n gives the shell, which is Bohr's shell in quantum form. A larger n means a shell that is larger and higher in energy, and n equals the period number of the element.
n=1,2,3,\ldots
Question 2. For a given n, l runs from 0 to n-1. For n=3:
l=0,\,1,\,2
So l can take three values, which are the 3s, 3p and 3d sub-shells.
3\ \text{values}
Question 3. m gives the orientation of an orbital in space. For each l it takes the values -l to +l, so it tells how many orbitals a sub-shell has (one s, three p, five d, seven f) and which of these degenerate orbitals an electron occupies.
m=-l,\ldots,+l

\text{(2) three values: } l=0,1,2

2.3Rules of electronic configuration: Aufbau principleOften examined

Electrons are distributed in orbitals and sub-shells in order of increasing energy: first the low energy orbitals are filled, and then electrons go into the higher energy orbitals. This is the Aufbau principle (Aufbau is German for building up). Four rules govern the electronic configuration of an element: the Aufbau principle, the n+l rule, Pauli's exclusion principle and Hund's rule.

The energies of orbitals are calculated using quantum mechanical principles. This calculation gives information about the allowed energy levels and the spatial distribution of electrons in an atom. The increasing energy order of the orbitals is

1s < 2s < 2p < 3s < 3p < 4s < 3d < \cdots

It is apparent that sometimes the electron is filled first in a higher shell (higher n) and then in a lower shell, as 4s is filled before 3d. This is explained by the n+l rule of the next section.

Figure 2.6: Filling order of atomic orbitals. Start at the top of each diagonal arrow and follow it down to the left, then move to the next arrow: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.

The relative energies of the orbitals up to 4p (not to scale). The 4s orbital lies below the 3d orbitals, so 4s is filled first.

Concept assessment exercise 2.4 (question 1): the 11th electron

In which orbital of the atom will the 11th electron be configured?

Fill the orbitals in Aufbau order and count the electrons: 1s holds 2, 2s holds 2 more, 2p holds 6 more.
2+2+6=10
Ten electrons fill 1s^2\,2s^2\,2p^6. The next orbital in the order is 3s, so the 11th electron goes there (this is sodium, 1s^2\,2s^2\,2p^6\,3s^1).
11\text{th electron}\to 3s

Concept assessment exercise 2.4 (question 2): increasing energy order

Place these orbitals in increasing energy order: 5s, 4p, 4s, 3d.

Work out n+l for each orbital (l=0 for s, 1 for p, 2 for d).
4s:4+0=4;\ 3d:3+2=5;\ 4p:4+1=5;\ 5s:5+0=5
Lowest n+l is lowest in energy, so 4s comes first.
4s
3d, 4p and 5s all have n+l=5. When n+l is equal, the orbital with the smaller n has the lower energy, so the order is 3d, then 4p, then 5s.
3d<4p<5s

4s < 3d < 4p < 5s

2.3The n + l ruleOften examined

The higher the n+l value of an orbital, the higher its energy.

The 4s and 3d orbitals

Which has the higher energy, the 4s or the 3d orbital?

Add the principal and azimuthal quantum numbers of each.
4s:\ n+l=4+0=4;\quad 3d:\ n+l=3+2=5
The 3d orbital has the larger n+l.
5>4

\text{The energy of 3d is greater than that of 4s, so 4s fills first}

The n+l values of the first orbitals
Orbital	n	l	n+l
1s	1	0	1
2s	2	0	2
2p	2	1	3
3s	3	0	3
3p	3	1	4
4s	4	0	4
3d	3	2	5
4p	4	1	5
5s	5	0	5

Worked example: ordering orbitals with the n+l rule

Arrange 3p, 4s, 3d, 4p and 5s in increasing order of energy. Which orbitals need the tie-break rule?

Find n+l for each orbital (l=0,1,2 for s, p, d).
3p:\ 3+1=4;\ 4s:\ 4+0=4;\ 3d:\ 3+2=5;\ 4p:\ 4+1=5;\ 5s:\ 5+0=5
The orbitals with n+l=4 (3p, 4s) are lower than those with n+l=5 (3d, 4p, 5s).
Within each group n+l is equal, so the orbital with the smaller n is lower in energy: 3p before 4s, and 3d before 4p before 5s.
3p<4s<3d<4p<5s

Energy ladder from the n+l rule (not to scale). Orbitals with equal n+l (accent brackets) are ordered by n.

3p<4s<3d<4p<5s

Concept assessment exercise 2.5 (n + l rule): which orbital has the greater energy?

Which of the following orbitals has the greater energy? (a) 5f or 6p, (b) 5s or 4d.

(a) Find n+l for each (l=3 for f, l=1 for p).
5f:5+3=8;\ 6p:6+1=7
5f has the larger n+l, so it has the greater energy.
5f>6p
(b) Find n+l for each (l=0 for s, l=2 for d).
5s:5+0=5;\ 4d:4+2=6
4d has the larger n+l, so it has the greater energy.
4d>5s

\text{(a) 5f; (b) 4d}

2.3Pauli exclusion principleOften examined

Pauli exclusion principle: No two electrons in an atom can have the same values of all four quantum numbers. At least the spin quantum number must be different.

If two electrons are present in the same orbital (n, l and m the same), their spins must be opposite. If one is clockwise (s=+\tfrac12, \uparrow), the other is anticlockwise (s=-\tfrac12, \downarrow).

Two electrons in a 3p_x orbital: they share n, l and m and differ only in spin
Quantum number	Electron A	Electron B
n	3	3
l	1	1
m	-1	-1
s	+\tfrac12	-\tfrac12

2.3Hund's ruleOften examined

Hund's rule: If degenerate orbitals (orbitals of equal energy) are available, the electrons tend to reside separately with the same spin, instead of pairing up in the same orbital with opposite spin.

For two electrons filling the p orbitals the configuration is p_x^{\uparrow}\,p_y^{\uparrow}. For three electrons it is p_x^{\uparrow}\,p_y^{\uparrow}\,p_z^{\uparrow}.

Hund's rule for the 2p orbitals: electrons occupy separate degenerate orbitals with parallel spins before any pairing takes place.

Half-filled and completely filled orbitals

The energy of an orbital is also related to the number of electrons present in it. Half-filled and completely filled orbitals are more stable, with less energy. This is obvious in the configuration of many elements, such as chromium (Cr) and copper (Cu).

Chromium and copper
Element	Expected from the Aufbau order	Actual configuration	Why
Chromium, Z=24	3d^4\,4s^2	3d^5\,4s^1	Half-filled 3d is more stable
Copper, Z=29	3d^9\,4s^2	3d^{10}\,4s^1	Completely filled 3d is more stable

2.4Electronic configuration of atoms and ionsOften examined

To find the electronic configuration of an atom or ion you need the number of protons and electrons it has. In a neutral atom the proton number (atomic number Z) equals the number of electrons. The configuration is found in three steps.

Step 1: identify the neutral atom's electron configuration

Each neutral atom has a specific number of electrons, which can be found from the periodic table. For example, a neutral fluorine atom has nine electrons.

Step 2: determine the charge of the ion

Ions can be positive or negative, depending on whether they have lost or gained electrons. Positive ions (cations) have a deficit of electrons, while negative ions (anions) have an excess of electrons.

Step 3: write the electron configuration of the ion

For a cation, start with the neutral atom's configuration and remove electrons equal to the magnitude of the positive charge from the outermost shell.
For an anion, start with the neutral atom's configuration and add electrons equal to the magnitude of the negative charge in the outermost shell.

Fluoride ion, \ce{F-}

Write the electronic configuration of the fluoride ion.

Fluorine has Z=9, so the neutral atom has nine electrons and the configuration is
1s^2\,2s^2\,2p^5
The ion has a charge of -1, so one electron is added to the outermost shell.
9+1=10\ \text{electrons}
The 2p sub-shell becomes full.
1s^2\,2s^2\,2p^6

\ce{F-}:\ 1s^2\,2s^2\,2p^6

Sodium ion, \ce{Na+}

Write the electronic configuration of the sodium ion.

Sodium has Z=11, so the neutral atom has 11 electrons.
1s^2\,2s^2\,2p^6\,3s^1
The ion has a charge of +1, so one electron is removed from the valence (outermost) orbital.
11-1=10\ \text{electrons}
The single 3s electron is the one removed.
1s^2\,2s^2\,2p^6

\ce{Na+}:\ 1s^2\,2s^2\,2p^6

Worked example: iron and the iron(III) ion

Write the electronic configuration of iron (Z=26) and of \ce{Fe^3+}, and draw the 4s and 3d boxes.

A neutral iron atom has 26 electrons. Fill in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d.
\ce{Fe}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6
\ce{Fe^3+} has lost three electrons. They come from the outermost shell first, so both 4s electrons leave, and then one 3d electron.
26-3=23\ \text{electrons}
Write the configuration of the ion and check that the superscripts add up to 23.
\ce{Fe^3+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\qquad 2+2+6+2+6+5=23
Draw the boxes. By Hund's rule the five 3d electrons of the ion go one in each box with parallel spins, a half-filled sub-shell, which is a stable arrangement.

4s and 3d boxes of Fe and \ce{Fe^3+}. The ion has a half-filled 3d sub-shell (accent) and an empty 4s.

\ce{Fe^3+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5

2.4Electronic configuration, energy and inter-electron repulsion

Inter-electron repulsion

Electrons tend to minimize repulsion by occupying their own separate orbitals, rather than sharing an orbital with another electron. Electrons repel each other because of their identical charges, and this repulsion is minimized when electrons occupy separate orbitals of the same sub-shell. The repulsion between electrons increases with the number of electrons in the same orbital, which leads to a higher energy for crowded orbitals.

Orbital energy

The energy of an orbital depends on its distance from the nucleus and on the electron-electron repulsion it experiences. Orbitals with lower energies are filled before those with higher energies.

Spin pairing repulsion

Electrons that share an orbital repel each other, and this is called spin pair repulsion. For this reason electrons occupy separate orbitals of the same degenerate sub-shell, with parallel spins, to minimize the repulsion (Hund's rule).

Energy and repulsion behind the filling rules
Idea	Effect on the configuration
Electrons repel each other	They spread out into separate orbitals
Lower energy orbitals are more stable	They are filled first (Aufbau principle)
Spin pair repulsion in a shared orbital	Degenerate orbitals each get one electron first (Hund's rule)

2.4Writing the electronic configurationOften examined

The electronic configuration of an element is written by filling electrons from the lower energy levels to the higher energy levels. The simple form of the configuration shows only the principal quantum number n and the azimuthal quantum number l (the sub-shell). The magnetic and spin quantum numbers are not shown. For example, the configuration 3p^6 means 3p_x^2, 3p_y^2 and 3p_z^2, and in each orbital the two electrons must have clockwise and anticlockwise spins.

An element with atomic number 25

Write the electronic configuration of the element with atomic number 25, showing every orbital.

Atomic number 25 means 25 electrons (manganese). Fill in Aufbau order: 1s, 2s, 2p, 3s, 3p, 4s, 3d.
2+2+6+2+6+2=20
Five electrons are left for the 3d sub-shell, which has five degenerate orbitals. By Hund's rule each orbital gets one electron with the same spin.
25-20=5
Written orbital by orbital:
1s^2,\ 2s^2,\ 2p_x^2\,2p_y^2\,2p_z^2,\ 3s^2,\ 3p_x^2\,3p_y^2\,3p_z^2,\ 4s^2,\ 3d_{xy}^{\uparrow}\,3d_{yz}^{\uparrow}\,3d_{zx}^{\uparrow}\,3d_{x^2-y^2}^{\uparrow}\,3d_{z^2}^{\uparrow}

The configuration of the element with Z=25 as an orbital box diagram: all sub-shells are full except 3d, which has five unpaired electrons (Hund's rule).

1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^5

The first eighteen elements

The electronic configuration of the first eighteen elements
Element	Atomic number	Electronic configuration
Hydrogen	1	1s^{1}
Helium	2	1s^{2}
Lithium	3	1s^{2}\,2s^{1}
Beryllium	4	1s^{2}\,2s^{2}
Boron	5	1s^{2}\,2s^{2}\,2p^{1}
Carbon	6	1s^{2}\,2s^{2}\,2p^{2}
Nitrogen	7	1s^{2}\,2s^{2}\,2p^{3}
Oxygen	8	1s^{2}\,2s^{2}\,2p^{4}
Fluorine	9	1s^{2}\,2s^{2}\,2p^{5}
Neon	10	1s^{2}\,2s^{2}\,2p^{6}
Sodium	11	1s^{2}\,2s^{2}\,2p^{6}\,3s^{1}
Magnesium	12	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}
Aluminium	13	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{1}
Silicon	14	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{2}
Phosphorus	15	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{3}
Sulphur	16	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{4}
Chlorine	17	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{5}
Argon	18	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}

2.4Concept assessment exercise 2.5: configurations of elements and ionsOften examined

The exercise has two questions: the configurations of four elements with all four quantum numbers, and the configurations of three ions. The m values below follow the usual convention of filling from m=-l to m=+l with spin +\tfrac12 first.

Exercise 2.5, 1 (a): scandium

Write down the electronic configuration of Scandium (atomic number 21) representing all four quantum numbers.

Scandium has Z=21, so it has 21 electrons. Fill them in Aufbau order.
1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{1}
Show the orbitals one by one. Each filled orbital holds two electrons of opposite spin.
1s^2,\ 2s^2,\ 2p_x^2\,2p_y^2\,2p_z^2,\ 3s^2,\ 3p_x^2\,3p_y^2\,3p_z^2,\ 4s^2,\ 3d_{xy}^{\uparrow}

The last electron to enter is the 3d electron, with n=3, l=2, m=-2 and s=+\tfrac12.

Sub-shell	n	l	m	s
1s^{2}	1	0	0	+\tfrac12 and -\tfrac12 in each orbital
2s^{2}	2	0	0	+\tfrac12 and -\tfrac12 in each orbital
2p^{6}	2	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
3s^{2}	3	0	0	+\tfrac12 and -\tfrac12 in each orbital
3p^{6}	3	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
4s^{2}	4	0	0	+\tfrac12 and -\tfrac12 in each orbital
3d^{1}	3	2	-2	+\tfrac12 in each

1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{1}

Exercise 2.5, 1 (b): chromium

Write down the electronic configuration of Chromium (atomic number 24) representing all four quantum numbers.

Chromium has Z=24, so it has 24 electrons. Fill them in Aufbau order.
1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}\,3d^{5}
Show the orbitals one by one. Each filled orbital holds two electrons of opposite spin.
1s^2,\ 2s^2,\ 2p^6,\ 3s^2,\ 3p^6,\ 4s^{\uparrow},\ 3d_{xy}^{\uparrow}\,3d_{yz}^{\uparrow}\,3d_{xz}^{\uparrow}\,3d_{x^2-y^2}^{\uparrow}\,3d_{z^2}^{\uparrow}

Chromium is an exception to the filling order. One electron moves from 4s to 3d, so that 3d is half-filled (five unpaired electrons) and 4s is half-filled, which is more stable. Hund's rule puts one \uparrow electron in each of the five 3d orbitals.

Sub-shell	n	l	m	s
1s^{2}	1	0	0	+\tfrac12 and -\tfrac12 in each orbital
2s^{2}	2	0	0	+\tfrac12 and -\tfrac12 in each orbital
2p^{6}	2	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
3s^{2}	3	0	0	+\tfrac12 and -\tfrac12 in each orbital
3p^{6}	3	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
4s^{1}	4	0	0	+\tfrac12 in each
3d^{5}	3	2	-2, -1, 0, +1, +2	+\tfrac12 in each

1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}\,3d^{5}

Exercise 2.5, 1 (c): manganese

Write down the electronic configuration of Manganese (atomic number 25) representing all four quantum numbers.

Manganese has Z=25, so it has 25 electrons. Fill them in Aufbau order.
1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{5}
Show the orbitals one by one. Each filled orbital holds two electrons of opposite spin.
1s^2,\ 2s^2,\ 2p^6,\ 3s^2,\ 3p^6,\ 4s^2,\ 3d_{xy}^{\uparrow}\,3d_{yz}^{\uparrow}\,3d_{xz}^{\uparrow}\,3d_{x^2-y^2}^{\uparrow}\,3d_{z^2}^{\uparrow}

After the 4s orbital is full, the five electrons of 3d occupy the five orbitals singly with parallel spins (Hund's rule).

Sub-shell	n	l	m	s
1s^{2}	1	0	0	+\tfrac12 and -\tfrac12 in each orbital
2s^{2}	2	0	0	+\tfrac12 and -\tfrac12 in each orbital
2p^{6}	2	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
3s^{2}	3	0	0	+\tfrac12 and -\tfrac12 in each orbital
3p^{6}	3	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
4s^{2}	4	0	0	+\tfrac12 and -\tfrac12 in each orbital
3d^{5}	3	2	-2, -1, 0, +1, +2	+\tfrac12 in each

1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{5}

Exercise 2.5, 1 (d): zinc

Write down the electronic configuration of Zinc (atomic number 30) representing all four quantum numbers.

Zinc has Z=30, so it has 30 electrons. Fill them in Aufbau order.
1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}
Show the orbitals one by one. Each filled orbital holds two electrons of opposite spin.
1s^2,\ 2s^2,\ 2p^6,\ 3s^2,\ 3p^6,\ 4s^2,\ 3d_{xy}^{\uparrow\downarrow}\,3d_{yz}^{\uparrow\downarrow}\,3d_{xz}^{\uparrow\downarrow}\,3d_{x^2-y^2}^{\uparrow\downarrow}\,3d_{z^2}^{\uparrow\downarrow}

All five 3d orbitals are completely filled, so every orbital holds an \uparrow\downarrow pair. The last electron to enter has n=3, l=2, m=+2 and s=-\tfrac12.

Sub-shell	n	l	m	s
1s^{2}	1	0	0	+\tfrac12 and -\tfrac12 in each orbital
2s^{2}	2	0	0	+\tfrac12 and -\tfrac12 in each orbital
2p^{6}	2	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
3s^{2}	3	0	0	+\tfrac12 and -\tfrac12 in each orbital
3p^{6}	3	1	-1, 0, +1	+\tfrac12 and -\tfrac12 in each orbital
4s^{2}	4	0	0	+\tfrac12 and -\tfrac12 in each orbital
3d^{10}	3	2	-2, -1, 0, +1, +2	+\tfrac12 and -\tfrac12 in each orbital

1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}

Exercise 2.5, 2: configurations of ions

Write down the electronic configuration of the ions (a) \ce{Na+}, (b) \ce{Cl-}, (c) \ce{Ca^2+}.

(a) \ce{Na+}: sodium has 11 electrons and the ion has lost one, so it has 10 electrons. The 3s electron is removed.
\ce{Na+}:\ 1s^2\,2s^2\,2p^6
(b) \ce{Cl-}: chlorine (Z=17) is 1s^2\,2s^2\,2p^6\,3s^2\,3p^5. The ion gains one electron, so it has 18 electrons and the 3p sub-shell becomes full.
\ce{Cl-}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6
(c) \ce{Ca^2+}: calcium (Z=20) is 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2. The two electrons are removed from the outermost shell (4s), leaving 18 electrons.
\ce{Ca^2+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6

\ce{Na+}:\ 1s^2 2s^2 2p^6;\ \ce{Cl-}:\ 1s^2 2s^2 2p^6 3s^2 3p^6;\ \ce{Ca^2+}:\ 1s^2 2s^2 2p^6 3s^2 3p^6

2.5Shapes of orbitalsOften examined

Orbital: The region in space where the probability of finding an electron is maximum.

Different orbitals have different shapes in three-dimensional space. A p orbital has two lobes, and a d orbital has four lobes except d_{z^2}. An s orbital is spherically symmetrical.

s orbital: spherical.
p orbital: dumbbell shaped (two lobes).
d orbital: cloverleaf shaped (four lobes, except d_{z^2}).
f orbitals: they have complicated shapes.

Figure 2.7: Shapes of orbitals. s is a sphere, each p orbital is a dumbbell along one axis, four of the d orbitals are cloverleafs and d_{z^2} has two lobes with a ring.

Shapes of the s, p, d and f orbitals
Sub-shell	Number of orbitals	Shape	Orbitals
s	1	Spherical	s
p	3	Dumbbell	p_x,\,p_y,\,p_z
d	5	Cloverleaf (four lobes; d_{z^2} is different)	d_{xy},\,d_{yz},\,d_{xz},\,d_{x^2-y^2},\,d_{z^2}
f	7	Complicated	seven f orbitals

2.5Free radicals

Free radical: An atom or molecule with one or more unpaired electrons in its outer shell. This makes it highly reactive and capable of immediate reactions. Free radicals are activated atoms or molecules.

The electronic configuration of a free radical is the same as that of the atom; the radical is shown with a dot on the symbol. For example, a chlorine free radical has the same configuration as a chlorine atom, 1s^2\,2s^2\,2p^6\,3s^2\,3p^5, but it is written with a dot. Examples are the chlorine radical, the methyl radical and the bromine radical:

\mathrm{Cl}\cdot \qquad \cdot\mathrm{CH_3} \qquad \mathrm{Br}\cdot

A chlorine atom has one unpaired 3p electron, so it is a free radical. The chloride ion has no unpaired electron, so it is not a free radical.

2.6Ionization energyOften examined

Ionization energy (I.E): The minimum energy needed to remove the outermost (most loosely bonded) electron from a gaseous isolated atom.

\begin{aligned}\ce{Na(g) &-> Na+(g) + e-} && \text{I.E}=496\,\mathrm{kJ\,mol^{-1}}\\ \ce{Mg(g) &-> Mg+(g) + e-} && \text{I.E}=738\,\mathrm{kJ\,mol^{-1}}\end{aligned}

The atom must be in the gaseous state so that other factors, such as the heat of fusion, the bond dissociation energy and the heat of vaporization, do not influence the measured value. The atoms are also isolated, so the electron removed is held only by its own nucleus.

2.6.1Periodic trends of ionization energyOften examined

Ionization energies generally increase from left to right in a period, with slight anomalies (exceptions).

The reason is the increasing number of protons at each step, which strengthens the attraction of the nucleus for the electrons.
Electronic configuration also affects the ionization energy. In group IIIA the ionization energy is lower, because there is only one electron in the p orbital, which is an unstable configuration. In group VA the ionization energy is greater, because of the half-filled p orbitals.
From left to right in the main group elements the shielding effect stays constant, so it has no effect on the ionization energy values across a period.
The atomic radius decreases from left to right, which brings the electrons nearer to the nucleus, so the attraction and the ionization energy increase.

Down a group the ionization energy decreases. As the atom becomes bigger, the valence electrons are farther from the nucleus and the shielding effect keeps increasing, so it is easier to remove the valence electron.

First ionization energies of period 2 (data of Figure 2.8)
Period 2	Li	Be	B	C	N	O	F	Ne
I_1 (kJ/mol)	520.2	899.4	800.6	1086.4	1402.3	1313.9	1681.0	2080.6

First ionization energies of period 3 (data of Figure 2.8)
Period 3	Na	Mg	Al	Si	P	S	Cl	Ar
I_1 (kJ/mol)	495.8	737.7	577.6	786.4	1011.7	999.6	1251.1	1520.5

First ionization energy decreases down group 1 (data of Figure 2.8)
Group 1	Li	Na	K	Rb	Cs
I_1 (kJ/mol)	520.2	495.8	418.8	403.0	375.7

First ionization energy across periods 2 and 3

First ionization energy rises across a period with two small dips in each period: B below Be and O below N in period 2, and Al below Mg and S below P in period 3. Place 1 is Li or Na, 2 is Be or Mg, 3 is B or Al, 4 is C or Si, 5 is N or P, 6 is O or S, 7 is F or Cl and 8 is Ne or Ar. Data from the book's Figure 2.8.

2.6.2Factors influencing the ionization energyOften examined

1. Nuclear charge

As the nuclear charge increases from left to right in the periodic table, the attraction on the outermost electrons also increases, which in turn increases the ionization energy. This follows Coulomb's law.

2. Atomic radii

Moving from left to right in the periodic table the radius decreases, which strengthens the attraction of the nucleus on the electrons and increases the ionization energy. From top to bottom the size increases and the ionization energy decreases.

3. Shielding effect

The shielding effect plays a significant role in the ionization energy. As you move down a group the number of electron shells increases. The inner electron shells act as a shield, which reduces the effective nuclear charge felt by the outer electrons. This makes it easier to remove the outer electrons, so the ionization energy decreases down a group. From left to right the shielding effect remains constant and has no effect on the ionization energy.

4. Spin pair repulsion

When electrons are paired in a p orbital after the p sub-shell is half-filled, there is electron pair repulsion, which raises the energy of the electron. The increased energy makes the electron easier to remove, so the ionization energy decreases. For example, there is a slight decrease in ionization energy from nitrogen to oxygen in the second period.

As you move across a period the effective nuclear charge increases and draws the electrons closer to the nucleus. But spin pair repulsion acts against the increased nuclear attraction, which makes it slightly easier to remove an electron.

Factors influencing the ionization energy
Factor	Change	Effect on ionization energy
Nuclear charge	Increases across a period	Increases
Atomic radius	Decreases across a period; increases down a group	Increases across a period; decreases down a group
Shielding by inner shells	Increases down a group; constant across a period	Decreases down a group; no effect across a period
Spin pair repulsion	Appears when a p electron pairs up (N to O, P to S)	Slight decrease

2.6.3Deduction of position of an element in the periodic table by using ionization energy data

To deduce the position of an element in the periodic table from ionization energy data, we compare its ionization energy with the known trends and values of the periodic table. Ionization energy is the amount of energy required to remove an electron from a neutral atom, which gives a positively charged ion. The key points are:

1. Trends in ionization energy

Ionization energy generally decreases from top to bottom in a group and increases from left to right across a period, with a few exceptions, such as the first ionization energy decreasing from beryllium to boron and from magnesium to aluminium.

2. Periodic table organization

The periodic table is arranged so that elements with similar ionization energies are grouped together in the same column. This is based on the electronic arrangement of the elements, which determines their chemical properties.

3. Comparing ionization energies

To find the position of an element, compare its ionization energy with the known values of the periodic table. If the values match those expected for a position, the position is confirmed. The element is placed sequentially between its higher value and its lower value, keeping the lower value element on its left and the higher value element on its right.

Ionization energy of 1251\,\mathrm{kJ\,mol^{-1}} (across a period)

An element X has a first ionization energy of 1251\,\mathrm{kJ\,mol^{-1}}. Where does it lie in the periodic table?

Across a period the ionization energy increases from left to right. Look for the two known elements with values on either side of 1251: S (1000\,\mathrm{kJ\,mol^{-1}}) and Ar (1520\,\mathrm{kJ\,mol^{-1}}).
1000<1251<1520
The lower value goes on the left and the higher value on the right, so X lies between S and Ar in the same period.
\ce{S} \ \text{(left)} \ \ \text{X} \ \ \ce{Ar}\ \text{(right)}
The element between sulphur and argon is chlorine, whose value in the book's Figure 2.8 is 1251.1.
\text{X}=\ce{Cl}

\text{X lies between S and Ar in period 3: it is chlorine (group 17)}

Ionization energy of 738\,\mathrm{kJ\,mol^{-1}} (down a group)

An element Y has a first ionization energy of 738\,\mathrm{kJ\,mol^{-1}}. Where does it lie in the periodic table?

Down a group the ionization energy decreases. Look for the two elements with values on either side of 738: Be (899\,\mathrm{kJ\,mol^{-1}}) and Ca (590\,\mathrm{kJ\,mol^{-1}}).
899>738>590
The higher value is above and the lower value is below, so Y lies below Be and above Ca in the same group (group 2).
\ce{Be}\ \text{(top)} \ \ \text{Y} \ \ \ce{Ca}\ \text{(below)}
The element between beryllium and calcium in group 2 is magnesium, whose value is 737.7.
\text{Y}=\ce{Mg}

\text{Y lies between Be and Ca in group 2: it is magnesium (period 3)}

2.6.4Electronic configuration and position in the periodic table from successive ionization energy dataOften examined

The successive ionization energy values are an index of the valence of an atom (the number of valence electrons). A big gap shows that the next electron is being removed from the lower shell.

Sodium

The first and second ionization energies of sodium are 495\,\mathrm{kJ\,mol^{-1}} and 4560\,\mathrm{kJ\,mol^{-1}}. What does the gap show?

There is a very large jump between I_1 and I_2.
495\ \to\ 4560\ \mathrm{kJ\,mol^{-1}}
The second electron comes from the inner shell, which is closer to the nucleus and not shielded, so it is held far more strongly.
I_2\gg I_1
Only one electron is in the valence shell, so sodium belongs to group 1.
\text{1 valence electron}\Rightarrow\text{group 1}

\text{Sodium has one valence electron: group 1}

Magnesium

The first, second and third ionization energies of magnesium are 735, 1445 and 7730\,\mathrm{kJ\,mol^{-1}}. What do they show?

The gap between I_1 and I_2 is small.
735\to1445
The gap between I_2 and I_3 is very large, so the third electron is removed from the inner shell.
1445\to7730
Two electrons are in the valence shell and are removed easily, so magnesium belongs to group 2.
\text{2 valence electrons}\Rightarrow\text{group 2}

\text{Magnesium has two valence electrons: group 2}

Successive ionization energies in kJ/mol for the elements of period 3 (the bold value is where core electrons start to be removed)
Element	I_1	I_2	I_3	I_4	I_5	I_6	I_7
Na	495	4560
Mg	735	1445	7730
Al	580	1815	2740	11600
Si	780	1575	3220	4350	16100
P	1060	1890	2905	4950	6270	21200
S	1005	2260	3375	4565	6950	8490	27000
Cl	1255	2295	3850	5160	6560	9360	11000
Ar	1527	2665	3945	5770	7230	8780	12000

Along every row the values rise from I_1 to I_7, and each element shows a big jump where the electron being removed comes from the inner (core) shell. Down a column, from Na to Ar, the values generally rise as the nuclear charge increases.

Successive ionization energies of sodium, magnesium and aluminium

Each element shows a very large jump when the electron removed comes from the inner shell: after the first electron for Na, after the second for Mg and after the third for Al. The number of electrons before the jump is the number of valence electrons.

Example: deducing the group and configuration

The first four ionization energies of an element in period 3 (in \mathrm{kJ\,mol^{-1}}) are 580, 1815, 2740 and 11600. Find its group and valence configuration.

Look for the first very large jump.
2740\to11600
Three electrons are removed before the jump, so there are three valence electrons.
3\ \text{valence electrons}\Rightarrow\text{group 13}
In period 3 the valence shell is n=3. Group 13 has the configuration ns^2\,np^1 in the valence shell.
3s^2\,3p^1
This is aluminium: 1s^2\,2s^2\,2p^6\,3s^2\,3p^1.
\ce{Al}

\text{group 13, valence configuration } 3s^2\,3p^1\ (\ce{Al})

Concept assessment exercise 2.6: ionization energy trends

(1) How does the trend in ionization energy across a period reflect changes in atomic structure and effective nuclear charge? (2) What factors contribute to the general decrease in ionization energy down a group? (3) In terms of ionization energy, why do noble gases possess the highest values among their respective periods, and how does this relate to their chemical reactivity?

Question 1. Across a period, protons are added to the nucleus while electrons enter the same shell. The effective nuclear charge increases and the atomic radius decreases, so the outer electron is held more tightly and the ionization energy increases. Small dips occur where the sub-shell structure changes (a lone p electron, or spin pair repulsion after the half-filled p).
Z_{\text{eff}}\uparrow,\ r\downarrow\ \Rightarrow\ \text{I.E}\uparrow
Question 2. Down a group each element has one more shell, so the atom is bigger and the valence electron is farther from the nucleus. The inner shells shield it, which lowers the effective nuclear charge it feels, so it is removed more easily.
\text{more shells, larger radius, more shielding}\Rightarrow\text{I.E}\downarrow
Question 3. A noble gas is at the end of a period, with the highest nuclear charge, the smallest radius and a completely filled, stable valence shell. Removing an electron is very hard, so the ionization energy is the highest of the period. For the same reason noble gases do not readily lose, gain or share electrons, so they are chemically unreactive.
\text{highest I.E}\Rightarrow\text{inert}

\text{(1) increases; (2) larger radius and more shielding; (3) filled stable shell, so unreactive}

Worked example: the group of an element from a graph

An element Y (atomic number 20) has successive ionization energies I_1 to I_5 of 590,\ 1145,\ 4912,\ 6491 and 8153\ \mathrm{kJ\,mol^{-1}}. Find its group and check the answer against its electronic configuration.

Given

I_1 = 590\ \mathrm{kJ\,mol^{-1}}
I_2 = 1145\ \mathrm{kJ\,mol^{-1}}
I_3 = 4912\ \mathrm{kJ\,mol^{-1}}
I_4 = 6491\ \mathrm{kJ\,mol^{-1}}
I_5 = 8153\ \mathrm{kJ\,mol^{-1}}

To find

group of Y
valence configuration

Divide each value by the one before it to find the biggest jump.
\dfrac{1145}{590}=1.94,\quad\dfrac{4912}{1145}=4.29,\quad\dfrac{6491}{4912}=1.32,\quad\dfrac{8153}{6491}=1.26
The biggest jump is from I_2 to I_3. The first two electrons come from the valence shell. The third has to come from the inner shell, which is closer to the nucleus and strongly held.
\text{2 valence electrons}\Rightarrow\text{group 2}
Check with the configuration. With 20 electrons, Z=20 gives 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2. The valence shell has two electrons (4s), as found.
4s^2

Successive ionization energies of element Y

The large jump comes after the second electron, so Y has two valence electrons.

\text{Y is in group 2, with valence configuration }4s^2

2.7Mass spectrometry and its principleOften examined

Mass spectrometry: A powerful analytical technique used to determine the molecular composition and structure of a sample by measuring the mass-to-charge ratio (m/z) of its ions. It is widely used in chemistry, biology and physics.

Principle of mass spectrometry

Mass spectrometry involves three fundamental steps: ionization, mass analysis and detection. The sample is first ionized, which converts its atoms or molecules into charged ions. The ions are then separated according to their mass-to-charge ratio (m/z) and detected, which gives a mass spectrum. The spectrum provides information about the composition and abundance of the sample's molecules, or about the isotopic composition of an element.

The three steps of mass spectrometry
Step	What happens
1. Ionization	The atoms or molecules of the sample become charged ions
2. Mass analysis	The ions are separated according to their m/z ratio
3. Detection	The separated ions are detected and a mass spectrum is produced

2.7.2Working of the mass spectrometerOften examined

The working of a mass spectrometer can be broken down into four steps.

Ionization

The sample is introduced into the ion source, where it is ionized. This can involve bombarding the sample with high-energy electrons (EI, electron ionization), introducing it to a high-voltage field (ESI, electrospray ionization), or using a laser to vaporize it (MALDI, Matrix Assisted Laser Desorption/Ionization).

Ion separation

The resulting ions are accelerated and directed into the mass analyzer. The mass analyzer separates the ions according to their m/z ratio, which allows ions of different masses to be focused and detected at different times.

Detection

As the ions leave the mass analyzer they hit the detector. The detector records the time it takes for each ion to reach it, and this data is used to calculate the m/z ratio of each ion.

Data analysis

The information collected by the detector is used to construct a mass spectrum. The x-axis of the spectrum represents the m/z ratio, while the y-axis represents the abundance of ions at each m/z ratio. The peaks in the spectrum correspond to the different ions present in the sample, and their positions reveal their masses.

Figure 2.9: Mass spectrometer. The sample is vaporized and ionized, the ions are accelerated, the magnetic field deflects the lightest ions most, and the detector records the ions that reach it.

Working of a mass spectrometer in four steps
Step	Process	Method or part
Ionization	Sample is turned into ions	EI (high-energy electrons), ESI (high-voltage field), MALDI (laser)
Ion separation	Ions are accelerated and separated by m/z	Mass analyzer (magnetic field in the figure)
Detection	Time for each ion to reach the detector is recorded, giving m/z	Detector
Data analysis	Mass spectrum is built	x-axis m/z, y-axis abundance

2.7.3Applications of mass spectrometry and relative atomic massOften examined

Mass spectrometry has a wide range of applications:

finding the relative abundance of the isotopes of an element,
identifying unknown compounds,
quantifying the amounts of specific substances,
elucidating (working out) molecular structures,
studying biomolecules such as proteins and nucleic acids.

Average atomic mass

The average atomic mass of an element can be calculated from the relative abundance of all its isotopes.

\text{Average atomic mass}=\dfrac{(\text{mass no.}\times\text{relative abundance})+(\text{mass no.}\times\text{relative abundance})+\cdots}{100}

Average atomic mass

Relative abundance is taken in percent, so the sum is divided by 100

Non-integer relative atomic masses

The atomic masses written in the periodic table are usually not whole numbers. These non-integer atomic masses result from the relative abundance of the different isotopes of an element. Relative abundance is usually taken in percentage units. The abundances are multiplied by the mass numbers, and the sum is divided by 100.

Mass spectrum of chlorine (abundance in %)

Figure of the chlorine isotope pattern: two peaks at m/z=35 and m/z=37. The book also marks the heights as relative intensities, 100 for the tallest peak and about 32 for the other.

Relative atomic mass of chlorine

Chlorine has two isotopes, chlorine-35 and chlorine-37, with abundances 75.76\% and 24.24\% respectively. Calculate the relative atomic mass of chlorine.

Multiply each mass number by its percentage abundance.
35\times75.76=2651.6\,;\quad 37\times24.24=896.88
Add the two products.
2651.6+896.88=3548.48
Divide by 100.
\dfrac{3548.48}{100}=35.4848\approx35.48

A_r(\ce{Cl})=35.48\ \mathrm{amu}

Concept assessment exercise 2.7: average atomic mass of magnesium

Calculate the average atomic mass of Mg using the mass spectrogram: ^{24}\mathrm{Mg} 79\%, ^{25}\mathrm{Mg} 10\%, ^{26}\mathrm{Mg} 11\%.

Read the mass numbers and abundances from the peaks.
24:79\%,\quad 25:10\%,\quad 26:11\%
Multiply each mass number by its abundance.
24\times79=1896,\quad 25\times10=250,\quad 26\times11=286
Add and divide by 100.
\dfrac{1896+250+286}{100}=\dfrac{2432}{100}=24.32

A_r(\ce{Mg})=24.32\ \mathrm{amu}

Mass spectrograph of magnesium (abundance in %)

Figure 2.10: Mass spectrograph of magnesium. Three isotopes appear at m/z=24, 25 and 26 with abundances 79\%, 10\% and 11\%.

Worked example: finding the abundance from the average atomic mass

Chlorine has two isotopes, ^{35}\ce{Cl} and ^{37}\ce{Cl}, and its average atomic mass is 35.5. Find the relative abundance of each isotope.

Let the abundance of ^{35}\ce{Cl} be x\%. The abundances add up to 100\%, so ^{37}\ce{Cl} is (100-x)\%.
Put these into the average atomic mass formula.
\dfrac{35x+37(100-x)}{100}=35.5
Multiply by 100 and expand.
35x+3700-37x=3550
Collect the terms and solve.
-2x=-150\ \Rightarrow\ x=75
Check: \dfrac{35\times75+37\times25}{100}=\dfrac{2625+925}{100}=35.5. The answer lies between 35 and 37 and is nearer to 35, which is the more abundant isotope.

^{35}\ce{Cl}:\ 75\%,\qquad ^{37}\ce{Cl}:\ 25\%

2.8Electronic configuration of modern electronic materials

Semiconductor: A material whose electrical conductivity lies between that of conductors (like metals) and insulators (like non-metals).

The electronic configuration of a semiconductor plays a crucial role in determining its behaviour. Silicon (Si) is the book's example.

The electronic configuration of silicon

Silicon has atomic number 14, so it has 14 protons and 14 electrons. The electronic configuration describes how these electrons are distributed in the energy levels (shells).

The first energy level holds up to 2 electrons (1s^2).
The second energy level holds up to 8 electrons (2s^2\,2p^6).
The third energy level holds 4 electrons (3s^2\,3p^2).

\ce{Si}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^2

Role of electronic configuration in the type of semiconductor

In silicon the outermost energy level is the third level. The 3s and 3p sub-levels together can hold a maximum of 8 electrons, but in its ground state silicon has only 4 valence electrons, the electrons in the outermost energy level. This configuration gives silicon the unique properties that make it a suitable material for semiconductors.

The arrangement of silicon's electrons creates a band structure. In the valence band the electrons are tightly bound to the atoms and cannot move freely. In the conduction band there are no electrons in the ground state. The energy gap between the valence and the conduction band, the band gap, is relatively small, so electrons can be promoted from the valence band to the conduction band. This creates electron-hole pairs and the process is called excitation.

Band structure of a semiconductor: the band gap is small, so an electron promoted from the valence band to the conduction band leaves a hole behind (an electron-hole pair).

Intrinsic and extrinsic semiconductors

Intrinsic semiconductors are pure materials like silicon or germanium.
Extrinsic semiconductors have controlled impurities added on purpose. These impurities are called dopants. They either introduce extra electrons (n-type) or create holes that behave like positive charge carriers (p-type). The process is called doping: some silicon atoms are substituted with other elements, such as phosphorus for n-type or boron for p-type.

Doped silicon
	n-type	p-type
Dopant (example)	Phosphorus	Boron
What the dopant adds	Excess electrons that move freely in the conduction band	Holes in the valence band that move and behave as positive charge carriers
Effect	Improves electrical conductivity	Improves electrical conductivity

By strategically combining n-type and p-type materials, engineers can create diodes, transistors and other electronic components that use the unique properties of semiconductors to control the flow of electric current. This forms the basis of modern electronics and technology.

Concept assessment exercise 2.8: semiconductors

(1) What is the key characteristic that defines a material as a semiconductor? (2) Can you name two commonly used semiconductor materials in electronic devices?

Question 1. Its electrical conductivity lies between that of a conductor (metal) and an insulator (non-metal). This comes from a small band gap, so electrons can be promoted from the valence band to the conduction band.
\text{conductivity between a metal and an insulator}
Question 2. The book names silicon and germanium (pure, or intrinsic, semiconductors).
\ce{Si},\ \ce{Ge}

\text{(1) conductivity between conductors and insulators; (2) silicon and germanium}

Formula sheet

Mass number

A = Z + N

A is the mass (nucleon) number, Z the atomic (proton) number and N the number of neutrons. Written in the book as P+N=A.

no unit

Electrons in a cation

\text{electrons}=Z-\text{charge on the cation}

The number of electrons is the atomic number minus the magnitude of the positive charge.

no unit

Electrons in an anion

\text{electrons}=Z+\text{charge on the anion}

The number of electrons is the atomic number plus the magnitude of the negative charge.

no unit

Values of the principal quantum number

n=1,\,2,\,3,\ldots

n numbers the shell and equals the period number.

no unit

Values of the azimuthal quantum number

l=0,\,1,\,2,\ldots,\,(n-1)

l=0,1,2,3 are the s, p, d and f sub-shells.

no unit

Values of the magnetic quantum number

m=-l,\ldots,0,\ldots,+l

m gives the orbitals of a sub-shell: 1 for s, 3 for p, 5 for d and 7 for f.

no unit

Values of the spin quantum number

s=+\tfrac12\ \text{or}\ -\tfrac12

The two opposite spins of the two electrons of an orbital.

no unit

The n + l rule

\text{higher } (n+l)\ \Rightarrow\ \text{higher energy}

For equal n+l the orbital with the smaller n has the lower energy.

no unit

Ionization energy

\ce{M(g) -> M+(g) + e-}

The minimum energy needed to remove the outermost electron from a gaseous isolated atom.

Unit: \mathrm{kJ\,mol^{-1}}

Average atomic mass

\text{Average atomic mass}=\dfrac{\sum(\text{mass no.}\times\text{relative abundance})}{100}

The relative abundance of each isotope is taken in percent.

atomic mass unit (amu)

Key terms

Electron: A negatively charged sub-atomic particle (charge -1.6022\times10^{-19}\,\mathrm{C}, relative charge -1) with a mass of 9.1095\times10^{-31}\,\mathrm{kg}; discovered by J. J. Thomson.
Proton: A positively charged sub-atomic particle (relative charge +1) that is 1836 times heavier than the electron, with a mass of 1.6727\times10^{-27}\,\mathrm{kg}.
Neutron: An uncharged particle of the nucleus, slightly heavier than a proton (1.6750\times10^{-27}\,\mathrm{kg}), not deflected by electric or magnetic fields.
Atomic number (proton number): The number of protons in the nucleus of an atom, Z; equal to the number of electrons in a neutral atom.
Mass number (nucleon number): The total number of protons and neutrons in the nucleus, A=Z+N.
Radius of an atom: The average distance of the nucleus from the outermost electrons.
Quantum numbers: A set of four numbers (n, l, m, s) that give the complete address of an electron in an atom.
Principal quantum number (n): The number of the shell, n=1,2,3,\ldots; it also gives the period number.
Azimuthal quantum number (l): Gives the sub-shell and the shape of the orbital; l=0 to n-1.
Magnetic quantum number (m): Gives the orientation of an orbital in space; m=-l to +l.
Spin quantum number (s): The spin of an electron, +\tfrac12 (clockwise, \uparrow) or -\tfrac12 (anticlockwise, \downarrow).
Orbital: The region of space around the nucleus where the probability of finding an electron is maximum.
Degenerate orbitals: Orbitals of the same sub-shell that have equal energy.
Electronic configuration: The distribution of electrons around the nucleus in orbits and orbitals.
Aufbau principle: Electrons fill the orbitals in order of increasing energy: low energy orbitals first.
The n + l rule: The higher the value of n+l, the higher the energy of the orbital.
Pauli exclusion principle: No two electrons in an atom can have the same four quantum numbers; two electrons in an orbital have opposite spins.
Hund's rule: Degenerate orbitals are filled singly, with parallel spins, before any pairing occurs.
Spin pair repulsion: The repulsion between two electrons paired in the same orbital, which raises their energy.
Free radical: An atom or molecule with one or more unpaired electrons, which makes it highly reactive.
Ionization energy: The minimum energy needed to remove the outermost electron from an isolated gaseous atom.
Shielding effect: The reduction of the nuclear attraction felt by the outer electrons because of the inner shells.
Successive ionization energies: The energies for removing the first, second, third... electron from an atom in turn; a very large jump shows that an inner shell electron is being removed.
Mass spectrometry: A technique that finds the composition of a sample by measuring the mass-to-charge ratio (m/z) of its ions.
Mass spectrum: A plot of ion abundance (y-axis) against m/z (x-axis); each peak is one kind of ion.
Relative atomic mass: The average mass of the isotopes of an element weighted by their abundance, usually not a whole number.
Emission spectrum: The set of spectral lines of light given out when excited electrons return to lower energy levels.
Semiconductor: A material whose electrical conductivity lies between that of conductors and insulators, such as silicon.
Band gap: The energy gap between the valence band and the conduction band of a semiconductor.
Doping: Adding controlled impurities (dopants) to a semiconductor: phosphorus gives n-type and boron gives p-type.

Multiple choice questionsp. 36

Choose the correct answer.

What does the electron configuration 1s^2\,2s^2\,2p^6 represent?

ACarbon (\ce{C})
BOxygen (\ce{O})
CNeon (\ce{Ne})Answer
DHelium (\ce{He})

Why

The configuration holds 2+2+6=10 electrons. A neutral atom with 10 electrons has atomic number 10, which is neon (\ce{Ne}). Carbon has 6 electrons and oxygen 8, and helium has only 1s^2 with 2 electrons, so options A, B and D are wrong.

Note. The book prints no key for this question; the answer was worked out from the book's text.

Which subshell can hold a maximum of 10 electrons?

As
Bp
CdAnswer
Df

Why

Each orbital holds two electrons, and a subshell has 1 orbital for s, 3 for p, 5 for d and 7 for f. The capacities are therefore 2, 6, 10 and 14 electrons, so only the d subshell holds 10.

Note. The book prints no key for this question; the answer was worked out from the book's text.

iii

How many unpaired electrons are there in the electron configuration 3d^5?

A0
B1
C2
D5Answer

Why

In 3d^5 the five electrons go into the five 3d orbitals one each, with parallel spins, before any pairing (Hund's rule). All five electrons are therefore unpaired.

Options A to C are wrong: 0 would mean that every electron is paired, and 1 or 2 would mean that pairs form while empty orbitals remain, which Hund's rule does not allow.

Note. The book prints no key for this question; the answer was worked out from the book's text.

Which element has the electron configuration 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^8?

AIron (\ce{Fe})
BZinc (\ce{Zn})
CNickel (\ce{Ni})Answer
DCopper (\ce{Cu})

Why

Add the superscripts: 2+2+6+2+6+2+8=28. A neutral atom with 28 electrons has atomic number 28, which is nickel.

Iron has 26 electrons, copper 29 and zinc 30 (with a full 3d^{10}), so none of them matches.

Note. The book prints no key for this question; the answer was worked out from the book's text.

What is the electron configuration of a chloride ion (\ce{Cl-})?

A1s^2\,2s^2\,2p^6\,3s^2\,3p^6Answer
B1s^2\,2s^2\,2p^6\,3s^2\,3p^5
C1s^2\,2s^2\,2p^6\,3s^2\,3p^7
D1s^2\,2s^2\,2p^6\,3s^2\,3p^8

Why

Chlorine has Z=17, so a neutral atom has 17 electrons. The ion \ce{Cl-} has gained one electron, giving 18 electrons: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6, the same configuration as argon.

Option B is the configuration of the neutral atom. Options C and D are impossible because the 3p subshell holds at most 6 electrons.

Note. The book prints no key for this question; the answer was worked out from the book's text.

Magnesium has how many isotopes?

A1
B2
C3Answer
D4

Why

Magnesium has three isotopes, \ce{^{24}Mg}, \ce{^{25}Mg} and \ce{^{26}Mg}. They appear as three peaks in the book's mass spectrum of magnesium (Concept Assessment Exercise 2.7). The isotopes differ in mass number but have the same atomic number 12, so the three peaks belong to one element.

Note. The book prints no key for this question. The answer follows from the three peaks in the magnesium spectrum of Concept Assessment Exercise 2.7 on page 32.

vii

What is the basis on which molecular ions are split in mass spectrometry?

AElectric field
BMagnetic fieldAnswer
CVelocity of ions
DGrid strength

Why

After ionization the ions are accelerated and then pass through a magnetic field. The field deflects each ion by an amount that depends on its mass-to-charge ratio, so lighter ions are deflected most (Figure 2.9).

The electric field only accelerates the ions, and velocity is not the basis of the separation. Grid strength is not part of the method.

Note. The book prints no key for this question; the answer was worked out from the book's text.

viii

Which orbital is sausage shaped?

As
BpAnswer
Cd
Df

Why

The p orbital has two lobes, the dumbbell shape that the book describes. An s orbital is spherical, a d orbital has four lobes (except d_{z^2}) and f orbitals have complicated shapes, so none of these is the elongated two-lobed shape.

Note. The book prints no key. 'Sausage' is not the book's word: it describes p orbitals as dumbbell shaped, which is the intended shape.

Electrons tend to reside separately in the degenerate orbitals. This is called:

AAufbau principle
BPauli exclusion principle
CHund's ruleAnswer
DFajan's rule

Why

Hund's rule says that when degenerate orbitals are available, electrons enter them singly with the same spin before any pairing takes place, which keeps the electrons apart and lowers their repulsion.

The Aufbau principle gives the order of filling and the Pauli principle limits each orbital to two electrons of opposite spin. Fajan's rule is not a rule of electronic configuration in this unit.

Note. The book prints no key for this question; the answer was worked out from the book's text.

Which quantum number explains the splitting of orbitals in three dimensional space?

APrincipal quantum number
BAzimuthal quantum number
CMagnetic quantum numberAnswer
DSpin quantum number

Why

The magnetic quantum number m gives the orientation of an orbital in space. For a subshell with azimuthal number l, it takes the values -l to +l, so a p subshell has three orientations (p_x, p_y and p_z).

The principal number gives the shell, the azimuthal number gives the subshell shape and the spin number gives the direction of the electron's spin, so none of options A, B or D explains the splitting in space.

Note. The book prints no key for this question; the answer was worked out from the book's text.

Short questionsp. 36–37

Give short answer.

i3 marks

What is the importance of Bohr's atomic model in modern atomic structure?

Solution

Bohr's model was the first to describe the atom with quantized energy levels, or shells, of fixed energy. This idea is the basis of modern atomic structure: an electron can have only certain energies, and it moves between shells by absorbing or releasing a definite amount of energy.

Bohr calculated the radius and the energy of the electron in each shell of the hydrogen atom and explained the emission spectrum of hydrogen by jumps between shells. The principal quantum number n is the expression of Bohr's shell in quantum form.

The model could not explain the fine spectral lines seen with high-resolution spectrophotometers, nor their splitting in an electric field (Stark effect) or a magnetic field (Zeeman effect). This led to subshells and orbitals, and then to the four quantum numbers.

ii3 marks

Explain the charge and mass of fundamental sub-atomic particles.

Solution

Charges and masses of the fundamental particles
Particle	Relative charge	Charge	Mass
Electron	-1	-1.6022\times10^{-19}\,\mathrm{C}	9.1095\times10^{-31}\,\mathrm{kg}
Proton	+1	+1.6022\times10^{-19}\,\mathrm{C}	1.6727\times10^{-27}\,\mathrm{kg}
Neutron	0	0	1.6750\times10^{-27}\,\mathrm{kg}

The electron and the proton carry charges of equal size and opposite sign, so a neutral atom has equal numbers of them. In an electric field the electron is deflected towards the positive plate and the proton towards the negative plate, while the neutron is not deflected.

\frac{m_\text{p}}{m_\text{e}}=\frac{1.6727\times10^{-27}\,\mathrm{kg}}{9.1095\times10^{-31}\,\mathrm{kg}}\approx 1836

The proton is therefore about 1836 times heavier than the electron, so almost all the mass of an atom is in its nucleus, which holds the protons and neutrons.

iii3 marks

Explain the periodic trends of atomic radius with justification.

Solution

The atomic radius is the average distance from the nucleus to the outermost electrons. Across a period, from left to right in the main groups, the radius decreases. The number of protons increases, which raises the nuclear charge and pulls the outer electrons closer. The shielding effect remains constant across a period, so nothing offsets this increased pull.

Down a group the radius increases. Each step down adds one shell, so the outermost electrons are further from the nucleus. The intervening inner electrons also increase the shielding, which reduces the pull of the nucleus on the outer electrons.

The book also states that in the transition metal series the radius increases as electrons enter the inner d-orbitals.

Note. The book states that the radius increases across the transition metal series. This is not the usual pattern, and the book gives no clear figures for it, so check it against the source before quoting it in an answer.

iv3 marks

How does shielding effect change the radius of an atom in a group from top to bottom?

Solution

Going down a group, each element has one more shell than the element above it, so the outermost electrons lie further from the nucleus and the atom becomes larger.

The shielding effect also increases. The inner electrons between the nucleus and the outer shell act as a screen, so the outer electrons feel a smaller effective nuclear charge. This weaker attraction lets the outer shell spread out, so the radius increases from top to bottom. Because the outer electrons are held less tightly, the ionization energy also decreases down a group.

v3 marks

Why is the cation always smaller than the parent atom, and anion is bigger than the parent atom?

Solution

A cation is smaller than its parent atom. When an atom loses its outer electron, the nuclear charge stays the same but the number of electrons falls. The electron-electron repulsion in the valence shell decreases and the effective nuclear charge on the remaining electrons increases, so the electron cloud is pulled inwards.

\ce{Na}\;(1s^2\,2s^2\,2p^6\,3s^1)\longrightarrow\ce{Na+}\;(1s^2\,2s^2\,2p^6)

An anion is bigger than its parent atom. An extra electron is added to the valence shell, so the electron-electron repulsion increases while the nuclear charge is unchanged. The electrons spread out and the ion is larger.

\ce{Cl}\;(1s^2\,2s^2\,2p^6\,3s^2\,3p^5)+e^-\longrightarrow\ce{Cl-}\;(1s^2\,2s^2\,2p^6\,3s^2\,3p^6)

vi3 marks

Explain how different spectral series originate in hydrogen spectrum.

Solution

The hydrogen spectrum is made of lines produced when electrons jump between shells. An electron that absorbs energy moves to a higher shell; when it falls back to a lower shell it emits the energy as light. Only certain jumps are possible because the shells have fixed energies, so only certain wavelengths appear as lines.

All the jumps that end in the same shell form one series of lines, and each series is named after the shell in which the electron ends.

Spectral series of the hydrogen atom
Series	Electron falls to	Region
Lyman	n=1	Ultraviolet
Balmer	n=2	Visible
Paschen	n=3	Infrared
Brackett	n=4	Infrared
Pfund	n=5	Infrared

For example, a jump from n=3 to n=2 gives a line of the Balmer series, and a jump from n=2 to n=1 gives a line of the Lyman series.

Note. The book spells the series 'Bracket' and 'p fund'. They are the Brackett and Pfund series.

vii3 marks

Explain magnetic quantum number in detail. Why do s orbital have only one value of the magnetic quantum number?

Solution

The magnetic quantum number m describes the orientation of an orbital in space, and it explains the splitting of spectral lines in a magnetic field (the Zeeman effect). Its values run from -l to +l through zero, where l is the azimuthal quantum number of the subshell.

Values of the magnetic quantum number for each subshell
Subshell	l	Values of m	Number of orbitals
s	0	0	1
p	1	-1,\;0,\;+1	3 (p_x, p_y, p_z)
d	2	-2,\;-1,\;0,\;+1,\;+2	5
f	3	-3 to +3	7

For an s orbital l=0, so the only possible value is m=0. The s orbital has only one orientation: it is spherically symmetrical, so its orientation makes no difference and a magnetic field does not split it. A p orbital has three orientations along the x, y and z axes.

viii3 marks

Why it is so that two electrons with same spin cannot reside in an orbital?

Solution

By the Pauli exclusion principle, no two electrons in an atom can have the same values of all four quantum numbers. The first three numbers (n, l and m) fix the orbital, so two electrons in the same orbital must differ in spin. There are only two spin values, +\tfrac12 and -\tfrac12, so an orbital holds at most two electrons and they must have opposite spins.

Two electrons with the same spin would have identical sets of four quantum numbers, which is not allowed. Electrons with similar spin also repel each other (spin pair repulsion), and a pair with opposite spins cancels its spin magnetic field.

(3,\,1,\,-1,\,+\tfrac12)\text{ and }(3,\,1,\,-1,\,-\tfrac12)\text{ allowed};\qquad (3,\,1,\,-1,\,+\tfrac12)\text{ twice not allowed}

ix3 marks

Why 3d orbital has greater energy than 4S orbital? Explain (n + l) rule.

Solution

The n+l rule gives the energy of an orbital. The orbital with the higher n+l value has the higher energy, and when two orbitals have the same n+l, the one with the lower n has the lower energy.

4s:\;\; n+l=4+0=4\qquad\qquad 3d:\;\; n+l=3+2=5

The 3d orbital has n+l=5 and the 4s orbital n+l=4, so 3d has the greater energy even though its n is smaller. The 4s orbital is therefore filled before 3d, following the Aufbau order 1s<2s<2p<3s<3p<4s<3d.

x2026 Annual I, Set B, Q.2 (x)3 marks

Calculate the average atomic mass of magnesium keeping in view the relative abundance of its isotopes.

Figure 2.10: Mass spectrum of magnesium, with relative abundance in percent (Concept Assessment Exercise 2.7)

Solution

Given

{}^{24}\text{Mg} = 79\,\%
{}^{25}\text{Mg} = 10\,\%
{}^{26}\text{Mg} = 11\,\%

To find

\bar{A}, the average atomic mass of Mg

Average atomic mass is the sum of each mass number multiplied by its relative abundance, divided by 100 when abundances are in percent.
\bar{A}=\dfrac{\sum(\text{mass number}\times\text{relative abundance})}{100}
Multiply each mass number by its percentage abundance.
24\times79=1896\qquad 25\times10=250\qquad 26\times11=286
Add the products and divide by 100.
\bar{A}=\dfrac{1896+250+286}{100}=\dfrac{2432}{100}=24.32\,\mathrm{amu}

\bar{A}_\text{Mg}=24.32\,\mathrm{amu}

Note. The question as printed gives no abundances. The data are the magnesium spectrum of Concept Assessment Exercise 2.7 (page 32). The book prints no answer for this question. The periodic table value is about 24.31 amu; the rounded percentages give 24.32 amu.

xi3 marks

What is effective nuclear charge?

Solution

The effective nuclear charge is the net positive charge that an outer electron actually feels. The full attraction of the nucleus is reduced by the inner electrons, which shield the outer electrons, so the effective charge is smaller than the number of protons.

Across a period the number of protons increases while the shielding stays about the same, so the effective nuclear charge increases. This pulls the outer electrons closer, which decreases the atomic radius and increases the ionization energy. It is also the reason why a cation is smaller than its parent atom.

xii3 marks

What is the importance of electronic configuration in semiconductor materials?

Solution

The electronic configuration decides whether a material can act as a semiconductor. Silicon has the configuration 1s^2\,2s^2\,2p^6\,3s^2\,3p^2, so its outermost (valence) shell holds four electrons in 3s^2\,3p^2. These electrons form a valence band of tightly bound electrons and a conduction band, separated by a small band gap. Electrons can be promoted across the gap, creating electron-hole pairs, so silicon conducts only partly, which is what makes it a semiconductor.

Doping changes the conductivity. Substituting phosphorus, which has five valence electrons, gives extra electrons, an n-type semiconductor in which phosphorus is the electron donor. Substituting boron, which has three valence electrons, creates holes that act as positive charge carriers, a p-type semiconductor in which boron is the electron acceptor.

Combining n-type and p-type materials gives diodes, transistors and other components that control the flow of electric current, which is the basis of modern electronics.

Note. The book reverses the terms: it calls n-type silicon an 'electron accepter' and p-type silicon an 'electron donor'. The answer uses the correct terms.

xiii3 marks

Why is there a large I.E gap between second and third values in Mg atoms?

Solution

Magnesium has the configuration 1s^2\,2s^2\,2p^6\,3s^2. The first two ionization energies remove the two 3s valence electrons, and they are small: 735 and 1445 \mathrm{kJ\,mol^{-1}}. The third electron has to come from the inner 2p shell, which is much closer to the nucleus and less shielded, so its ionization energy is 7730 \mathrm{kJ\,mol^{-1}}.

\begin{aligned}\text{I.E}_3-\text{I.E}_2&=7730-1445=6285\,\mathrm{kJ\,mol^{-1}}\\ \text{compared with}\quad\text{I.E}_2-\text{I.E}_1&=1445-735=710\,\mathrm{kJ\,mol^{-1}}\end{aligned}

The large jump between the second and third values shows that the third electron is removed from an inner shell. Magnesium has only two valence electrons, so it belongs to group 2 of the periodic table.

Long answer questionsp. 37–38

38 marks

Explain quantum numbers in detail.

Solution

Introduction

Four quantum numbers together give the complete address of an electron in an atom. n, l and m describe the orbital and s describes the spin of the electron. No two electrons in an atom can have the same four values.

Principal quantum number (n)

The principal quantum number is the expression of Bohr's shell in quantum form. It takes the values n=1,2,3,\dots, which give the shells K, L, M and N. The energy and the size of a shell increase with n. It also gives the period number of an element, because each period starts a new shell.

Azimuthal quantum number (l)

The azimuthal quantum number describes the shape of an orbital and specifies the subshell within a shell. It takes the values l=0,1,2,\dots,n-1, where l=0,1,2,3 denote the s, p, d and f subshells. It gives the angular momentum of the electron and is related to the blocks of the periodic table.

Magnetic quantum number (m)

The magnetic quantum number gives the orientation of an orbital in space and explains the splitting of spectral lines in a magnetic field. Its values run from -l to +l through zero: one value for s, three for p, five for d and seven for f.

Spin quantum number (s)

The spin quantum number describes the spin of the electron about its own axis. Its values are +\tfrac12 (clockwise, symbol \uparrow) and -\tfrac12 (anticlockwise, symbol \downarrow). An orbital holds two electrons only if their spins are opposite, so that their spin magnetic fields cancel.

Summary

Possible quantum numbers and electron capacity of the first four shells
Shell n	Subshells (l)	Values of m	Orbitals and electrons	Total electrons
1 (K)	l=0 (1s)	0	1 orbital, 2 electrons	2
2 (L)	l=0,1 (2s, 2p)	0; -1,0,+1	1 + 3 orbitals; 2 + 6 electrons	8
3 (M)	l=0,1,2 (3s, 3p, 3d)	0; -1,0,+1; -2,\dots,+2	1 + 3 + 5 orbitals; 2 + 6 + 10 electrons	18
4 (N)	l=0,1,2,3 (4s, 4p, 4d, 4f)	0; -1,0,+1; -2,\dots,+2; -3,\dots,+3	1 + 3 + 5 + 7 orbitals; 2 + 6 + 10 + 14 electrons	32

Example

An electron in a 3p orbital with m=-1 and spin +\tfrac12 has the set n=3, l=1, m=-1, s=+\tfrac12. The two electrons of one 3p orbital in the book differ only in spin: electron A has s=+\tfrac12 and electron B has s=-\tfrac12.

Note. The book's table labels n = 4 as the L-shell. The shell with n = 4 is the N-shell, so the answer uses N.

42025 Annual I, Set B, Q.38 marks

What is mass spectrometry? Explain its working and tell how the data is analysed.

Solution

Principle

Mass spectrometry separates ions according to their mass-to-charge ratio, m/z. The sample is ionized, the ions are separated by their m/z and then detected. The result is a mass spectrum, which gives the composition and the isotopic abundance of the sample.

Working

Ionization: the sample is introduced into the ion source and ionized, for example by bombarding it with high-energy electrons (EI), with a high-voltage field (ESI) or with a laser (MALDI).
Ion separation: the ions are accelerated and directed into the mass analyser, where a magnetic field deflects them. Ions of different m/z are deflected by different amounts, so lighter ions are deflected most.
Detection: as the ions leave the analyser they hit the detector, which records the time each ion takes to arrive. This time is used to calculate the m/z of each ion.

Figure 2.9: Working of a mass spectrometer, from ionization to detection

Data analysis

The detector data are used to build a mass spectrum. The x-axis shows the m/z ratio and the y-axis shows the abundance of ions at each m/z. Each peak corresponds to an ion in the sample, its position gives the mass of that ion and its height gives the relative abundance.

Applications: relative atomic mass

The spectrum gives the relative abundance of the isotopes of an element. The average atomic mass is found by a weighted average:

\begin{aligned}&\text{Average atomic mass}\\ &=\dfrac{\begin{aligned}&(\text{mass number}\times\text{relative abundance})\\ &+(\text{mass number}\times\text{relative abundance})\\ &+\dots\end{aligned}}{100}\end{aligned}

Chlorine has \ce{^{35}Cl} at 75.76% and \ce{^{37}Cl} at 24.24%, so:

\dfrac{(35\times75.76)+(37\times24.24)}{100}=\dfrac{2651.6+896.88}{100}=35.48\,\mathrm{amu}

Other applications

The book also lists the identification of unknown compounds, the quantification of specific substances, the elucidation of molecular structures and the study of biomolecules such as proteins and nucleic acids.

58 marks

Describe different rules adopted for electronic configuration of elements.

Solution

Aufbau principle

Electrons fill orbitals in order of increasing energy, low-energy orbitals first. The order used is 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d\dots. The 4s orbital is filled before the 3d orbital.

n + l rule

The energy of an orbital increases with the value of n+l. When two orbitals have the same n+l, the one with the lower n has the lower energy. For example, 4s has n+l=4 and 3d has n+l=5, so 3d is filled after 4s. Sometimes an electron enters a higher shell before a lower one for this reason.

Pauli exclusion principle

No two electrons in an atom can have the same four quantum numbers. An orbital holds at most two electrons, and they must have opposite spins, +\tfrac12 and -\tfrac12.

Hund's rule

When degenerate orbitals are available, electrons occupy them singly with the same spin before any pairing takes place. For two electrons in a p subshell the configuration is p_x^{\uparrow}\,p_y^{\uparrow}, and for three it is p_x^{\uparrow}\,p_y^{\uparrow}\,p_z^{\uparrow}. Half-filled and completely filled subshells are more stable, as in the configurations of chromium and copper.

Example: manganese (Z = 25)

Writing the configuration

A neutral manganese atom has 25 electrons, so the configuration must hold 25 electrons. Fill the orbitals in Aufbau order.
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^5
Check the total: 2+2+6+2+6+2+5=25. The five 3d electrons go into five different orbitals with parallel spins (Hund's rule).
3d^5:\quad 3d_{xy}^{\uparrow}\,3d_{yz}^{\uparrow}\,3d_{zx}^{\uparrow}\,3d_{x^2-y^2}^{\uparrow}\,3d_{z^2}^{\uparrow}

1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^5

68 marks

Explain ionization energy trends in the periodic table with justifications of these trends and anomalies.

Solution

Definition

The ionization energy is the minimum energy needed to remove the outermost, most loosely bound electron from an isolated gaseous atom. The atom must be in the gaseous state so that heat of fusion, heat of vaporization and bond dissociation energy do not interfere.

\ce{Na(g) -> Na+(g) + e-}\qquad \text{I.E}=496\,\mathrm{kJ\,mol^{-1}}

Trend across a period

Ionization energy generally increases from left to right across a period. Each step adds a proton, which raises the nuclear charge, and the atomic radius decreases, so the outer electrons are held more strongly. The shielding stays constant across a period, so it does not oppose this increase.

Ionization energies of period 2 elements in kJ/mol, from the book's Figure 2.8

The graph shows two dips. Boron (800.6 \mathrm{kJ\,mol^{-1}}) is lower than beryllium (899.4 \mathrm{kJ\,mol^{-1}}), and oxygen (1313.9 \mathrm{kJ\,mol^{-1}}) is lower than nitrogen (1420.3 \mathrm{kJ\,mol^{-1}}).

Anomalies

Boron is lower than beryllium, and aluminium is lower than magnesium (577.8 and 737.7 \mathrm{kJ\,mol^{-1}}). The electron removed from boron or aluminium is a p electron. It lies at a higher energy and is better shielded than the s electrons of beryllium or magnesium, so it is removed more easily. The book describes this group IIIA effect as the single electron in the p orbital.
Nitrogen is higher than oxygen. Nitrogen has a half-filled 2p subshell, which is stable. In oxygen one p orbital holds a pair of electrons, and the electron-pair repulsion in that orbital makes one electron easier to remove.

Trend down a group

Ionization energy decreases down a group, for example Li 520.2, Na 495.8, K 418.8, Rb 403.0 and Cs 375.7 \mathrm{kJ\,mol^{-1}}. Each step adds a shell, so the outer electron is further from the nucleus, and the shielding by the inner shells increases. The outermost electron is therefore easier to remove.

78 marks

Explore the practical applications of atomic emission spectroscopy.

Solution

Principle

When atoms absorb energy, for example from heat or electricity, electrons move to higher energy levels. When they return to lower levels they release the energy as light of definite wavelengths. The emitted light forms a line spectrum that is characteristic of each element.

Applications

Identification of elements: each element has a characteristic line spectrum, so the lines identify the element. This is why emission spectroscopy is widely used to identify elements.
Quantized energy levels: distinct lines support Bohr's idea that electrons occupy specific energy levels, or shells.
Electronic configuration: the number and position of the lines show how many shell electrons there are and how they are distributed among the energy levels. The hydrogen spectrum gives the energy levels and configuration of hydrogen.
Sodium: the characteristic yellow emission line of sodium is analysed to determine its configuration, including the distribution of its electrons in the energy levels.

Summary

Emission spectroscopy is therefore used to identify elements and to deduce their electronic configurations from the lines they emit.

Note. The book names no other applications, so the answer is limited to the uses it gives.

88 marks

Explore the impact of isotopes on atomic mass calculation.

Solution

Isotopes

Isotopes are atoms of the same element with the same atomic (proton) number but different mass numbers, because they have different numbers of neutrons. Magnesium has three isotopes: \ce{^{24}Mg}, \ce{^{25}Mg} and \ce{^{26}Mg}. Isotopes have the same number of electrons and the same electronic configuration.

Why atomic masses are not whole numbers

The atomic mass in the periodic table is an average over the isotopes, weighted by how abundant each isotope is. For this reason it is usually not an integer.

Formula

Worked example: chlorine

Write the abundances: \ce{^{35}Cl} 75.76% and \ce{^{37}Cl} 24.24%.
Multiply each mass number by its abundance and add the products.
(35\times75.76)+(37\times24.24)=2651.6+896.88=3548.48
Divide the sum by 100.
\dfrac{3548.48}{100}=35.48\,\mathrm{amu}

The relative atomic mass of chlorine is 35.48 amu, which lies between the two mass numbers, closer to 35 because chlorine-35 is more abundant.

Magnesium

The same method applied to the magnesium spectrum of Concept Assessment Exercise 2.7 (79% of \ce{^{24}Mg}, 10% of \ce{^{25}Mg} and 11% of \ce{^{26}Mg}) gives 24.32 amu (short question 10).

Impact of isotopes

Because the average depends on the abundances, an element's atomic mass depends on its isotopic composition. A sample with a different mix of isotopes would give a slightly different average. Mass spectrometry measures these abundances, so the mass of each element is found from its isotopic composition.

Projectp. 38

Express the atomic structure through art.

Solution

This is an open task with no single answer. The artwork should show the main ideas of the unit, with a label for each feature.

A nucleus with protons (positive) and neutrons (neutral), with the nucleus much smaller than the whole atom.
Electrons on shells (orbits) with the shell number n, as in Bohr's model, and the energy increasing outwards.
The shapes of the s (sphere), p (dumbbell) and d (cloverleaf) orbitals, each labelled with its quantum numbers l and m.

With the help of clay try to form shapes of 5 d orbitals.

Solution

The five d orbitals are d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2} and d_{z^2}. A model should show the following shapes.

d_{xy}, d_{yz} and d_{xz}: four lobes each, lying between the two axes named in the subscript (cloverleaf shape).
d_{x^2-y^2}: four lobes lying along the x and y axes.
d_{z^2}: the exception, with two lobes along the z axis and a ring of charge around the middle.

Draw the electronic configuration of all elements till Xe.

Solution

Each configuration follows the Aufbau order with the Pauli and Hund rules, as in the book's table for the first eighteen elements. Between Z = 24 and Z = 47 eight elements (Cr, Cu, Nb, Mo, Ru, Rh, Pd and Ag) differ from the simple order because a half-filled or completely filled d subshell is more stable.

Electronic configurations of the elements from hydrogen to xenon
Z	Element	Electronic configuration
1	\ce{H} Hydrogen	1s^{1}
2	\ce{He} Helium	1s^{2}
3	\ce{Li} Lithium	1s^{2}\,2s^{1}
4	\ce{Be} Beryllium	1s^{2}\,2s^{2}
5	\ce{B} Boron	1s^{2}\,2s^{2}\,2p^{1}
6	\ce{C} Carbon	1s^{2}\,2s^{2}\,2p^{2}
7	\ce{N} Nitrogen	1s^{2}\,2s^{2}\,2p^{3}
8	\ce{O} Oxygen	1s^{2}\,2s^{2}\,2p^{4}
9	\ce{F} Fluorine	1s^{2}\,2s^{2}\,2p^{5}
10	\ce{Ne} Neon	1s^{2}\,2s^{2}\,2p^{6}
11	\ce{Na} Sodium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{1}
12	\ce{Mg} Magnesium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}
13	\ce{Al} Aluminium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{1}
14	\ce{Si} Silicon	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{2}
15	\ce{P} Phosphorus	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{3}
16	\ce{S} Sulphur	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{4}
17	\ce{Cl} Chlorine	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{5}
18	\ce{Ar} Argon	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}
19	\ce{K} Potassium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}
20	\ce{Ca} Calcium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}
21	\ce{Sc} Scandium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{1}
22	\ce{Ti} Titanium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{2}
23	\ce{V} Vanadium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{3}
24	\ce{Cr} Chromium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}\,3d^{5}
25	\ce{Mn} Manganese	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{5}
26	\ce{Fe} Iron	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{6}
27	\ce{Co} Cobalt	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{7}
28	\ce{Ni} Nickel	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{8}
29	\ce{Cu} Copper	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}\,3d^{10}
30	\ce{Zn} Zinc	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}
31	\ce{Ga} Gallium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{1}
32	\ce{Ge} Germanium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{2}
33	\ce{As} Arsenic	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{3}
34	\ce{Se} Selenium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{4}
35	\ce{Br} Bromine	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{5}
36	\ce{Kr} Krypton	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}
37	\ce{Rb} Rubidium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{1}
38	\ce{Sr} Strontium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}
39	\ce{Y} Yttrium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{1}
40	\ce{Zr} Zirconium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{2}
41	\ce{Nb} Niobium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{1}\,4d^{4}
42	\ce{Mo} Molybdenum	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{1}\,4d^{5}
43	\ce{Tc} Technetium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{5}
44	\ce{Ru} Ruthenium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{1}\,4d^{7}
45	\ce{Rh} Rhodium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{1}\,4d^{8}
46	\ce{Pd} Palladium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,4d^{10}
47	\ce{Ag} Silver	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{1}\,4d^{10}
48	\ce{Cd} Cadmium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}
49	\ce{In} Indium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}\,5p^{1}
50	\ce{Sn} Tin	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}\,5p^{2}
51	\ce{Sb} Antimony	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}\,5p^{3}
52	\ce{Te} Tellurium	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}\,5p^{4}
53	\ce{I} Iodine	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}\,5p^{5}
54	\ce{Xe} Xenon	1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{2}\,3d^{10}\,4p^{6}\,5s^{2}\,4d^{10}\,5p^{6}

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 11 mark

The third ionization potential value of an element is significantly high as compared to its 1st and 2nd ionization potentials. The element belongs to:

AGroup-1 (I-A)
BGroup-2 (II-A)Answer
CGroup-17 (VII-A)
DGroup-18 (VIII-A)

Why

The first and second ionization energies are low, so the first two electrons are valence electrons that are easy to remove. A very large jump at the third value means the third electron comes from the inner shell, which is close to the nucleus and strongly held. Two valence electrons means Group 2 (II-A), so B.

Group 1 would show the big jump between the first and second values, and Group 17 and Group 18 have no early small values at all, since every ionization energy of these elements is already large.

2025 Annual I, Set B, Q.2 (i)3 marks

Write down electronic configuration of: \ce{^{29}Cu^{+2}}, \ce{^{33}As}, \ce{^{35}Br^{-1}}

Solution

\ce{^{29}Cu^2+} (1 mark)

Neutral copper has 29 electrons. Copper is an exception: a completely filled 3d sub-shell is more stable, so one 4s electron moves to 3d.
\ce{Cu}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\,3d^{10}
The charge is +2, so two electrons are removed from the outermost shell: the 4s electron first, then one from 3d. The ion has 27 electrons.
29-2=27

\ce{Cu^2+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{9}

\ce{^{33}As} (1 mark)

Arsenic has 33 electrons. After 3d^{10} the remaining 3 electrons go into 4p, one in each of the three orbitals (Hund's rule).

\ce{As}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^3

\ce{^{35}Br-} (1 mark)

Neutral bromine has 35 electrons, 4p^5 at the end. The charge -1 adds one electron to the outermost shell, giving 36 electrons and a full 4p sub-shell.

\ce{Br-}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6

\ce{Cu^2+}:\,\ldots3p^6\,3d^9;\ \ce{As}:\,\ldots4s^2\,3d^{10}\,4p^3;\ \ce{Br-}:\,\ldots4s^2\,3d^{10}\,4p^6

2026 Annual I, Set B, Q.2 (i) OR1x3 = 3 marks

Compare s and p orbitals. (Three points)

Solution

Compare the two orbitals on three points: shape, azimuthal quantum number and the number of orbitals (with the electrons they hold).

Point	s orbital	p orbital
Shape	Spherical, the same in every direction	Dumbbell shaped, two lobes on opposite sides of the nucleus
Azimuthal quantum number	l=0	l=1
Number of orbitals (m values)	One orbital, m=0	Three degenerate orbitals p_x, p_y, p_z, m=-1,0,+1, along the x, y and z axes
Electrons held by the sub-shell	2	6 (two in each orbital)

Any three of these points earn full marks. Both kinds of orbital can hold a maximum of two electrons each, and the first shell (n=1) has only an s orbital, while the p sub-shell starts from n=2.

2025 Annual I, Set B, Q.2 (v) OR3 marks

Briefly describe the terms 'shell', 'sub-shell' and 'orbital'.

Solution

Shell: Bohr's energy level around the nucleus, given by the principal quantum number n=1,2,3,\dots and named K, L, M, N. A bigger n means a larger and higher-energy shell.
Sub-shell: A part of a shell with the same shape of orbital, given by the azimuthal quantum number l=0 to n-1 and named s, p, d, f. The shell n has n sub-shells.
Orbital: The region of space around the nucleus where the probability of finding an electron is maximum. Its orientation is given by the magnetic quantum number m, and it holds at most two electrons of opposite spin.

Shell	Sub-shells	Orbitals in the shell
K (n=1)	1s	1
L (n=2)	2s, 2p	1 + 3 = 4
M (n=3)	3s, 3p, 3d	1 + 3 + 5 = 9

Example: the L shell contains the 2s and 2p sub-shells; the 2p sub-shell contains three orbitals 2p_x, 2p_y, 2p_z.

2026 Annual I, Set B, Q.2 (viii)3 marks

Write electronic configuration of following: (a) \ce{^{52}_{24}Cr} (b) \ce{^{35}_{17}Cl^{-1}}

Solution

Chromium and the chloride ion: count the electrons, then fill the sub-shells in the order 1s,2s,2p,3s,3p,4s,3d. The mass number is not needed for the configuration.

(a) Chromium, Z=24 (1.5 marks)

A neutral atom has as many electrons as its atomic number, so \ce{Cr} has 24 electrons. Filling in Aufbau order would give
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^4
Chromium is an exception. A half-filled 3d sub-shell (five orbitals with one electron each) is more stable and has less energy, so one electron moves from 4s to 3d.
4s^2\,3d^4\ \to\ 4s^1\,3d^5

\ce{Cr}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\,3d^5

(b) Chloride ion, \ce{Cl-}, Z=17 (1.5 marks)

Neutral chlorine has 17 electrons.
1s^2\,2s^2\,2p^6\,3s^2\,3p^5
The charge is -1, so one electron is added to the outermost shell, giving 18 electrons. The 3p sub-shell becomes full.
17+1=18

\ce{Cl-}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6

\ce{Cr}:\,1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\,3d^5;\quad \ce{Cl-}:\,1s^2\,2s^2\,2p^6\,3s^2\,3p^6

2025 Annual I, Set B, Q.2 (ix)3 marks

Determine the principal energy level and number of valence electrons in the different sub-shells and orbitals of an element present in 3rd period and group-VI A of the periodic table.

Solution

The period number is the principal quantum number of the valence shell, so the third period gives the principal energy level.
n=3
Group VI A has six valence electrons (ns^2\,np^4), so the element is sulphur, Z=16.
1s^2\,2s^2\,2p^6\,3s^2\,3p^4
Distribute the six valence electrons: two fill the 3s orbital, and the four 3p electrons go into three degenerate orbitals by Hund's rule (one electron each, then one pair).
3s^{\uparrow\downarrow}\ \ 3p_x^{\uparrow\downarrow}\,3p_y^{\uparrow}\,3p_z^{\uparrow}

Sub-shell	Orbital	Valence electrons
3s	3s	2
3p	3p_x	2
	3p_y	1
	3p_z	1
Total		6

Sulphur has two unpaired electrons, in 3p_y and 3p_z.

n=3;\quad 3s^2\,3p^4\ (6\ \text{valence electrons})

2026 Annual I, Set B, Q.2 (x)3 marks

Magnesium has three isotopes, Mg-24, Mg-25 and Mg-26 with abundances of 79%, 10% and 11% respectively. Calculate its average atomic mass.

Solution

Given

{}^{24}\text{Mg} = 79\,\%
{}^{25}\text{Mg} = 10\,\%
{}^{26}\text{Mg} = 11\,\%

To find

average atomic mass \bar{A} of magnesium

Average atomic mass is the sum of each mass number multiplied by its abundance, divided by 100 because the abundances are percentages.
\bar{A}=\dfrac{\sum(\text{mass number}\times\text{abundance})}{100}
Check that the abundances add up to 100%, so no isotope is missing.
79+10+11=100\,\%
Multiply each mass number by its abundance.
24\times79=1896\qquad 25\times10=250\qquad 26\times11=286
Add the products and divide by 100.
\bar{A}=\dfrac{1896+250+286}{100}=\dfrac{2432}{100}=24.32\,\mathrm{amu}

The answer lies between 24 and 26 and is close to 24 because \ce{^{24}Mg} is by far the most abundant isotope.

\bar{A}_\text{Mg}=24.32\,\mathrm{amu}

2026 Annual I, Set B, Q.2 (xi)1x3 = 3 marks

For 3d orbital, give allowed values of following quantum numbers: (a) Principal (b) Azimuthal (c) Magnetic

Solution

The letter d means l=2. Each quantum number is then fixed by the rules for allowed values.

(a) Principal quantum number: the number in front of the letter.
n=3
(b) Azimuthal quantum number: d is the sub-shell with l=2. This is allowed because l runs from 0 to n-1=2.
l=2
(c) Magnetic quantum number: m runs from -l to +l, giving 2l+1=5 values, one for each of the five 3d orbitals.
m=-2,\,-1,\,0,\,+1,\,+2

n=3;\quad l=2;\quad m=-2,-1,0,+1,+2

2025 Annual I, Set B, Q.31+4+2 = 7 marks

What is mass spectrometry? Explain the construction and working of a mass spectrometer. How can average atomic mass of an element be calculated from data obtained?

Solution

Mass spectrometry (1 mark)

Mass spectrometry is a technique used to determine the molecular composition and structure of a sample by measuring the mass-to-charge ratio (m/z) of its ions. It has three steps: ionization, mass analysis (separation of the ions by m/z) and detection, which together give a mass spectrum.

Construction and working (4 marks)

A mass spectrometer: the ion source makes ions, an electric field accelerates them, the magnetic field bends their paths (lighter ions more sharply) and the detector records where and how many ions arrive.

A mass spectrometer is made of an ion source, an accelerating electric field, a mass analyser (the magnetic field in the figure) and a detector, joined to a system that records the spectrum. It works in four stages.

Ionization. The sample is vaporized and ionized in the ion source, for example by bombarding it with high-energy electrons (EI), with a high-voltage field (ESI) or with a laser (MALDI).
Ion separation. The ions are accelerated and enter the mass analyser. The magnetic field deflects them, and ions of different m/z follow paths of different radius: the lightest ions are deflected most.
Detection. The ions that reach the detector are recorded. The detector notes the time each ion takes to arrive, and this is used to calculate its m/z.
Data analysis. The record is plotted as a mass spectrum: m/z on the x-axis and the abundance of ions on the y-axis. Each peak is one kind of ion, its position gives the mass and its height gives the relative abundance.

Average atomic mass from the data (2 marks)

The peaks of an element give the mass number and the relative abundance of each isotope. The average atomic mass is the weighted average:

\text{Average atomic mass}=\dfrac{(\text{mass no.}\times\text{abundance})+(\text{mass no.}\times\text{abundance})+\cdots}{100}

Example: chlorine has two peaks, \ce{^{35}Cl} at 75.76% and \ce{^{37}Cl} at 24.24%.

Mass spectrum of chlorine: two peaks at m/z = 35 and 37 with abundances 75.76% and 24.24%.

\bar{A}=\dfrac{35\times75.76+37\times24.24}{100}=\dfrac{2651.6+896.88}{100}=35.48\,\mathrm{amu}

This explains why the atomic masses in the periodic table are not whole numbers: each is the weighted average of the isotopes of the element.

2025 Annual I, Set D, MCQ 11 mark

The electronic configuration of an ion with mass number 63 and charge +2 is [\ce{Ne}]\ 3s^{2}, 3p^{6}, 3d^{9}. The number of electrons, protons and neutrons present in the atom are:

Ae=27 p=27 n=36
Be=29 p=27 n=36
Ce=29 p=29 n=34Answer
De=34 p=29 n=63

Why

The configuration [\ce{Ne}]\,3s^2\,3p^6\,3d^9 is that of the ion, and it has 10+2+6+9=27 electrons. The ion has a charge of +2, so it has lost two electrons and the neutral atom has 27+2=29 electrons. A neutral atom has equal protons and electrons, so p=29, and neutrons n=63-29=34. The answer is C (e=29, p=29, n=34).

Option A counts the electrons of the ion (27) and wrongly takes this as the proton number, and option B mixes the two. The question asks for the atom, not for the ion.

2025 Annual I, Set D, Q.2 (i)1x3 = 3 marks

Write down the electronic configuration of an element with atomic number 24 and also of its ion with charge +3. Describe the reason for any discrepancy in electronic configuration of element.

Solution

Chromium, Z=24 (1 mark)

Chromium has 24 electrons. Its configuration is

\ce{Cr}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1\,3d^5

Chromium ion with charge +3 (1 mark)

The ion \ce{Cr^3+} has 24-3=21 electrons. The electrons are removed from the outermost shell: the 4s electron first, then two 3d electrons.

\ce{Cr^3+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^3

The discrepancy (1 mark)

By the Aufbau order the atom would be expected to be 4s^2\,3d^4. Instead it is 4s^1\,3d^5, because half-filled and completely filled sub-shells are more stable and lower in energy. One 4s electron moves into 3d, so each of the five 3d orbitals holds one electron of the same spin (Hund's rule). The energies of 4s and 3d are close, so this shift costs little.

\ce{Cr}:\,4s^1\,3d^5;\quad \ce{Cr^3+}:\,3d^3

2026 Annual I, Set D, Q.2 (i) OR1x3 = 3 marks

Draw general shapes of following orbitals: p_x, p_y, d_{z^2}

Solution

The p_x and p_y orbitals are dumbbells (two lobes) that lie along the x axis and the y axis. The d_{z^2} orbital has two lobes along the z axis and a ring (collar) around the nucleus in the xy plane.

Shapes of the px and py orbitals (dumbbells along the x and y axes) and the dz² orbital (two lobes along the z axis with a ring around the nucleus).

Orbital	Shape	Direction
p_x	Dumbbell, two lobes	Along the x axis
p_y	Dumbbell, two lobes	Along the y axis
d_{z^2}	Two lobes and a ring (the exception among the d orbitals)	Lobes along the z axis, ring in the xy plane

2025 Annual I, Set D, Q.2 (v)1+2 = 3 marks

What is azimuthal quantum number? Describe its significance.

Solution

Definition (1 mark)

The azimuthal quantum number l describes the shape of the orbitals of a shell, that is, it specifies the sub-shell within a principal energy level. It decides the angular momentum of the electron. Its values run from 0 to n-1.

l=0,\,1,\,2,\,3,\ldots,(n-1)

Significance (2 marks)

l	Sub-shell	Shape of orbital
0	s	Spherical
1	p	Dumbbell
2	d	Cloverleaf (four lobes)
3	f	Complicated

It fixes the number of sub-shells in a shell: the shell n has n sub-shells (for n=3, l=0,1,2, so 3s, 3p, 3d).
It is related to the fine lines of the emission spectrum, which come from the presence of sub-shells.
Together with n it decides the energy of the electron, and the s, p, d and f sub-shells correspond to the blocks of the periodic table.

2026 Annual I, Set D, Q.2 (viii)1.5+1.5 = 3 marks

Write electronic configuration of following: (a) \ce{^{65}_{30}Zn} (b) \ce{^{40}_{20}Ca^{+2}}

Solution

(a) Zinc, Z=30 (1.5 marks)

Zinc has 30 electrons. Fill in Aufbau order: 4s before 3d.
2+2+6+2+6+2=20\ \text{electrons so far}
The remaining 30-20=10 electrons fill the 3d sub-shell completely.
3d^{10}

\ce{Zn}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}

(b) Calcium ion, \ce{Ca^2+}, Z=20 (1.5 marks)

Neutral calcium has 20 electrons.
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2
A charge of +2 means two electrons are removed from the outermost shell, the 4s electrons, leaving 18 electrons.
20-2=18

\ce{Ca^2+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6

\ce{Zn}:\,\ldots4s^2\,3d^{10};\quad \ce{Ca^2+}:\,1s^2\,2s^2\,2p^6\,3s^2\,3p^6

2026 Annual I, Set D, Q.2 (x)3 marks

A naturally occurring sample of chlorine contains only two isotopes Cl-35 and Cl-37, Cl-35 with 75.76% abundance. Average atomic mass of chlorine is 35.48 amu. Calculate percentage abundance of Cl-37.

Solution

Given

{}^{35}\text{Cl} = 75.76\,\%
\bar{A} = 35.48\,\mathrm{amu}

To find

percentage abundance of \ce{^{37}Cl}

The sample has only two isotopes, so their abundances add up to 100%.
\%\,{}^{37}\text{Cl}=100-75.76=24.24\,\%
Check with the average atomic mass formula.
\bar{A}=\dfrac{35\times75.76+37\times24.24}{100}=\dfrac{2651.6+896.88}{100}=35.48\,\mathrm{amu}

The check reproduces the given average of 35.48 amu. Solving the formula directly for the unknown, 37x=3548-2651.6 gives x=24.23\%, the same to the rounding of the given average.

\%\,\ce{^{37}Cl}=24.24\,\%

2026 Annual I, Set D, Q.2 (xi)1x3 = 3 marks

For 4f orbital, give allowed values of following quantum numbers: (a) Principal (b) Azimuthal (c) Magnetic

Solution

The letter f means l=3 and the number 4 is n.

(a) Principal quantum number.
n=4
(b) Azimuthal quantum number: f is the sub-shell with l=3, allowed because l runs from 0 to n-1=3.
l=3
(c) Magnetic quantum number: from -l to +l, which is 2l+1=7 values, one for each of the seven 4f orbitals.
m=-3,\,-2,\,-1,\,0,\,+1,\,+2,\,+3

n=4;\quad l=3;\quad m=-3,\dots,+3\ (7\ \text{values})

2025 Annual I, Set D, Q.2 (xii)3 marks

Describe how (n+l) rule and Hund's rule are applied in writing the electronic and orbital configuration of \ce{Mn} (Z=25).

Solution

The (n+l) rule

The orbital with the larger value of n+l has the higher energy. When two orbitals have the same n+l, the one with the smaller n is lower. This fixes the order in which orbitals are filled.

Orbital	1s	2s	2p	3s	3p	4s	3d
n+l	1	2	3	3	4	4	5

Orbital	n+l
1s	1
2s	2
2p	3
3s	3
3p	4
4s	4
3d	5

The order is therefore 1s<2s<2p<3s<3p<4s<3d. The 4s (n+l=4) is below 3d (n+l=5), so 4s fills first.

Applying it to manganese, Z=25

Fill the sub-shells in this order up to the capacity of each (2,2,6,2,6,2) until 25 electrons are used.
2+2+6+2+6+2=20
The remaining 5 electrons go into 3d, which has five degenerate orbitals.
25-20=5

Hund's rule

Degenerate orbitals are occupied singly with the same spin before any pairing takes place. The five 3d electrons therefore occupy the five 3d orbitals separately, with parallel spins (a half-filled, stable sub-shell). The two 4s electrons are paired in one orbital with opposite spins.

\begin{aligned}\ce{Mn}:\ &1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\\ &3d_{xy}^{\uparrow}\,3d_{yz}^{\uparrow}\,3d_{xz}^{\uparrow}\,3d_{x^2-y^2}^{\uparrow}\,3d_{z^2}^{\uparrow}\end{aligned}

\ce{Mn}:\,1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^5

Exam focus

Write the electronic configuration of atoms and ions, including the exceptions Cr and Cu (for example \ce{Cr}, \ce{Cl-}, \ce{Zn}, \ce{Ca^2+}, \ce{Cu^2+}, \ce{As}, \ce{Br-}). (short)
Asked in every set: 2026 Set B Q.2 (viii) and Set D Q.2 (viii), 2025 Set B Q.2 (i) and Set D Q.2 (i), plus 2025 Set D MCQ 1; also book exercise MCQs 1 to 5.
Go to answer
Give the allowed values of n, l and m for a given orbital (for example 3d and 4f), and explain the azimuthal quantum number and its significance. (short)
2026 Set B Q.2 (xi) and Set D Q.2 (xi), 2025 Set D Q.2 (v) and Set B Q.2 (v) OR (shell, sub-shell, orbital); book long question iii.
Go to answer
Calculate the average atomic mass from isotopic abundances, or an unknown abundance from the average (Mg, Cl). (numerical)
2026 Set B Q.2 (x) (Mg) and Set D Q.2 (x) (Cl); also the third part of 2025 Set B Q.3. Book short question x and long question viii.
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What is mass spectrometry? Explain the construction and working of a mass spectrometer and how the data give the average atomic mass. (long)
A 7-mark Section C question in 2025 Set B Q.3; same as book long question iv.
Go to answer
Apply the n+l rule, Hund's rule, Pauli's exclusion principle and the Aufbau principle to write a configuration (for example \ce{Mn}). (short)
2025 Set D Q.2 (xii) (Mn); book long question v and short questions viii and ix.
Go to answer
Use successive ionization energies to find the group of an element, and explain the periodic trends of ionization energy. (short)
2025 Set B MCQ 1 (third ionization energy jump); book short question xiii and long question vi.
Go to answer
Draw the shapes of the s, p and d orbitals (p_x, p_y, d_{z^2}) and compare s and p orbitals. (short)
2026 Set D Q.2 (i) OR (draw p_x, p_y, d_{z^2}) and Set B Q.2 (i) OR (compare s and p orbitals).
Go to answer
Explain the trends of atomic and ionic radius, and the effect of shielding. (short)
Book short questions iii, iv and v. Not asked in the 2025 or 2026 papers.
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Find the number of electrons, protons and neutrons in an atom or an ion from its mass number, atomic number and charge. (mcq)
2025 Set D MCQ 1; an SLO of the unit.
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