Atomic Structure & Periodicity

Subatomic Particles

Atoms consist of three principal subatomic particles. Their properties are summarised below.

Property	Proton	Neutron	Electron
Symbol	$p$ or $p^+$	$n$ or $n^0$	$e^-$
Relative charge	$+1$	$0$	$-1$
Charge (C)	$+1.602 \times 10^{-19}$	$0$	$-1.602 \times 10^{-19}$
Relative mass	$1.00728$	$1.00867$	$0.00055$ ($\approx 1/1836$)
Location	Nucleus	Nucleus	Electron shells / orbitals
Spin	$+\tfrac{1}{2}$	$+\tfrac{1}{2}$	$+\tfrac{1}{2}$

The atomic number (proton number) $Z$ defines the element. The mass number $A = Z + N$, where $N$ is the neutron number. The notation is:

$${}^{A}_{Z}\mathrm{X}$$

Isotopes

Isotopes are atoms of the same element (same $Z$) with different neutron numbers (different $A$). Chemical properties are virtually identical because they share the same electron configuration. Physical properties (melting point, density, reaction rate) differ slightly due to mass effects.

Relative Atomic Mass ($A_r$)

The relative atomic mass is the weighted mean of the isotopic masses, weighted by their natural abundances:

$$A_r(\mathrm{X}) = \frac◆LB◆\sum_i m_i \cdot a_i◆RB◆◆LB◆\sum_i a_i◆RB◆$$

where $m_i$ is the isotopic mass and $a_i$ is the relative abundance of isotope $i$.

Worked Example. Chlorine has two stable isotopes: ${}^{35}\mathrm{Cl}$ (75.77%, $m = 34.969\,\mathrm{u}$) and ${}^{37}\mathrm{Cl}$ (24.23%, $m = 36.966\,\mathrm{u}$).

$$A_r(\mathrm{Cl}) = \frac{(34.969)(75.77) + (36.966)(24.23)}{100} = \frac{2650.4 + 895.3}{100} = 35.45$$

Relative Molecular Mass ($M_r$)

For a compound with formula $\mathrm{X}_a\mathrm{Y}_b$:

$$M_r = a \cdot A_r(\mathrm{X}) + b \cdot A_r(\mathrm{Y})$$

Mass Spectrometry

Mass spectrometry separates ions by their mass-to-charge ratio ($m/z$). The stages are:

1. Vaporisation -- sample converted to gaseous state.
2. Ionisation -- typically by electron impact (EI): a high-energy electron beam ejects an electron from the sample molecule, producing a molecular ion $\mathrm{M}^{+\bullet}$.
3. Acceleration -- ions accelerated through a potential difference $V$, gaining kinetic energy $\tfrac{1}{2}mv^2 = zVe$.
4. Deflection -- a magnetic field $B$ deflects ions into a curved path of radius $r$:

$$r = \frac◆LB◆\sqrt{2mV/z}◆RB◆◆LB◆B◆RB◆$$

Lighter ions (lower $m/z$) are deflected more. The detector records the abundance at each $m/z$.

5. Detection -- ions strike the detector, generating a current proportional to their abundance.

Interpreting Mass Spectra

• The molecular ion peak ($\mathrm{M}^{+\bullet}$) gives the relative molecular mass.
• Fragmentation produces characteristic peaks. For example, $\mathrm{CH}_4^{+\bullet}$ at $m/z = 16$; $\mathrm{CH}_3^+$ at $m/z = 15$.
• The base peak is the most intense signal (assigned 100% relative abundance).

Worked Example. A compound shows a molecular ion peak at $m/z = 78$ and a base peak at $m/z = 77$. The M+1 peak at $m/z = 79$ has 6.4% of the molecular ion intensity. This is consistent with benzene ($\mathrm{C}_6\mathrm{H}_6$, $M_r = 78$). The M+1 peak intensity ($\approx 6 \times 1.1\% = 6.6\%$) confirms six carbon atoms.

Electron Configuration

Quantum Numbers

Each electron in an atom is described by four quantum numbers:

Quantum Number	Symbol	Values	Describes
Principal	$n$	$1, 2, 3, \ldots$	Energy level / shell
Azimuthal	$\ell$	$0, 1, \ldots, n-1$	Subshell type ($0 = s$, $1 = p$, $2 = d$)
Magnetic	$m_\ell$	$-\ell, -\ell+1, \ldots, +\ell$	Orbital orientation
Spin	$m_s$	$+\tfrac{1}{2}, -\tfrac{1}{2}$	Electron spin direction

Subshell Capacities

Subshell	$\ell$	Orbitals	Max Electrons
$s$	$0$	$1$	$2$
$p$	$1$	$3$	$6$
$d$	$2$	$5$	$10$
$f$	$3$	$7$	$14$

The maximum number of electrons in shell $n$ is $2n^2$.

Aufbau Principle

Electrons occupy the lowest energy subshells first. The filling order is:

$$1s \lt 2s \lt 2p \lt 3s \lt 3p \lt 4s \lt 3d \lt 4p \lt 5s \lt 4d \lt 5p \lt 6s \lt 4f \lt 5d \lt 6p \lt 7s$$

The $4s$ subshell fills before $3d$ because its energy is lower for $Z \le 20$. For $Z \gt 20$, the energies shift and $3d$ becomes lower -- this is important for transition metals (see Transition Metals).

Pauli Exclusion Principle

No two electrons in the same atom can have the same set of four quantum numbers. Consequently, each orbital holds at most two electrons, with opposite spins.

Hund's Rule

Within a given subshell, electrons occupy degenerate orbitals singly first, with parallel spins, before pairing. This minimises electron-electron repulsion and maximises total spin.

Writing Electron Configurations

Use the notation $n\ell^x$ where $x$ is the number of electrons in that subshell.

Element	$Z$	Configuration	Shorthand
H	1	$1s^1$	$1s^1$
C	6	$1s^2 2s^2 2p^2$	$[\mathrm{He}]\,2s^2 2p^2$
Na	11	$1s^2 2s^2 2p^6 3s^1$	$[\mathrm{Ne}]\,3s^1$
Fe	26	$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$	$[\mathrm{Ar}]\,4s^2 3d^6$
Cu	29	$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}$	$[\mathrm{Ar}]\,4s^1 3d^{10}$

Note on Cr and Cu. Chromium ($Z = 24$) is $[\mathrm{Ar}]\,4s^1 3d^5$ and copper ($Z = 29$) is $[\mathrm{Ar}]\,4s^1 3d^{10}$. The half-filled $d^5$ and fully-filled $d^{10}$ configurations are stabilised by exchange energy -- the extra stability gained from maximising parallel spins outweighs the energy cost of promoting an electron from $4s$ to $3d$.

d-block Electron Configurations

For transition metals, when forming cations, the $4s$ electrons are removed before the $3d$ electrons. For example:

• $\mathrm{Fe}$: $[\mathrm{Ar}]\,4s^2 3d^6$
• $\mathrm{Fe}^{2+}$: $[\mathrm{Ar}]\,3d^6$
• $\mathrm{Fe}^{3+}$: $[\mathrm{Ar}]\,3d^5$

This is because once the $3d$ subshell begins to populate, it drops below $4s$ in energy.

Ionisation Energy

Definition

The first ionisation energy of an element is the enthalpy change when one mole of gaseous atoms each loses one electron to form one mole of gaseous $1+$ ions:

$$\mathrm{X}(g) \to \mathrm{X}^+(g) + e^- \quad \Delta H = \mathrm{IE}_1$$

The second ionisation energy is:

$$\mathrm{X}^+(g) \to \mathrm{X}^{2+}(g) + e^- \quad \Delta H = \mathrm{IE}_2$$

Ionisation energies are always endothermic (positive $\Delta H$).

General Trend Across a Period

Ionisation energy generally increases across a period (left to right) because:

1. Nuclear charge increases -- each successive element adds one proton to the nucleus.
2. Shielding increases negligibly -- additional electrons enter the same shell, so the shielding effect of inner electrons remains approximately constant.
3. Atomic radius decreases -- the increased effective nuclear charge pulls electron shells closer.
4. Net effect -- the outer electron is held more tightly, requiring more energy to remove.

Element	$Z$	Configuration	$\mathrm{IE}_1$ (kJ/mol)
Na	11	$[\mathrm{Ne}]\,3s^1$	496
Mg	12	$[\mathrm{Ne}]\,3s^2$	738
Al	13	$[\mathrm{Ne}]\,3s^2 3p^1$	578
Si	14	$[\mathrm{Ne}]\,3s^2 3p^2$	786
P	15	$[\mathrm{Ne}]\,3s^2 3p^3$	1012
S	16	$[\mathrm{Ne}]\,3s^2 3p^4$	1000
Cl	17	$[\mathrm{Ne}]\,3s^2 3p^5$	1251
Ar	18	$[\mathrm{Ne}]\,3s^2 3p^6$	1521

Anomalies: Al vs Mg and S vs P

Al (578) < Mg (738). The electron removed from Al is a $3p$ electron, which is in a higher energy subshell than the $3s$ electrons of Mg. The $3p$ electron is further from the nucleus on average and experiences greater shielding from the $3s$ electrons.

S (1000) < P (1012). In P ($3p^3$), each $3p$ orbital contains one electron. In S ($3p^4$), one orbital must contain two paired electrons. Pairing introduces electron-electron repulsion within the same orbital, which slightly reduces the energy required to remove one of the paired electrons.

Trend Down a Group

Ionisation energy decreases down a group because:

1. Principal quantum number increases -- outer electrons occupy shells further from the nucleus.
2. Shielding increases -- additional inner shells screen the nuclear charge.
3. Atomic radius increases -- the outer electron is further from the nucleus.
4. Net effect -- despite increasing nuclear charge, the increased distance and shielding dominate, making electron removal easier.

| Element | $\mathrm{IE}_1$ (kJ/mol) | |---| | Li | 520 | | Na | 496 | | K | 419 | | Rb | 403 |

Successive Ionisation Energies

Plotting successive ionisation energies ($\mathrm{IE}_1$, $\mathrm{IE}_2$, $\mathrm{IE}_3$, ...) against ionisation number reveals the electron configuration. Large jumps occur when electrons are removed from a new inner shell.

Worked Example. For aluminium ($Z = 13$), successive ionisation energies are:

$n$	$\mathrm{IE}_n$ (kJ/mol)	Shell
1	578	$n = 3$
2	1817	$n = 3$
3	2745	$n = 3$
4	11577	$n = 2$ (jump)

The large jump between $\mathrm{IE}_3$ and $\mathrm{IE}_4$ indicates that the first three electrons were removed from the third shell (valence), and the fourth is removed from the second shell (core).

Atomic and Ionic Radii

Atomic Radius

The atomic radius is the distance from the nucleus to the outermost electrons. It cannot be measured directly; instead, half the internuclear distance in a covalent bond or metallic lattice is used (covalent radius or metallic radius).

Trend across a period: Atomic radius decreases. Increasing nuclear charge pulls electrons closer without a compensating increase in shielding (same shell).

Trend down a group: Atomic radius increases. Each successive element adds a new electron shell.

Ionic Radius

Cations are smaller than their parent atoms because removing electrons reduces electron-electron repulsion and the remaining electrons are drawn closer by the unchanged nuclear charge. Higher charge cations are even smaller: $\mathrm{Fe}^{3+} \lt \mathrm{Fe}^{2+} \lt \mathrm{Fe}$.

Anions are larger than their parent atoms because adding electrons increases electron-electron repulsion, causing the electron cloud to expand. For isoelectronic species (same number of electrons), the ionic radius decreases with increasing nuclear charge:

$$\mathrm{O}^{2-} \gt \mathrm{F}^- \gt \mathrm{Na}^+ \gt \mathrm{Mg}^{2+} \gt \mathrm{Al}^{3+}$$

All have the neon electron configuration ($1s^2 2s^2 2p^6$), but nuclear charge increases from 8 to 13.

Shielding and Effective Nuclear Charge

Shielding Constant ($\sigma$)

The shielding constant $\sigma$ quantifies the extent to which inner electrons screen the nuclear charge from a valence electron. A simplified model (Slater's rules) assigns:

• Electrons in the same group ($ns, np$): shield with efficiency $0.35$ each (except $1s$: $0.30$).
• Electrons in the $(n-1)$ shell: shield with efficiency $0.85$.
• Electrons in shells $\le (n-2)$: shield with efficiency $1.00$.

Effective Nuclear Charge ($Z_\mathrm{eff}$)

$$Z_\mathrm{eff} = Z - \sigma$$

Worked Example. For potassium ($Z = 19$, configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$), the shielding experienced by the $4s$ electron:

$$\sigma = (8 \times 0.85) + (10 \times 1.00) = 6.8 + 10.0 = 16.8$$ $$Z_\mathrm{eff} = 19 - 16.8 = 2.2$$

This low effective nuclear charge explains why the $4s$ electron is so easily lost (low first ionisation energy of K).

For calcium ($Z = 20$, $4s^2$), each $4s$ electron experiences:

$$\sigma = (1 \times 0.35) + (8 \times 0.85) + (10 \times 1.00) = 0.35 + 6.8 + 10.0 = 17.15$$ $$Z_\mathrm{eff} = 20 - 17.15 = 2.85$$

The higher $Z_\mathrm{eff}$ for Ca compared to K explains the higher first ionisation energy of Ca (590 vs 419 kJ/mol).

Periodic Trends Summary

Property	Across a period	Down a group
Atomic radius	Decreases	Increases
Ionic radius (same charge)	Decreases	Increases
First ionisation energy	Generally increases	Decreases
Electronegativity	Increases	Decreases
Metallic character	Decreases	Increases
Melting point (metals)	Increases	Decreases
Melting point (non-metals)	Variable	Variable

Practice Problems

Problem 1

Bromine has two stable isotopes: ${}^{79}\mathrm{Br}$ (50.69%) and ${}^{81}\mathrm{Br}$ (49.31%). Calculate the relative atomic mass of bromine.

Solution:

$$A_r(\mathrm{Br}) = \frac◆LB◆(79 \times 50.69) + (81 \times 49.31)◆RB◆◆LB◆100◆RB◆ = \frac{4004.5 + 3994.1}{100} = 79.99 \approx 80.0$$

Problem 2

The first five ionisation energies of an element $X$ are (in kJ/mol): 590, 1145, 4912, 6491, 8153. Identify the group of element $X$.

Solution:

The large jump between $\mathrm{IE}_2$ (1145) and $\mathrm{IE}_3$ (4912) indicates that the third electron is being removed from a new, inner shell. This means $X$ has two valence electrons and belongs to Group 2. The first ionisation energy (590 kJ/mol) is consistent with calcium.

Problem 3

Write the electron configuration of $\mathrm{Cr}^{3+}$ and explain why it has a $d^3$ configuration.

Solution:

Chromium ($Z = 24$) has the ground-state configuration $[\mathrm{Ar}]\,4s^1 3d^5$. When forming $\mathrm{Cr}^{3+}$, the $4s$ electrons are removed first (as they are higher in energy once the $3d$ subshell is populated), followed by one $3d$ electron:

$$\mathrm{Cr}^{3+}: [\mathrm{Ar}]\,3d^3$$

The $d^3$ configuration has one electron in each of three $d$ orbitals with parallel spins, which is stabilised by exchange energy (Hund's rule). This half-filled-like arrangement is relatively stable.

Problem 4

Explain why the first ionisation energy of oxygen is less than that of nitrogen.

Solution:

Nitrogen ($1s^2 2s^2 2p^3$) has three unpaired $2p$ electrons, one in each $2p$ orbital (Hund's rule). Oxygen ($1s^2 2s^2 2p^4$) has four $2p$ electrons, so one orbital must contain a pair. Removing an electron from oxygen means removing one of the paired electrons from the doubly-occupied orbital. The pairing repulsion in that orbital makes the electron less tightly held than a nitrogen $2p$ electron, so oxygen has a lower first ionisation energy.

Problem 5

The mass spectrum of an element shows three peaks at $m/z = 52$, 54, and 56 with relative abundances 17.4%, 67.8%, and 14.8% respectively. (a) Identify the element. (b) Explain the pattern.

Solution:

(a) The element is chromium ($\mathrm{Cr}$, $Z = 24$, $A_r \approx 52$). The peaks correspond to ${}^{52}\mathrm{Cr}$, ${}^{54}\mathrm{Cr}$, and ${}^{56}\mathrm{Cr}$.

(b) Chromium has four stable isotopes: ${}^{50}\mathrm{Cr}$ (4.3%), ${}^{52}\mathrm{Cr}$ (83.8%), ${}^{53}\mathrm{Cr}$ (9.5%), and ${}^{54}\mathrm{Cr}$ (2.4%). However, the data shows three peaks at 52, 54, and 56, which is more consistent with iron ($\mathrm{Fe}$, $Z = 26$): ${}^{54}\mathrm{Fe}$ (5.8%), ${}^{56}\mathrm{Fe}$ (91.7%), ${}^{57}\mathrm{Fe}$ (2.1%), and ${}^{58}\mathrm{Fe}$ (0.3%).

Re-examining: the $m/z$ values of 52, 54, 56 with the given abundances most closely match chromium: ${}^{52}\mathrm{Cr}$ ($A_r = 51.94$, abundance $\approx 83.8\%$), ${}^{53}\mathrm{Cr}$ ($A_r = 52.94$, $\approx 9.5\%$), ${}^{54}\mathrm{Cr}$ ($A_r = 53.94$, $\approx 2.4\%$). However, the relative abundances in the problem (17.4%, 67.8%, 14.8%) do not match chromium's known isotope pattern.

The correct identification requires the data to be internally consistent: $A_r = 52 \times 0.174 + 54 \times 0.678 + 56 \times 0.148 = 9.05 + 36.61 + 8.29 = 53.95$. This value ($\approx 54$) is closest to chromium, but the isotope pattern does not match natural abundance. The element is therefore most likely chromium with the mass spectrum showing a simplified or experimental dataset.

Mass Spectrometry in Detail

High-Resolution Mass Spectrometry

Low-resolution mass spectrometry gives $m/z$ to the nearest integer. High-resolution MS gives $m/z$ to several decimal places, allowing distinction between species with the same nominal mass:

Species	Exact mass ($m/z$)	Nominal mass
$\mathrm{C}_3\mathrm{H}_8$	44.063	44
$\mathrm{CO}_2$	43.990	44
$\mathrm{N}_2\mathrm{O}$	44.001	44
$\mathrm{C}_2\mathrm{H}_4\mathrm{O}$	44.026	44

High-resolution MS can distinguish $\mathrm{C}_3\mathrm{H}_8$ ($44.063$) from $\mathrm{CO}_2$ ($43.990$) -- a difference of $0.073\,\mathrm{amu}$, easily resolved by modern instruments.

The Rule of 13

The rule of 13 provides possible molecular formulas from the molecular ion peak:

$$M_r = 13n + r$$

where $n$ is the number of carbon atoms and $r$ is the number of hydrogen atoms (plus halogens counted as hydrogen equivalents: F = H, Cl = H + 35, Br = H + 79, I = H + 127).

Example. For $M_r = 92$: $92 = 13 \times 7 + 1$. Base formula: $\mathrm{C}_7\mathrm{H}_1$. This is unrealistic (too few H for 7 C), so subtract 6 H and add one degree of unsaturation: $\mathrm{C}_7\mathrm{H}_8$. Possible structures: toluene ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3$) or methylcyclohexene.

Isotope Peaks in Mass Spectrometry

The M+1 peak provides a way to estimate the number of carbon atoms:

$$\text{Number of carbons} \approx \frac◆LB◆\%\,\text{intensity of M+1}◆RB◆◆LB◆1.1◆RB◆$$

Worked Example. A compound has a molecular ion at $m/z = 150$ (100%) and M+1 at $m/z = 151$ (9.9%). Estimate the number of carbons.

$$n(\mathrm{C}) \approx \frac{9.9}{1.1} = 9$$

The compound likely contains 9 carbon atoms. $12 \times 9 = 108$. Remaining mass = $150 - 108 = 42$. This could correspond to $\mathrm{C}_2\mathrm{H}_2\mathrm{O}$ (or other combinations). The rule of 13 confirms: $150 = 13 \times 11 + 7$, so the base formula is $\mathrm{C}_{11}\mathrm{H}_7$. With 9 carbons, the formula becomes $\mathrm{C}_9\mathrm{H}_{10}\mathrm{O}$ ($M_r = 9 \times 12 + 10 + 16 = 138$, which does not match). The calculation illustrates the method but requires additional information (like the M+2 peak for halogens) for unambiguous determination.

Electron Configuration and Periodic Trends

Exceptions to the Aufbau Principle

The following elements have ground-state configurations that deviate from the simple Aufbau prediction:

Element	Expected	Actual	Reason
Cr ($Z=24$)	$[\mathrm{Ar}]\,4s^2 3d^4$	$[\mathrm{Ar}]\,4s^1 3d^5$	Half-filled $d$ subshell is stabilised by exchange energy
Cu ($Z=29$)	$[\mathrm{Ar}]\,4s^2 3d^9$	$[\mathrm{Ar}]\,4s^1 3d^{10}$	Fully-filled $d$ subshell is stabilised

The exchange energy arises because electrons with parallel spins (Hund's rule) cannot occupy the same region of space (Pauli exclusion), which reduces electron-electron repulsion. The energy gained from exchange stabilisation can exceed the energy difference between $4s$ and $3d$ orbitals.

Ionisation Energy Trends Explained

Across a period: IE generally increases because:

1. Nuclear charge increases (stronger attraction to electrons).
2. Atomic radius decreases (electrons are closer to the nucleus).
3. Shielding increases only slightly (inner electrons screen imperfectly).

Down a group: IE decreases because:

1. Atomic radius increases significantly (outermost electron is further from the nucleus).
2. Shielding increases (more inner electron shells).

Exceptions across Period 2:

• Be ($\mathrm{IE} \gt \mathrm{B}$): Be has a filled $2s$ subshell ($2s^2$), which is relatively stable. B has $2s^2 2p^1$, and the $2p$ electron is higher in energy and easier to remove.
• N ($\mathrm{IE} \gt \mathrm{O}$): N has a half-filled $2p$ subshell ($2p^3$), which is stabilised by exchange energy. O has $2p^4$ with pairing repulsion in one orbital.

Successive Ionisation Energies

Successive ionisation energies ($\mathrm{IE}_1$, $\mathrm{IE}_2$, $\mathrm{IE}_3$, ...) always increase because each electron is removed from an increasingly positive ion. Large jumps indicate that the electron is being removed from a new, inner shell closer to the nucleus.

Worked Example. The first five ionisation energies of an element are (in $\mathrm{kJ/mol}$): $578$, $1817$, $2745$, $11578$, $14842$.

The large jump between $\mathrm{IE}_3$ and $\mathrm{IE}_4$ indicates that the first three electrons are removed from the outer shell, and the fourth is removed from an inner shell. The element is in Group 13 (three valence electrons). This is aluminium.

Using Ionisation Energy Data to Identify Elements

A plot of $\log(\mathrm{IE})$ vs ionisation number shows clear steps corresponding to the removal of electrons from different shells. The number of electrons in each step gives the group number.

Element	$\mathrm{IE}_1$ ($\mathrm{kJ/mol}$)	Group	Evidence
Na	$496$, $4562$, ...	1	Large jump after 1st
Mg	$738$, $1451$, $7733$, ...	2	Large jump after 2nd
Al	$578$, $1817$, $2745$, $11578$, ...	13	Large jump after 3rd
Si	$789$, $1577$, $3228$, $4356$, $16091$, ...	14	Large jump after 4th

Trends Across Period 3 in Detail

Property	Na	Mg	Al	Si	P	S	Cl	Ar
$\mathrm{IE}_1$ ($\mathrm{kJ/mol}$)	496	738	578	789	1012	1000	1251	1521
Atomic radius ($\mathrm{pm}$)	186	160	143	118	110	104	99	--
Melting point ($^\circ\mathrm{C}$)	98	650	661	1414	44	115	$-101$	$-189$
Electrical conductivity	Good	Good	Good	Poor (semiconductor)	Poor	Poor	Poor	Poor

Melting point anomaly: Silicon has the highest melting point because it has a giant covalent (macromolecular) structure with strong covalent bonds throughout. Sulphur has a higher melting point than phosphorus because $S_8$ molecules are larger and have more London forces than $P_4$ molecules.

Metallic vs Non-Metallic Character

Metallic character decreases across a period (increasing nuclear charge pulls electrons closer, making them harder to lose) and increases down a group (increasing atomic radius makes electron loss easier).

Non-metallic character follows the opposite trend: it increases across a period and decreases down a group.

The diagonal relationship: Elements diagonally adjacent in the periodic table (e.g. Li and Mg, Be and Al) show similar properties due to similar charge-to-radius ratios.

Additional Practice Problems

Problem 4

The first four ionisation energies of an element X are: $738$, $1451$, $7733$, and $10540\,\mathrm{kJ/mol}$. Identify the element and explain the pattern.

Solution:

The large jump between the second ($1451$) and third ($7733$) ionisation energies indicates that removing the third electron requires breaking into a new, inner electron shell. The element has two valence electrons and is in Group 2.

The element is magnesium ($\mathrm{Mg}$, $Z = 12$, electron configuration $[\mathrm{Ne}]\,3s^2$). The first two electrons are removed from the $3s$ subshell (valence); the third must be removed from the $2p$ subshell (core), which is much closer to the nucleus and more tightly held.

Problem 5

A mass spectrum shows the following peaks for an organic compound: $m/z = 78$ (M+, base peak), 79 (M+1, 6.7% of M+), 80 (M+2, 0.5% of M+). No significant peaks at $m/z > 80$.

(a) Suggest the molecular formula. (b) Identify the compound.

Solution:

(a) M+1 = 6.7% suggests approximately $6.7/1.1 = 6$ carbons. $\mathrm{C}_6\mathrm{H}_6$: $M = 72$ (too low). $\mathrm{C}_6\mathrm{H}_6$... $M = 6(12) + 6(1) = 78$. Check: M+2 = 0.5%, consistent with approximately 6 carbons (each $^{13}\mathrm{C}$ contributes 1.1%, and $6 \times 1.1 = 6.6\%$; M+2 from two $^{13}\mathrm{C}$ atoms: $C(6,2) \times (0.011)^2 \approx 15 \times 0.000121 = 0.18\%$, plus other contributions). The data are consistent with $\mathrm{C}_6\mathrm{H}_6$.

(b) $\mathrm{C}_6\mathrm{H}_6$ is benzene ($M = 78$). The mass spectrum of benzene characteristically shows the molecular ion as the base peak (highly stable aromatic ring).

Problem 6

Explain why the first ionisation energy of oxygen is lower than that of nitrogen, even though oxygen has a higher nuclear charge.

Solution:

Nitrogen has the electron configuration $1s^2 2s^2 2p^3$ with a half-filled $2p$ subshell. Half-filled subshells are stabilised by exchange energy (electrons with parallel spins minimise repulsion due to the Pauli exclusion principle). Removing an electron from this stable configuration requires extra energy.

Oxygen has $1s^2 2s^2 2p^4$. The fourth electron in the $2p$ subshell must pair with another electron in one of the $p$ orbitals. The pairing repulsion between the two electrons in the same orbital partially offsets the increased nuclear charge. This makes the electron easier to remove from oxygen than from nitrogen, despite the higher nuclear charge.

Data: $\mathrm{IE}_1(\mathrm{N}) = 1402\,\mathrm{kJ/mol}$, $\mathrm{IE}_1(\mathrm{O}) = 1314\,\mathrm{kJ/mol}$.

Advanced Ionisation Energy Problems

Problem 7

The successive ionisation energies of element X are shown below:

$n$	1	2	3	4	5	6	7	8	9	10	11
IE ($\mathrm{kJ/mol}$)	578	1817	2745	11578	14842	18377	23327	27466	31862	37216	42640

(a) In which group of the periodic table is element X? (1 mark)

(b) Identify element X. (1 mark)

(c) Explain the large jump between the third and fourth ionisation energies. (2 marks)

(d) Predict the formula of the oxide formed by element X. (1 mark)

Solution:

(a) Group 13. The large jump between the 3rd and 4th ionisation energies indicates that the first three electrons are in the outer shell (valence electrons), and the 4th electron is removed from an inner shell.

(b) Aluminium ($\mathrm{Al}$, $Z = 13$). The first ionisation energy (578 kJ/mol) matches the known value for aluminium.

(c) The first three electrons are removed from the $n = 3$ shell ($3s^2 3p^1$). The fourth electron must be removed from the $n = 2$ shell ($2p^6$), which is much closer to the nucleus and more strongly held. The large increase in ionisation energy reflects the transition from valence to core electrons (1 mark for identifying the shell transition, 1 mark for explaining that inner shell electrons experience less shielding and are closer to the nucleus).

(d) $\mathrm{Al}_2\mathrm{O}_3$. Aluminium is in Group 13 and forms a $3+$ ion. Oxygen forms a $2-$ ion. The formula requires charge balance: $2 \times 3+ = 3 \times 2-$, giving $\mathrm{Al}_2\mathrm{O}_3$.

Problem 8

A mass spectrum of an organic compound shows the following peaks:

• $m/z = 92$ (M+, 100% relative intensity)
• $m/z = 93$ (M+1, 7.7% relative intensity)
• $m/z = 94$ (M+2, 0.3% relative intensity)
• $m/z = 65$ (base fragment)
• $m/z = 91$ (major fragment, tropylium ion)

(a) Determine the molecular formula. (3 marks)

(b) Suggest the identity of the compound. (1 mark)

(c) Explain the origin of the peak at $m/z = 91$. (2 marks)

Solution:

(a) M+1 = 7.7% suggests approximately $7.7/1.1 = 7$ carbon atoms. $7 \times 12 = 84$. Remaining mass = $92 - 84 = 8$, which corresponds to 8 hydrogen atoms. The molecular formula is $\mathrm{C}_7\mathrm{H}_8$. Verification: $\text{DoU} = \frac{2(7) + 2 - 8}{2} = 4$. This is consistent with a benzene ring (4 degrees of unsaturation: one ring + three double bonds in the aromatic system).

M+2 = 0.3% is consistent with two $^{13}\mathrm{C}$ atoms: $C(7,2) \times (0.011)^2 = 21 \times 0.000121 = 0.25\%$, close to the observed value.

(b) $\mathrm{C}_7\mathrm{H}_8$ is consistent with toluene ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3$).

(c) The peak at $m/z = 91$ corresponds to the tropylium ion ($\mathrm{C}_7\mathrm{H}_7^+$), formed by loss of one hydrogen atom from the methyl group followed by ring expansion of the resulting benzyl cation. The tropylium ion has a cyclic, aromatic structure ($\mathrm{C}_7\mathrm{H}_7^+$, $6\,\pi$ electrons satisfying Huckel's rule) that makes it exceptionally stable, explaining its high intensity in the mass spectrum.

Electron Configuration and Periodicity in Detail

Shielding and Penetration Effects

Not all electrons in the same principal shell shield equally. The $s$ orbital penetrates closer to the nucleus than the $p$ orbital, and $p$ penetrates closer than $d$. This means:

• $s$ electrons experience less shielding (higher $Z_\mathrm{eff}$) than $p$ electrons in the same shell.
• The $4s$ orbital is lower in energy than $3d$ for $Z \le 20$ because the $4s$ electron penetrates through the inner shell electron cloud more effectively.
• For $Z > 20$, the increasing nuclear charge pulls the $3d$ electrons closer, and $3d$ drops below $4s$ in energy.

Effective Nuclear Charge Calculations for Period 3

Element	$Z$	Configuration	$\sigma$	$Z_\mathrm{eff}$ (valence)
Na	11	$[\mathrm{Ne}]3s^1$	$10 \times 0.85 = 8.5$	$11 - 8.5 = 2.5$
Mg	12	$[\mathrm{Ne}]3s^2$	$1 \times 0.35 + 10 \times 0.85 = 8.85$	$12 - 8.85 = 3.15$
Al	13	$[\mathrm{Ne}]3s^2 3p^1$	$2 \times 0.35 + 10 \times 0.85 = 9.20$	$13 - 9.20 = 3.80$
Si	14	$[\mathrm{Ne}]3s^2 3p^2$	$3 \times 0.35 + 10 \times 0.85 = 9.55$	$14 - 9.55 = 4.45$
P	15	$[\mathrm{Ne}]3s^2 3p^3$	$4 \times 0.35 + 10 \times 0.85 = 9.90$	$15 - 9.90 = 5.10$
S	16	$[\mathrm{Ne}]3s^2 3p^4$	$5 \times 0.35 + 10 \times 0.85 = 10.25$	$16 - 10.25 = 5.75$
Cl	17	$[\mathrm{Ne}]3s^2 3p^5$	$6 \times 0.35 + 10 \times 0.85 = 10.60$	$17 - 10.60 = 6.40$
Ar	18	$[\mathrm{Ne}]3s^2 3p^6$	$7 \times 0.35 + 10 \times 0.85 = 10.95$	$18 - 10.95 = 7.05$

The steady increase in $Z_\mathrm{eff}$ across the period explains the general increase in first ionisation energy.

Anomalous Trends: Quantitative Treatment

The drop from Be to B:

• $\mathrm{IE}_1(\mathrm{Be}) = 899\,\mathrm{kJ/mol}$ ($2s^2$, filled subshell)
• $\mathrm{IE}_1(\mathrm{B}) = 801\,\mathrm{kJ/mol}$ ($2s^2 2p^1$)

The $2p$ electron in boron is shielded by the two $2s$ electrons (which have the same $n$ value but different $\ell$). The $2s$ orbital penetrates more effectively than $2p$, so the $2s$ electrons are held more tightly and shield the $2p$ electron more effectively than $2p$ electrons shield each other. This reduces $Z_\mathrm{eff}$ for the $2p$ electron, making it easier to remove.

The drop from N to P:

• $\mathrm{IE}_1(\mathrm{N}) = 1402\,\mathrm{kJ/mol}$ ($2p^3$, half-filled)
• $\mathrm{IE}_1(\mathrm{O}) = 1314\,\mathrm{kJ/mol}$ ($2p^4$)

In nitrogen ($2p^3$), each $p$ orbital has one electron (Hund's rule). In oxygen ($2p^4$), one orbital must contain two paired electrons. The pairing energy (exchange energy lost + Coulombic repulsion between paired electrons) makes the fourth electron less tightly held.

Exchange energy stabilisation for half-filled subshells: A half-filled $p^3$ configuration has three electrons with parallel spins. The number of exchange interactions is $\binom{3}{2} = 3$. For $p^4$, the number of parallel spin pairs is $\binom{3}{2} = 3$ (among the three unpaired electrons) plus the exchange between one of the unpaired electrons and the paired electron if spins are parallel. The net exchange stabilisation decreases slightly when the subshell goes from half-filled to more-than-half-filled.

Mass Spectrometry: Fragmentation Patterns

Understanding fragmentation patterns is essential for identifying organic compounds from mass spectra.

Common fragmentation pathways:

Fragment	$m/z$	Origin
$\mathrm{CH}_3^+$	15	Loss of alkyl group
$\mathrm{OH}^+$	17	Alcohols, carboxylic acids
$\mathrm{C}_2\mathrm{H}_5^+$	29	Ethyl group
$\mathrm{CHO}^+$	29	Aldehydes
$\mathrm{CH}_3\mathrm{CO}^+$	43	Methyl ketones
$\mathrm{COOH}^+$	45	Carboxylic acids
$\mathrm{C}_6\mathrm{H}_5^+$	77	Aromatic compounds
$\mathrm{C}_7\mathrm{H}_7^+$	91	Benzyl/tropylium ion

Alpha cleavage: The bond adjacent to the carbonyl group breaks, producing an acylium ion. For example, in propanone ($\mathrm{CH}_3\mathrm{COCH}_3$), alpha cleavage gives $\mathrm{CH}_3\mathrm{CO}^+$ at $m/z = 43$ (base peak) and $\mathrm{CH}_3^+$ at $m/z = 15$.

McLafferty rearrangement: A hydrogen atom six atoms away from a carbonyl group transfers to the carbonyl oxygen, with simultaneous cleavage of the bond between the $\alpha$- and $\beta$-carbons. This produces an enol radical cation and a neutral alkene. For butanal ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}$), McLafferty rearrangement gives $\mathrm{C}_2\mathrm{H}_4\mathrm{O}^{+\bullet}$ at $m/z = 44$.

Exam-Style Questions with Full Mark Schemes

Q1 (4 marks)

Explain why the first ionisation energy of sodium is lower than that of magnesium, and why the first ionisation energy of aluminium is lower than that of magnesium.

Mark Scheme:

Na < Mg (2 marks): Both have electrons in the $3s$ subshell. Magnesium has a higher nuclear charge ($Z = 12$ vs $Z = 11$) with similar shielding, so the outer electrons are held more tightly, giving a higher ionisation energy (1 mark). The $3s^2$ configuration of Mg also has the added stability of a filled subshell (1 mark).

Al < Mg (2 marks): The electron removed from Al is a $3p$ electron, which is higher in energy than the $3s$ electrons of Mg (1 mark). The $3p$ electron is further from the nucleus on average and is shielded by the $3s^2$ electrons, making it easier to remove (1 mark).

Q2 (5 marks)

The mass spectrum of a compound shows a molecular ion peak at $m/z = 78$ (100%) and a base peak at $m/z = 77$ (M-1). The M+1 peak at $m/z = 79$ has 6.7% relative intensity. A significant peak also appears at $m/z = 51$.

(a) Suggest the molecular formula. (2 marks)

(b) Identify the compound. (1 mark)

(c) Explain the peak at $m/z = 77$. (1 mark)

(d) Suggest the identity of the fragment at $m/z = 51$. (1 mark)

Mark Scheme:

(a) M+1 = 6.7% gives approximately $6.7/1.1 \approx 6$ carbon atoms (1 mark). $\mathrm{C}_6\mathrm{H}_6$: $M = 72 + 6 = 78$. Molecular formula: $\mathrm{C}_6\mathrm{H}_6$ (1 mark).

(b) Benzene ($\mathrm{C}_6\mathrm{H}_6$) (1 mark).

(c) The peak at $m/z = 77$ is $\mathrm{C}_6\mathrm{H}_5^+$ (phenyl cation), formed by loss of one hydrogen atom from the molecular ion (1 mark).

(d) The peak at $m/z = 51$ is $\mathrm{C}_4\mathrm{H}_3^+$, formed by further fragmentation of the phenyl cation (loss of $\mathrm{C}_2\mathrm{H}_2$ from $\mathrm{C}_6\mathrm{H}_5^+$: $77 - 26 = 51$) (1 mark).

Q3 (6 marks)

Define the term first ionisation energy. The first five ionisation energies of an element are shown below:

$n$	1	2	3	4	5
IE ($\mathrm{kJ/mol}$)	738	1451	7733	10540	13630

(a) Identify the element. (1 mark)

(b) Explain why there is a large increase between the second and third ionisation energies. (2 marks)

(c) Write the electron configuration of the ion formed after the second ionisation. (1 mark)

(d) Explain why the ion in part (c) has a smaller ionic radius than the atom. (2 marks)

Mark Scheme:

(a) Magnesium ($\mathrm{Mg}$, $Z = 12$) (1 mark).

(b) The first two electrons are removed from the $3s$ subshell (valence electrons) (1 mark). The third electron is removed from the $2p$ subshell (core electrons), which is closer to the nucleus and experiences less shielding, so much more energy is required (1 mark).

(c) $\mathrm{Mg}^{2+}$: $1s^2 2s^2 2p^6$ or $[\mathrm{Ne}]$ (1 mark).

(d) When electrons are removed, the remaining electrons experience less electron-electron repulsion and are pulled closer to the nucleus by the unchanged nuclear charge (1 mark). The increased effective nuclear charge per electron reduces the ionic radius (1 mark).

Q4 (5 marks)

High-resolution mass spectrometry gives the molecular ion peak of a compound at $m/z = 60.0575$. Two possible molecular formulas are $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$ and $\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2$.

(a) Calculate the exact mass of each formula. (2 marks)

(b) Identify the correct formula. (1 mark)

(c) The compound gives a positive iodoform test. Suggest the identity of the compound. (2 marks)

Mark Scheme:

(a) $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$: $3(12.0000) + 8(1.0078) + 15.9949 = 36.0000 + 8.0625 + 15.9949 = 60.0574$ (1 mark).

$\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2$: $2(12.0000) + 4(1.0078) + 2(15.9949) = 24.0000 + 4.0313 + 31.9898 = 60.0211$ (1 mark).

(b) The measured value ($60.0575$) matches $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$ (60.0574) much more closely than $\mathrm{C}_2\mathrm{H}_4\mathrm{O}_2$ (60.0211). The correct formula is $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$ (1 mark).

(c) A positive iodoform test indicates a $\mathrm{CH}_3\mathrm{CO}-$ group. With the formula $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$, the compound is propan-2-ol ($\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_3$). Under the iodoform test conditions, propan-2-ol is oxidised to propanone, which contains the $\mathrm{CH}_3\mathrm{CO}-$ group (1 mark for propan-2-ol, 1 mark for explanation).

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