Atomic Structure — Diagnostic Tests

Unit Tests

UT-1: Atomic Structure and Electronic Configuration

Question:

(a) Describe the four quantum numbers ($n$, $l$, $m_l$, $m_s$) used to describe electrons in atoms. State what each quantum number defines and the possible values it can take.

(b) Write the full electronic configuration (using the $1s^2\, 2s^2\, 2p^6$ notation) for the following atoms: potassium ($Z = 19$), iron ($Z = 26$), and copper ($Z = 29$). Note any exceptions to the expected filling order.

(c) Explain the terms “first ionisation energy” and “electron affinity.” Describe the general trend in first ionisation energy across Period 3.

(d) The first ionisation energies of sodium, magnesium, and aluminium are $496$, $738$, and $578\,\text{kJ mol}^{-1}$ respectively. Explain why there is a drop from magnesium to aluminium.

Solution:

(a)

• Principal quantum number ($n$): defines the energy level (shell) of the electron. Values: $n = 1, 2, 3, 4, \ldots$
• Azimuthal (orbital angular momentum) quantum number ($l$): defines the subshell (shape of orbital). Values: $l = 0, 1, 2, \ldots, n-1$, corresponding to $s, p, d, f$ subshells.
• Magnetic quantum number ($m_l$): defines the orientation of the orbital in space. Values: $m_l = -l, \ldots, 0, \ldots, +l$.
• Spin quantum number ($m_s$): defines the spin direction of the electron. Values: $+\frac{1}{2}$ or $-\frac{1}{2}$.

(b)

• Potassium ($Z = 19$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^1$
• Iron ($Z = 26$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^6\, 4s^2$
• Copper ($Z = 29$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^{10}\, 4s^1$

Copper is an exception: the expected configuration would be $3d^9\, 4s^2$, but a completely filled $3d$ subshell ($3d^{10}$) is more stable, so one electron is promoted from the $4s$ orbital to the $3d$ orbital.

(c) First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous $+1$ ions: $\text{X}(g) \rightarrow \text{X}^+(g) + e^-$. Electron affinity is the energy change when one mole of electrons is added to one mole of gaseous atoms.

Across Period 3 (Na to Ar), first ionisation energy generally increases. This is because the nuclear charge increases across the period while the atomic radius decreases (additional electrons are added to the same shell, increasing the effective nuclear charge on the outer electrons). Stronger attraction means more energy is required to remove an electron.

(d) Magnesium has the electronic configuration $[\text{Ne}]\, 3s^2$ and aluminium has $[\text{Ne}]\, 3s^2\, 3p^1$. The $3s$ electrons in magnesium are in a subshell closer to the nucleus (lower energy) and experience less shielding than the $3p$ electron in aluminium. The $3p$ orbital is slightly higher in energy and further from the nucleus (penetrates less effectively), so the $3p$ electron is more easily removed. This results in a drop in first ionisation energy from Mg ($738\,\text{kJ mol}^{-1}$) to Al ($578\,\text{kJ mol}^{-1}$).

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UT-2: Periodic Trends

Question:

(a) Describe and explain the trend in atomic radius down Group 2 and across Period 3.

(b) Define electronegativity. Describe the trend in electronegativity across Period 3 and down Group 17, and explain the reasons for each trend.

(c) Explain the trend in melting points across Period 3 from sodium to argon. Refer to the types of bonding present in each element.

(d) State what is meant by “metallic character” and describe its trend across a period and down a group.

Solution:

(a) Down Group 2: Atomic radius increases. Each successive element has an additional electron shell, so the outer electrons are further from the nucleus. Although nuclear charge increases, the effect of additional shells (and increased shielding from inner electrons) outweighs the increased nuclear attraction.

Across Period 3: Atomic radius decreases. The nuclear charge increases (more protons) while electrons are added to the same outer shell (same principal quantum number). The increased nuclear charge attracts outer electrons more strongly, and there is no additional shielding to offset this, so the radius decreases.

(b) Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself.

Across Period 3 (Na to Cl): Electronegativity increases. Nuclear charge increases and atomic radius decreases, so the ability to attract bonding electrons increases. Argon is a noble gas and does not typically form covalent bonds, so it has no electronegativity value.

Down Group 17 (F to I): Electronegativity decreases. Although nuclear charge increases, the outer electrons are further from the nucleus in larger shells and are increasingly shielded by inner electron shells. The attraction of the nucleus for bonding electrons weakens.

(c) Across Period 3:

• Na, Mg, Al: metallic bonding (delocalised electrons and cations). Melting points increase from Na ($98\,^\circ\text{C}$) to Mg ($650\,^\circ\text{C}$) to Al ($660\,^\circ\text{C}$) due to increasing charge on the cation ($+1$ to $+3$) and decreasing radius, strengthening metallic bonds.
• Si: giant covalent structure (each Si atom bonded to 4 others). Very high melting point ($1414\,^\circ\text{C}$) due to the strength and extent of covalent bonds.
• P, S, Cl, Ar: molecular structures (simple covalent molecules) held together by weak intermolecular forces (London dispersion forces). Melting points are low and vary due to differences in molecular size and polarity. Argon (monatomic) has the lowest melting point as it has the weakest London forces.

(d) Metallic character is the tendency of an element to lose electrons and form positive ions (the ease with which an element exhibits metallic properties such as conductivity, malleability, and lustre).

Across a period: Metallic character decreases. Elements become less willing to lose electrons (ionisation energy increases) and more likely to gain electrons. Non-metallic character increases.

Down a group: Metallic character increases. Outer electrons are further from the nucleus, more shielded, and easier to lose (ionisation energy decreases).

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UT-3: Chemical Bonding

Question:

(a) Explain the difference between a sigma ($\sigma$) bond and a pi ($\pi$) bond. How many sigma and pi bonds are present in a molecule of nitrogen, $\text{N}_2$?

(b) Draw the Lewis structure (electron dot diagram) for the carbonate ion, $\text{CO}_3^{2-}$, showing all bonds and lone pairs. State the shape and bond angle of the carbonate ion.

(c) Explain VSEPR theory. Use VSEPR to predict the shape and bond angle of: (i) $\text{BF}_3$, (ii) $\text{NH}_3$, (iii) $\text{H}_2\text{O}$.

(d) Explain what is meant by electronegativity difference and how it determines whether a bond is ionic, polar covalent, or non-polar covalent.

Solution:

(a) A sigma ($\sigma$) bond is formed by the head-on (axial) overlap of two atomic orbitals, with the electron density concentrated along the axis between the two nuclei. A sigma bond allows free rotation. A pi ($\pi$) bond is formed by the sideways (lateral) overlap of two p-orbitals, with the electron density concentrated above and below (or in front of and behind) the internuclear axis. A pi bond restricts rotation.

In $\text{N}_2$: the triple bond consists of one sigma bond and two pi bonds. Total: 1 sigma, 2 pi bonds (plus the sigma bond formed by overlap of the $s$ orbitals or $sp$ hybrid orbitals, giving a total of 1 sigma + 2 pi = 3 bonds).

(b) Lewis structure of $\text{CO}_3^{2-}$:

Carbon (Group 4, 4 valence electrons) is central. Total valence electrons: $4 + 3 \times 6 + 2 = 24$ electrons. The structure has one $\text{C}=\text{O}$ double bond and two $\text{C}-\text{O}$ single bonds, with the negative charge delocalised (resonance structures exist). Each oxygen has lone pairs to complete octets.

The carbonate ion has trigonal planar shape with bond angles of $120^\circ$. The carbon has three bonding pairs and no lone pairs, giving an electron domain geometry of trigonal planar.

(c) VSEPR theory (Valence Shell Electron Pair Repulsion): electron pairs in the valence shell of the central atom repel each other and arrange themselves as far apart as possible to minimise repulsion. Lone pairs repel more strongly than bonding pairs because they are closer to the central atom.

(i) $\text{BF}_3$: Boron has 3 bonding pairs, 0 lone pairs. Electron domain geometry: trigonal planar. Shape: trigonal planar, bond angle $120^\circ$.

(ii) $\text{NH}_3$: Nitrogen has 3 bonding pairs, 1 lone pair. Electron domain geometry: tetrahedral. Shape: trigonal pyramidal, bond angle $107^\circ$ (less than $109.5^\circ$ due to lone pair repulsion pushing bonding pairs together).

(iii) $\text{H}_2\text{O}$: Oxygen has 2 bonding pairs, 2 lone pairs. Electron domain geometry: tetrahedral. Shape: bent (V-shaped), bond angle $104.5^\circ$ (greater lone pair repulsion compresses the bond angle further).

(d) The electronegativity difference ($\Delta\chi$) between two bonded atoms determines bond type:

• Non-polar covalent: $\Delta\chi \approx 0$ to $0.4$. Electrons are shared equally (e.g., $\text{H}-\text{H}$, $\text{Cl}-\text{Cl}$).
• Polar covalent: $\Delta\chi \approx 0.4$ to $1.7$. Electrons are shared unequally; the more electronegative atom carries a partial negative charge ($\delta-$), the less electronegative a partial positive charge ($\delta+$) (e.g., $\text{H}-\text{Cl}$, $\text{C}-\text{O}$).
• Ionic: $\Delta\chi > 1.7$. The difference is so large that the electron is effectively transferred from the less electronegative to the more electronegative atom, forming ions (e.g., $\text{Na}-\text{Cl}$ becomes $\text{Na}^+\text{Cl}^-$).

These thresholds are approximate guidelines, not strict boundaries.

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Integration Tests

IT-1: Periodic Trends and Bonding

Question:

(a) Explain why the noble gases are inert. Use electronic configuration and ionisation energy in your answer.

(b) Silicon tetrachloride ($\text{SiCl}_4$) is a liquid at room temperature, while sodium chloride ($\text{NaCl}$) is a solid with a very high melting point. Explain this difference in terms of the bonding and structure of each compound.

(c) The element astatine (At) is in Group 17, Period 6. Predict the following properties of astatine: (i) its state at room temperature, (ii) whether it is more or less reactive than iodine, (iii) the type of bonding it would form with sodium.

(d) Explain why the second ionisation energy of sodium ($4562\,\text{kJ mol}^{-1}$) is much higher than its first ionisation energy ($496\,\text{kJ mol}^{-1}$).

Solution:

(a) Noble gases have a complete outer electron shell (e.g., He: $1s^2$; Ne: $[\text{He}]\, 2s^2\, 2p^6$; Ar: $[\text{Ne}]\, 3s^2\, 3p^6$). A complete outer shell is very stable and requires a very large amount of energy to disturb. The first ionisation energies of noble gases are the highest in their respective periods, confirming that removing an electron from a noble gas requires exceptional energy. As a result, noble gases have little tendency to gain, lose, or share electrons, making them essentially inert (although some heavier noble gases can form compounds under extreme conditions).

(b) $\text{SiCl}_4$ consists of simple covalent molecules with strong Si—Cl covalent bonds within each molecule but weak London dispersion forces between molecules. The weak intermolecular forces require little energy to overcome, so $\text{SiCl}_4$ is a liquid at room temperature. $\text{NaCl}$ is an ionic compound with a giant ionic lattice of $\text{Na}^+$ and $\text{Cl}^-$ ions held together by strong electrostatic forces of attraction in all directions. A very large amount of energy is required to overcome these forces, giving $\text{NaCl}$ a high melting point ($801\,^\circ\text{C}$).

(c) Based on trends down Group 17:

(i) Astatine would be a solid at room temperature. The elements progress from gas (F, Cl) to liquid (Br) to solid (I) down the group as London dispersion forces increase with molecular size. (ii) Astatine is less reactive than iodine. Reactivity decreases down Group 17 because atoms become larger, shielding increases, and the ability to attract an extra electron decreases (electron affinity decreases). (iii) Astatine would form an ionic bond with sodium, producing $\text{NaAt}$. The electronegativity difference between Na ($0.9$) and At ($2.2$, estimated) is large enough for electron transfer to occur.

(d) Sodium has the configuration $[\text{Ne}]\, 3s^1$. The first ionisation energy removes the single $3s$ electron, producing $\text{Na}^+$ with the stable configuration of neon. The second ionisation energy would remove an electron from the $2p$ subshell (a complete, stable octet), which requires much more energy. This is why the second ionisation energy ($4562\,\text{kJ mol}^{-1}$) is approximately 9 times higher than the first ($496\,\text{kJ mol}^{-1}$).

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IT-2: Applications of Atomic Structure

Question:

(a) Using the concepts of effective nuclear charge and shielding, explain why the atomic radius of potassium ($Z = 19$) is greater than that of argon ($Z = 18$), even though potassium has one more proton.

(b) The bond lengths of the hydrogen halides are: $\text{HF} = 91.8\,\text{pm}$, $\text{HCl} = 127.4\,\text{pm}$, $\text{HBr} = 141.4\,\text{pm}$, $\text{HI} = 160.9\,\text{pm}$. Explain the trend in bond lengths.

(c) Explain why carbon and silicon are in the same group but carbon dioxide ($\text{CO}_2$) is a gas at room temperature while silicon dioxide ($\text{SiO}_2$) is a solid with a very high melting point.

(d) A scientist proposes that element 119 (predicted to be in Group 1, Period 8) would be the most reactive metal. Evaluate this claim using your knowledge of periodic trends.

Solution:

(a) Argon has the configuration $[\text{Ne}]\, 3s^2\, 3p^6$ with electrons in the $n = 3$ shell. Potassium has the configuration $[\text{Ne}]\, 3s^2\, 3p^6\, 4s^1$ with its outermost electron in the $n = 4$ shell. Although potassium has one more proton than argon, the outer electron is in a new, larger shell that is further from the nucleus. The additional shell outweighs the increased nuclear charge because the inner shells provide substantial shielding. Therefore potassium has a larger atomic radius than argon.

(b) The bond length increases from HF to HI because the halogen atom increases in size down Group 17. Each successive halogen has one more electron shell, making the atom larger. The H—X bond length is determined largely by the size of the halogen atom (hydrogen is very small and contributes little to the bond length). Fluorine is the smallest halogen, so HF has the shortest bond; iodine is the largest, so HI has the longest bond.

(c) Although carbon and silicon are in Group 4, carbon can form stable double bonds ($\text{C}=\text{O}$) while silicon generally cannot form stable pi bonds due to its larger atomic size and the poor overlap of its 3p orbitals. Carbon dioxide exists as discrete $\text{O}=\text{C}=\text{O}$ molecules with only weak London dispersion forces between them, making it a gas. Silicon dioxide forms a giant covalent lattice where each silicon atom is bonded to four oxygen atoms (and each oxygen to two silicon atoms) via single bonds only ($\text{Si}-\text{O}-\text{Si}$). The extensive covalent bonding throughout the structure gives $\text{SiO}_2$ a very high melting point.

(d) Element 119 would be in Group 1, Period 8. Its outermost electron would be in the $8s$ orbital, very far from the nucleus and heavily shielded by seven filled inner shells. Following the trend of increasing reactivity down Group 1 (due to decreasing ionisation energy), element 119 would be expected to lose its outer electron very easily, making it highly reactive. The claim is plausible based on periodic trends, but uncertainties exist because relativistic effects (where electrons move at speeds approaching the speed of light) become significant for very heavy elements and can alter expected properties. The element has not yet been synthesised, so the claim remains theoretical.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Writing electronic configurations without accounting for exceptions (e.g., chromium is $[\text{Ar}]\, 3d^5\, 4s^1$ and copper is $[\text{Ar}]\, 3d^{10}\, 4s^1$).
• Forgetting that effective nuclear charge increases across a period, not just nuclear charge alone.
• Confusing atomic radius trends: radius decreases across a period but increases down a group.
• Misapplying VSEPR by not counting lone pairs as electron domains that affect molecular shape.
• Stating that noble gases have the highest electronegativity — noble gases are typically excluded from electronegativity scales.