Atomic Structure and Periodicity -- Diagnostic Tests

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    A[Diag Atomic Structure] --> B[Key Concepts]
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Atomic structure is like a solar system, electrons orbit the nucleus, but with quantum rules instead of gravity.

Question:

(a) Write the full electron configuration of $\text{Cr}^{3+}$ and $\text{Cu}^+$ Explaining any deviation from the expected $4s$-before-$3d$ filling order.

(b) The third ionisation energy of chromium is $2987\,\text{kJ mol}^{-1}$ While the fourth ionisation energy is $4743\,\text{kJ mol}^{-1}$. Explain why this jump occurs and identify which electron is being removed at the fourth ionisation.

Solution:

(a) $\text{Cr}$ has atomic number 24. Its ground state configuration is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\,4s^1$ (not $3d^4\,4s^2$). The half-filled $3d$ subshell and singly-occupied $4s$ orbital provide extra stability due to exchange energy and symmetrical charge distribution.

$\text{Cr}^{3+}$: Remove the $4s^1$ electron first, then two $3d$ electrons:

$\text{Cu}$ has atomic number 29. Its ground state configuration is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1$ (not $3d^9\,4s^2$). A fully-filled $3d$ subshell provides extra stability.

$\text{Cu}^+$: Remove the $4s^1$ electron:

Key point: When forming cations, $4s$ electrons are removed before $3d$ electrons, despite $4s$ filling before $3d$. This is because once the $3d$ subshell is occupied, the $3d$ orbital energy drops below $4s$.

(b) The jump from third to fourth ionisation energy corresponds to breaking into a new shell. After removing three electrons ($4s^1$ and two $3d$ electrons), $\text{Cr}^{3+}$ has the configuration $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^3$. The fourth electron removed comes from the $3p$ subshell (closer to the nucleus, experiencing greater effective nuclear charge and less shielding). This is a significant jump because the electron transitions from the $n = 3$ valence shell to being removed from a lower-energy subshell with significantly higher nuclear attraction.

Question:

The first eight successive ionisation energies of an element $X$ are shown below:

(a) Identify the group of element $X$ in the periodic table, explaining your reasoning.

(b) Explain why the increase from the 4th to 5th ionisation energy is so much larger than the increase from the 3rd to 4th.

(c) A student claims that the general increasing trend across successive ionisation energies is solely due to increasing effective nuclear charge. Evaluate this claim.

Solution:

(a) The large jump occurs between the 4th and 5th ionisation energies (from $4356$ to $16092\,\text{kJ mol}^{-1}$). This indicates that the first four electrons are relatively easy to remove (from the outermost shell), while the fifth electron comes from a shell closer to the nucleus. Therefore, element $X$ is in Group 4 of the periodic table.

(b) The first four electrons are removed from the same principal energy level (valence shell, $n$). Although each successive removal is harder (because the remaining electrons experience greater effective nuclear charge with fewer electrons repelling each other), the differences are relatively modest. The 5th electron must be removed from the next inner shell, which is much closer to the nucleus and has significantly greater nuclear attraction. The large jump reflects the transition from removing valence electrons to removing core electrons.

(c) The student”s claim is partially correct but incomplete. While increasing effective nuclear charge (more protons attracting fewer remaining electrons) does explain the general increasing trend within a given shell, there are additional factors:

• Electron-electron repulsion: As electrons are removed, the remaining electrons are pulled closer together, increasing repulsion. This partially counteracts the effect of increased effective nuclear charge.
• Shell transitions: The very large jumps (like 4th to 5th) cannot be explained by gradual changes in effective nuclear charge alone; they reflect removal from an entirely different shell with different $n$ value and much lower energy.
• Subshell effects: Small variations within the same shell (e.g., $s$ vs $p$ electrons) also contribute to non-uniform increases.

Question:

Across Period 3 (Na to Ar), the first ionisation energy generally increases but with two notable dips: between Mg and Al, and between P and S.

(a) Explain the general increasing trend across Period 3.

(b) Explain why the first ionisation energy of Al is lower than that of Mg.

(c) Explain why the first ionisation energy of S is lower than that of P.

(d) Predict which has the higher first ionisation energy: Si or P? Justify your answer using both effective nuclear charge and electronic configuration arguments.

Solution:

(a) Across Period 3, the nuclear charge increases by one proton each element, while electrons are added to the same principal shell ($n = 3$). The additional inner shell electrons ($n = 1$ and $n = 2$) provide approximately constant shielding. Therefore, the effective nuclear charge experienced by the outer electron increases steadily, pulling electrons closer to the nucleus and making them harder to remove.

(b) Mg has the configuration $[\text{Ne}]\,3s^2$ (a filled $3s$ subshell), while Al has $[\text{Ne}]\,3s^2\,3p^1$. The $3p$ electron in Al is:

• At a slightly higher energy level than the $3s$ electrons
• Further from the nucleus on average (the $p$ orbital has a nodal plane at the nucleus)
• More effectively shielded by the $3s$ electrons

So the $3p^1$ electron in Al is easier to remove than a $3s$ electron from Mg, despite the higher nuclear charge of Al.

(c) P has the configuration $[\text{Ne}]\,3s^2\,3p^3$ (half-filled $3p$ subshell). A half-filled subshell has extra stability due to exchange energy (parallel spins in each $p$ orbital minimise electron-electron repulsion). S has $[\text{Ne}]\,3s^2\,3p^4$ Where the fourth $3p$ electron must pair with an electron in an already-occupied orbital. The pairing introduces additional electron-electron repulsion, making this electron slightly easier to remove despite the higher nuclear charge.

(d) P has the higher first ionisation energy. Although Si ($[\text{Ne}]\,3s^2\,3p^2$) has one less proton than P, the key factor is that P has a half-filled $3p^3$ configuration with maximum exchange energy and no paired electrons in the $p$ subshell. The electron removed from Si is one of the paired electrons in a $3p$ orbital (since $3p^2$ means one orbital is doubly occupied), experiencing greater repulsion. The combined effect of higher nuclear charge in P and the extra stability of the half-filled subshell means P $\gt$ Si.

Question:

Element $Y$ has the following first five ionisation energies:

(a) Identify element $Y$ and explain your reasoning.

(b) Write the formula of the most likely ionic compound formed between $Y$ and oxygen. Explain the type of bonding present.

(c) The oxide of $Y$ has a melting point of $2072\,^\circ\text{C}$. Explain this property in terms of the bonding and structure.

Solution:

(a) The large jump occurs between the 3rd and 4th ionisation energies ($2745 \to 11577\,\text{kJ mol}^{-1}$), indicating $Y$ is in Group 3. Referring to the periodic table, the Group 3 element with a first ionisation energy of approximately $577\,\text{kJ mol}^{-1}$ is aluminium ($Z = 13$).

(b) Aluminium forms $\text{Al}^{3+}$ ions (losing its three valence electrons: $3s^2\,3p^1$). Oxygen gains two electrons to form $\text{O}^{2-}$ ($2s^2\,2p^4 \to 2s^2\,2p^6$). To balance charges, the formula is $\text{Al}_2\text{O}_3$.

The bonding is ionic: aluminium donates electrons to oxygen, forming $\text{Al}^{3+}$ and $\text{O}^{2-}$ ions held together by strong electrostatic attraction in a giant ionic lattice.

(c) The very high melting point of $\text{Al}_2\text{O}_3$ is due to the strong electrostatic attraction between $\text{Al}^{3+}$ and $\text{O}^{2-}$ ions in the giant ionic lattice. The high charges on both ions ($3+$ and $2-$) result in particularly strong ionic bonds (ionic bond strength is proportional to the product of the charges). A large amount of thermal energy is required to overcome these strong forces and break the lattice, hence the high melting point.

Question:

An electron in a hydrogen atom is excited from the $n = 1$ ground state to the $n = 4$ energy level, then relaxes back to the ground state.

(a) Calculate the energy, in $\text{kJ mol}^{-1}$ Of a photon emitted when the electron transitions directly from $n = 4$ to $n = 2$. Use $E_n = -1312/n^2\,\text{kJ mol}^{-1}$.

(b) Calculate the wavelength of this photon in nm. Use c = 3.00 \times 10^8\,\text{m s}^{-1}$$h = 6.63 \times 10^{-34}\,\text{J s}$$N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}.

(c) Identify the region of the electromagnetic spectrum in which this photon lies, and explain why the transition from $n = 4$ to $n = 1$ produces a photon in a different region.

Solution:

(a) Energy at $n = 4$: $E_4 = -1312/16 = -82.0\,\text{kJ mol}^{-1}$

Energy at $n = 2$: $E_2 = -1312/4 = -328.0\,\text{kJ mol}^{-1}$

The negative sign indicates energy is released. The photon energy is $246.0\,\text{kJ mol}^{-1}$.

(b) Converting to energy per photon:

(c) $486.8\,\text{nm}$ lies in the visible region of the electromagnetic spectrum (specifically, blue-green light). This is part of the Balmer series.

The transition from $n = 4$ to $n = 1$ involves a much larger energy change:

This corresponds to a much shorter wavelength ($\approx 97\,\text{nm}$), placing it in the ultraviolet region (Lyman series). The energy difference between $n = 1$ and $n = 4$ is much larger than between $n = 2$ and $n = 4$ Because the energy levels converge (get closer together) as $n$ increases.

Question:

Consider the elements Na, Mg, Al, Si, P, S, Cl, and Ar.

(a) Arrange these elements in order of increasing electrical conductivity of their solid state. Explain the trend, identifying which element(s) are conductors, semiconductors, and insulators.

(b) The melting points of Na ($98\,^\circ\text{C}$), Mg ($650\,^\circ\text{C}$), Al ($660\,^\circ\text{C}$), Si ($1414\,^\circ\text{C}$), P ($44\,^\circ\text{C}$ White), S ($115\,^\circ\text{C}$), Cl ($-101\,^\circ\text{C}$), and Ar ($-189\,^\circ\text{C}$) show a complex pattern. Explain this pattern in terms of bonding and structure, accounting for why Si has the highest melting point.

(c) A sample of silicon is doped with phosphorus. Explain, in terms of electron configuration, how this affects the electrical conductivity of silicon.

Solution:

(a) Increasing electrical conductivity: $\text{Ar} \lt \text{P} \lt \text{S} \lt \text{Cl} \lt \text{Si} \lt \text{Na} \lt \text{Mg} \lt \text{Al}$

• Insulators (no delocalised electrons): Ar (noble gas, monatomic), P (simple molecular), S (simple molecular), Cl (simple molecular)
• Semiconductor: Si (giant covalent structure; electrons can be promoted across the band gap at higher temperatures)
• Conductors: Na, Mg, Al (metallic bonding with delocalised electrons). Conductivity increases from Na to Al due to increasing charge density and greater number of delocalised electrons per atom.

(b) The pattern reflects the changing types of bonding and structure across Period 3:

• Na, Mg, Al: Giant metallic structures. Melting points increase Na $\lt$ Mg $\lt$ Al due to increasing ionic charge ($+$, $2+$, $3+$) and decreasing ionic radius, leading to stronger metallic bonding.
• Si: Giant covalent (macromolecular) structure with a continuous network of strong covalent bonds in all directions. Each Si atom forms four strong covalent bonds, requiring enormous energy to break the entire network. This is the highest melting point.
• P, S, Cl: Simple molecular structures held together by weak intermolecular forces (van der Waals). Only these weak forces need to be overcome to melt, hence low melting points.
• Ar: Monatomic with only very weak van der Waals forces between atoms. Lowest melting point.

(c) Phosphorus ($[\text{Ne}]\,3s^2\,3p^3$) has five valence electrons, while silicon ($[\text{Ne}]\,3s^2\,3p^2$) has four. When P substitutes for Si in the lattice, four of P’s valence electrons form covalent bonds with neighbouring Si atoms, but the fifth electron is not required for bonding and becomes a delocalised electron. This creates an n-type semiconductor (negative charge carriers), significantly increasing electrical conductivity at room temperature because the extra electrons can move freely through the lattice.

Question: The first three ionisation energies of aluminium are $578$, $1817$ And $2745\,\mathrm{kJ\,mol^{-1}}$. Explain why the second ionisation energy is so much larger than the first, and why the third is larger still but the jump from second to third is smaller than from first to second.

Solution:

The first ionisation removes a $3p$ electron from Al ($[\text{Ne}]\,3s^2\,3p^1$). This is a relatively easy removal because the $3p$ electron is shielded by the $[\text{Ne}]\,3s^2$ core and is far from the nucleus (1 mark).

The second ionisation removes a $3s$ electron from $\mathrm{Al}^+$ ($[\text{Ne}]\,3s^2$). The jump ($1817 - 578 = 1239\,\mathrm{kJ/mol}$) is large because: (a) the remaining $3s$ electrons are closer to the nucleus than the $3p$ electron (lower energy subshell), (b) there is less shielding after removing the $3p$ electron, and (c) the effective nuclear charge experienced by the $3s$ electrons is higher (1 mark).

The third ionisation removes the second $3s$ electron from $\mathrm{Al}^{2+}$ ($[\text{Ne}]\,3s^1$). The jump ($2745 - 1817 = 928\,\mathrm{kJ/mol}$) is smaller than the first jump because both electrons being removed are from the same subshell ($3s$), so the change in effective nuclear charge and shielding is less dramatic (1 mark).

Question: occurring boron has two isotopes: $\mathrm{^{10}B}$ and $\mathrm{^{11}B}$. The relative atomic mass of boron is $10.81$. Calculate the percentage abundance of each isotope.

Solution:

Let $x$ be the fraction of $\mathrm{^{10}B}$ and $(1-x)$ be the fraction of $\mathrm{^{11}B}$.

$\mathrm{^{10}B}$: $19.0\%$, $\mathrm{^{11}B}$: $81.0\%$

Question: Explain why the first ionisation energy of oxygen ($1314\,\mathrm{kJ/mol}$) is lower than that of nitrogen ($1402\,\mathrm{kJ/mol}$), despite oxygen having a higher nuclear charge. Relate this to the electronic configurations.

Solution:

Nitrogen has the electron configuration $1s^2\,2s^2\,2p^3$ with one electron in each of the three $2p$ orbitals (Hund’s rule). Each electron in a separate orbital experiences minimal electron-electron repulsion within the subshell (1 mark).

Oxygen has the configuration $1s^2\,2s^2\,2p^4$ Meaning one of the $2p$ orbitals must contain a pair of electrons. The paired electrons in the same orbital repel each other, making it easier to remove one of them compared to removing an unpaired electron from nitrogen (1 mark).

Although oxygen has a higher nuclear charge ($+8$ vs $+7$), the extra electron-electron repulsion from the paired configuration outweighs the increased nuclear attraction, resulting in a lower first ionisation energy (1 mark).

Removing 3d electrons before 4s when forming cations: Electrons are removed from the highest principal quantum number shell first. Even though 4s fills before 3d, once the 3d subshell is occupied its energy drops below 4s. So when forming cations, always remove 4s electrons before 3d electrons (e.g., Fe$^{2+}$ is $[\text{Ar}]\,3d^6$, not $[\text{Ar}]\,3d^4\,4s^2$).

Assuming the ionisation energy always increases steadily across a period: The trend generally increases, but there are two dips: between Mg and Al (the 3p electron in Al is higher energy and more shielded than the 3s electrons in Mg) and between P and S (the fourth 3p electron in S pairs up, introducing electron-electron repulsion). These exceptions are frequently tested.

Confusing electron configuration of Cr and Cu with the expected filling order: Chromium is $[\text{Ar}]\,3d^5\,4s^1$ (not $3d^4\,4s^2$) and copper is $[\text{Ar}]\,3d^{10}\,4s^1$ (not $3d^9\,4s^2$). The half-filled and fully-filled d subshells have extra stability from exchange energy. These are the two most commonly tested exceptions.

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