Periodicity — Diagnostic Tests

Unit Tests

UT-1: Ionisation Energy Trend Explanation

Question: The first ionisation energies of the period 3 elements (in $\text{kJ mol}^{-1}$) are: Na (496), Mg (738), Al (578), Si (786), P (1012), S (1000), Cl (1251), Ar (1521). Identify the two points where the trend decreases rather than increases and explain each anomaly in terms of electron configuration.

Solution: The trend decreases at Al (578) compared to Mg (738), and at S (1000) compared to P (1012).

At Al: Mg has configuration $[Ne]\, 3s^2$ (fully filled $3s$, stable). Al has $[Ne]\, 3s^2\, 3p^1$. The electron removed from Al is in the higher-energy $3p$ subshell, which experiences greater shielding and less penetration than $3s$, so less energy is needed despite the extra proton.

At S: P has $[Ne]\, 3s^2\, 3p^3$ (half-filled $3p$ with three unpaired electrons -- extra exchange energy stability). S has $[Ne]\, 3s^2\, 3p^4$ where the fourth $3p$ electron must pair with an existing electron. The electron-electron repulsion from pairing makes this electron easier to remove.

---

UT-2: Atomic and Ionic Radii Comparison

Question: Arrange the following species in order of increasing ionic/atomic radius and justify each placement: $\text{O}^{2-}$, $\text{F}^-$, $\text{Na}^+$, $\text{Mg}^{2+}$. All are isoelectronic (10 electrons).

Solution: Increasing radius: $\text{Mg}^{2+} \lt \text{Na}^+ \lt \text{F}^- \lt \text{O}^{2-}$.

All four species have the electron configuration $1s^2\, 2s^2\, 2p^6$ (10 electrons). For isoelectronic species, the radius is determined entirely by nuclear charge. Higher nuclear charge pulls electrons closer to the nucleus.

• $\text{Mg}^{2+}$: $Z = 12$, highest effective nuclear charge, smallest radius.
• $\text{Na}^+$: $Z = 11$, less pull than $\text{Mg}^{2+}$.
• $\text{F}^-$: $Z = 9$, fewer protons attracting the same 10 electrons.
• $\text{O}^{2-}$: $Z = 8$, lowest effective nuclear charge, largest radius.

---

UT-3: Electronegativity and Bond Character Prediction

Question: The Pauling electronegativities are: H (2.20), Cl (3.16), Na (0.93), O (3.44). Use electronegativity differences to classify the bonds in NaCl, HCl, and $\text{H}_2\text{O}$ as non-polar covalent, polar covalent, or ionic. Calculate the percentage ionic character for each using the empirical formula $\%\ \text{ionic} = 1 - \exp\left(-\frac{(\Delta\chi)^2}{4}\right)$.

Solution:

NaCl: $\Delta\chi = 3.16 - 0.93 = 2.23$. $\%\ \text{ionic} = 1 - \exp\left(-\frac{2.23^2}{4}\right) = 1 - \exp(-1.243) = 1 - 0.289 = 71.1\%$. Classification: ionic ($\Delta\chi \gt 1.7$).

HCl: $\Delta\chi = 3.16 - 2.20 = 0.96$. $\%\ \text{ionic} = 1 - \exp\left(-\frac{0.96^2}{4}\right) = 1 - \exp(-0.230) = 1 - 0.795 = 20.5\%$. Classification: polar covalent ($0.4 \lt \Delta\chi \lt 1.7$).

$\text{H}_2\text{O}$: $\Delta\chi = 3.44 - 2.20 = 1.24$. $\%\ \text{ionic} = 1 - \exp\left(-\frac{1.24^2}{4}\right) = 1 - \exp(-0.384) = 1 - 0.681 = 31.9\%$. Classification: polar covalent.

Integration Tests

IT-1: Periodicity and Chemical Bonding (with Chemical Bonding)

Question: Explain why silicon dioxide ($\text{SiO}_2$) is a solid with a very high melting point ($1713\ ^\circ\text{C}$), while carbon dioxide ($\text{CO}_2$) is a gas at room temperature, despite both being group 14/16 compounds. Use periodic trends in electronegativity and atomic size in your explanation.

Solution: Si and C are both in group 14, but Si is in period 3 while C is in period 2. The larger atomic radius of Si means it forms longer, weaker $\pi$ bonds. Therefore, Si uses $sp^3$ hybridisation to form four single $\sigma$ bonds to four oxygen atoms in a continuous macromolecular (giant covalent) network -- each Si is tetrahedrally bonded, and each O bridges two Si atoms. Breaking this structure requires breaking strong covalent bonds throughout the entire lattice, hence the very high melting point.

C, being much smaller, can form strong $\pi$ bonds. In $\text{CO}_2$, carbon uses $sp$ hybridisation to form two double bonds ($\text{O}=\text{C}=\text{O}$). These are discrete molecules held together only by weak London dispersion forces (instantaneous dipole-induced dipole). Very little energy is needed to overcome these intermolecular forces, so $\text{CO}_2$ sublimes at $-78.5\ ^\circ\text{C}$.

The key periodic trend is that $\pi$ bonding becomes progressively weaker down a group as atomic orbitals become more diffuse and overlap less effectively. This explains the fundamental structural difference.

---

IT-2: Ionisation Energy and Reactivity (with Redox)

Question: The first ionisation energies of group 1 elements decrease from Li (520) to Cs (376 $\text{kJ mol}^{-1}$). Explain this trend using nuclear charge and shielding. Then explain why Cs is a stronger reducing agent than Li in aqueous solution, despite both forming $+1$ ions.

Solution: Down group 1, each successive element adds a full electron shell. Although nuclear charge increases by one proton each time, the additional inner shell provides significant shielding. The valence electron is further from the nucleus and experiences a much lower effective nuclear charge. Therefore, less energy is required to remove it -- first ionisation energy decreases.

A reducing agent donates electrons. Lower ionisation energy means the element more readily loses its valence electron. Cs has the lowest first ionisation energy in group 1, so it loses its electron most easily: $\text{Cs} \to \text{Cs}^+ + e^-$. In aqueous solution, the reaction is $\text{Cs}(s) + \text{H}_2\text{O}(l) \to \text{Cs}^+(aq) + \text{OH}^-(aq) + \tfrac{1}{2}\text{H}_2(g)$, which occurs extremely vigorously. The standard electrode potential $E^\circ(\text{Cs}^+/\text{Cs}) = -3.03\ \text{V}$ is more negative than $E^\circ(\text{Li}^+/\text{Li}) = -3.04\ \text{V}$; the values are very close (hydration energy partially compensates for the ionisation energy difference), but thermodynamically Cs is still a marginally stronger reducing agent, and kinetically it reacts much faster due to its lower ionisation energy and lower melting point.

---

IT-3: Electronegativity Trends and Acid Strength (with Acids and Bases)

Question: Across period 3, the hydrides are: $\text{NaH}$, $\text{MgH}_2$, $\text{AlH}_3$, $\text{SiH}_4$, $\text{PH}_3$, $\text{H}_2\text{S}$, HCl. Use electronegativity trends to predict and explain which of these behave as acids, which as bases, and which are essentially neutral in water.

Solution: Electronegativity increases across period 3 from Na (0.93) to Cl (3.16). The polarity of the E--H bond reverses direction across the period.

Left side (Na, Mg, Al): These elements have electronegativity less than H (2.20). The E--H bond is polarised E$^\delta$--H$^\delta+$, making H partially negative. In water, the hydride ion ($\text{H}^-$) acts as a strong base: $\text{H}^- + \text{H}_2\text{O} \to \text{H}_2 + \text{OH}^-$. So $\text{NaH}$ and $\text{MgH}_2$ are strongly basic.

Middle (Si, P): Electronegativity is close to H (Si: 1.90, P: 2.19). The bonds are nearly non-polar. $\text{SiH}_4$ is essentially neutral. $\text{PH}_3$ is a very weak base (lone pair on P) but practically insoluble and unreactive in water.

Right side (S, Cl): These have electronegativity greater than H (S: 2.58, Cl: 3.16). The bond is polarised E$^\delta+$--H$^\delta-$, making H partially positive. In water, the H can be donated: $\text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^-$ ($K_a = 9.1 \times 10^{-8}$) and $\text{HCl} \to \text{H}^+ + \text{Cl}^-$ ($K_a \gg 1$, strong acid). So $\text{H}_2\text{S}$ is a weak acid and HCl is a strong acid.