Periodicity

1. Periodic Trends Recap

Effective Nuclear Charge

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

$Z_{\mathrm{eff}}$ increases across a period (shielding increases slowly, $Z$ increases by 1 per element) and changes only slightly down a group (new shells increase shielding proportionally).

Summary of Trends

Property	Across a period	Down a group
Atomic radius	Decreases	Increases
Ionic radius	Decreases	Increases
Ionization energy	Increases	Decreases
Electron affinity	More negative	Less negative
Electronegativity	Increases	Decreases
Metallic character	Decreases	Increases

Common Pitfalls

• Noble gas atomic radii (van der Waals) are significantly larger than covalent radii of adjacent halogens.
• The $d$-block contraction causes period 5 and 6 elements to have similar radii.

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2. Group 1: Alkali Metals

Physical Properties

Property	Trend down the group
Melting point	Decreases
Boiling point	Decreases
Density	Generally increases (K anomaly)
Atomic radius	Increases
Hardness	Decreases (softer)

Chemical Properties

All have outer configuration $ns^1$. They lose one electron to form $\mathrm{M}^+$ ions.

Reaction with Water

$$2\mathrm{M}(s) + 2\mathrm{H}_2\mathrm{O}(l) \to 2\mathrm{MOH}(aq) + \mathrm{H}_2(g)$$

Reactivity increases down the group as $IE_1$ decreases.

Metal	Observation
Li	Fizzes steadily
Na	Melts into a ball, rapid fizzing
K	Ignites with lilac flame
Rb, Cs	Explosive

Oxides

Metal	Limited O$_2$	Excess O$_2$
Li	$\mathrm{Li}_2\mathrm{O}$	$\mathrm{Li}_2\mathrm{O}$
Na	$\mathrm{Na}_2\mathrm{O}$	$\mathrm{Na}_2\mathrm{O}_2$ (peroxide)
K	$\mathrm{K}_2\mathrm{O}_2$	$\mathrm{KO}_2$ (superoxide)
Rb, Cs	Superoxides	Superoxides

The trend from oxide to peroxide to superoxide down the group reflects the decreasing charge density of the $\mathrm{M}^+$ ion, which stabilizes the larger anions ($\mathrm{O}_2^{2-}$, $\mathrm{O}_2^-$).

Flame Tests

Ion	Colour
$\mathrm{Li}^+$	Crimson red
$\mathrm{Na}^+$	Yellow
$\mathrm{K}^+$	Lilac (view through cobalt glass)

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3. Group 2: Alkaline Earth Metals

Physical Properties

Property	Trend down the group
Melting point	Generally decreases
Boiling point	Generally decreases
Density	Increases
Atomic radius	Increases

Group 2 metals are harder and have higher melting points than Group 1 metals due to the release of two delocalised electrons per atom, producing stronger metallic bonding.

Chemical Properties

Outer configuration: $ns^2$. They lose two electrons to form $\mathrm{M}^{2+}$ ions.

Reactivity with Water

Reactivity increases down the group. Beryllium does not react with water. Magnesium reacts slowly with steam. Calcium, strontium, and barium react with cold water.

$$\mathrm{M}(s) + 2\mathrm{H}_2\mathrm{O}(l) \to \mathrm{M(OH)}_2(aq) + \mathrm{H}_2(g)$$

Thermal Decomposition of Nitrates and Carbonates

Group 2 carbonates decompose on heating:

$$\mathrm{MCO}_3(s) \to \mathrm{MO}(s) + \mathrm{CO}_2(g)$$

Group 2 nitrates decompose differently:

$$2\mathrm{M(NO}_3)_2(s) \to 2\mathrm{MO}(s) + 4\mathrm{NO}_2(g) + \mathrm{O}_2(g)$$

Thermal stability increases down the group. This is because the larger $\mathrm{M}^{2+}$ ion has lower polarising power, distorting the carbonate/nitrate ion less and making it harder to decompose.

Compound	Ease of decomposition (decreasing)
$\mathrm{MgCO}_3$	Decomposes at low temperature
$\mathrm{CaCO}_3$	Decomposes at high temperature
$\mathrm{BaCO}_3$	Very stable, requires strong heat

Solubility Trends

Species	Trend down the group
Hydroxides	Solubility increases
Sulfates	Solubility decreases
Carbonates	Generally insoluble

$$\mathrm{BaSO}_4 \mathrm{ is very insoluble — used in barium meals (X-ray contrast).}$$

Uses

Compound	Use
$\mathrm{Mg(OH)}_2$	Antacids, laxatives
$\mathrm{CaCO}_3$	Limestone, cement, antacids
$\mathrm{CaSO}_4 \cdot 2\mathrm{H}_2\mathrm{O}$	Plaster of Paris
$\mathrm{BaSO}_4$	Barium meals (radiocontrast)

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4. Group 17: Halogens

Physical Properties

Property	Trend down the group
Melting point	Increases
Boiling point	Increases
Atomic radius	Increases
State at RTP	F$_2$, Cl$_2$ (gas); Br$_2$ (liq); I$_2$ (solid)

Chemical Properties

Outer configuration: $ns^2\, np^5$. Reactivity decreases down the group as atomic radius increases and electron affinity becomes less favourable.

Displacement Reactions

A more reactive halogen displaces a less reactive halogen from its halide:

$$\mathrm{Cl}_2(aq) + 2\mathrm{KBr}(aq) \to 2\mathrm{KCl}(aq) + \mathrm{Br}_2(aq)$$ $$\mathrm{Br}_2(aq) + 2\mathrm{KI}(aq) \to 2\mathrm{KBr}(aq) + \mathrm{I}_2(aq)$$

$\mathrm{Br}_2$ cannot displace $\mathrm{Cl}^-$; $\mathrm{I}_2$ cannot displace $\mathrm{Br}^-$ or $\mathrm{Cl}^-$.

Halide Ion Tests

Add dilute $\mathrm{HNO}_3$ first (to remove carbonate/hydroxide), then $\mathrm{AgNO}_3$:

Halide	Precipitate colour	Solubility in $\mathrm{NH}_3$(aq)
$\mathrm{Cl}^-$	White ($\mathrm{AgCl}$)	Soluble in dilute $\mathrm{NH}_3$
$\mathrm{Br}^-$	Cream ($\mathrm{AgBr}$)	Partially soluble in conc. $\mathrm{NH}_3$
$\mathrm{I}^-$	Yellow ($\mathrm{AgI}$)	Insoluble

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5. Period 3 Properties

Physical Trends Across Period 3

Element	Na	Mg	Al	Si	P	S	Cl	Ar
Structure	Metallic	Metallic	Metallic	Giant covalent	Molecular (P$_4$)	Molecular (S$_8$)	Molecular (Cl$_2$)	Monatomic
Melting point	Low	Higher	Higher	High ($1414\degree\mathrm{C}$)	Low	Low	Low	Very low
Conductivity	Good	Good	Good	Poor (semiconductor)	Poor	Poor	Poor	Poor

The melting point peaks at silicon (giant covalent network) and drops sharply at phosphorus (simple molecular).

Oxides of Period 3

Element	Oxide	Structure	Acid/Base	pH of solution
Na	$\mathrm{Na}_2\mathrm{O}$	Ionic	Basic	$\gt 7$
Mg	$\mathrm{MgO}$	Ionic	Basic	$\gt 7$
Al	$\mathrm{Al}_2\mathrm{O}_3$	Ionic/covalent	Amphoteric	$7$
Si	$\mathrm{SiO}_2$	Giant covalent	Acidic	Insoluble
P	$\mathrm{P}_4\mathrm{O}_{10}$	Molecular	Acidic	$1$--$2$
S	$\mathrm{SO}_3$	Molecular	Acidic	$0$--$1$
Cl	$\mathrm{Cl}_2\mathrm{O}_7$	Molecular	Acidic	Strongly acidic

Trend: Oxides become more acidic across the period.

Amphoteric Oxides

Aluminium oxide dissolves in both acids and bases:

$$\mathrm{Al}_2\mathrm{O}_3(s) + 6\mathrm{HCl}(aq) \to 2\mathrm{AlCl}_3(aq) + 3\mathrm{H}_2\mathrm{O}(l)$$ $$\mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{NaOH}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \to 2\mathrm{NaAl(OH)}_4(aq)$$

Highest Oxidation States of Period 3

Element	Highest oxidation state	Oxide
Na	$+1$	$\mathrm{Na}_2\mathrm{O}$
Mg	$+2$	$\mathrm{MgO}$
Al	$+3$	$\mathrm{Al}_2\mathrm{O}_3$
Si	$+4$	$\mathrm{SiO}_2$
P	$+5$	$\mathrm{P}_4\mathrm{O}_{10}$
S	$+6$	$\mathrm{SO}_3$
Cl	$+7$	$\mathrm{Cl}_2\mathrm{O}_7$

The trend of increasing highest oxidation state across the period reflects the increasing number of valence electrons available for bonding.

Chlorides of Period 3

Element	Chloride	Bonding	Reaction with water
Na	$\mathrm{NaCl}$	Ionic	Dissolves, neutral
Mg	$\mathrm{MgCl}_2$	Ionic	Dissolves, slightly acidic
Al	$\mathrm{AlCl}_3$	Covalent (layer)	Hydrolyses: $\mathrm{AlCl}_3 + 3\mathrm{H}_2\mathrm{O} \to \mathrm{Al(OH)}_3 + 3\mathrm{HCl}$
Si	$\mathrm{SiCl}_4$	Covalent (mol.)	Hydrolyses: $\mathrm{SiCl}_4 + 2\mathrm{H}_2\mathrm{O} \to \mathrm{SiO}_2 + 4\mathrm{HCl}$
P	$\mathrm{PCl}_3/\mathrm{PCl}_5$	Covalent	Hydrolyses violently
S	$\mathrm{SCl}_2/\mathrm{S}_2\mathrm{Cl}_2$	Covalent	Hydrolyses
Cl	—	—	N/A

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6. Transition Metals ($d$-Block)

Definition

A transition metal has a partially filled $d$-subshell in its atom or in any of its common oxidation states.

By this definition, $\mathrm{Sc}$ (always $\mathrm{Sc}^{3+}$, $d^0$) and $\mathrm{Zn}$ (always $\mathrm{Zn}^{2+}$, $d^{10}$) are not transition metals.

Variable Oxidation States

Element	Common oxidation states
Ti	$+2, +3, +4$
V	$+2, +3, +4, +5$
Cr	$+2, +3, +6$
Mn	$+2, +3, +4, +6, +7$
Fe	$+2, +3$
Co	$+2, +3$
Cu	$+1, +2$

The maximum oxidation state increases across the period to $\mathrm{Mn}$ ($+7$), then decreases.

Complex Ions

A complex ion consists of a central metal ion surrounded by ligands:

$$[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}, \quad [\mathrm{Ag}(\mathrm{NH}_3)_2]^+, \quad [\mathrm{Fe}(\mathrm{CN})_6]^{3-}$$

Ligand type	Examples	Denticity
Monodentate	$\mathrm{H}_2\mathrm{O}$, $\mathrm{NH}_3$, $\mathrm{Cl}^-$	$1$
Bidentate	en, oxalate	$2$
Hexadentate	EDTA	$6$

Colour

Transition metal complexes are coloured due to $d$-$d$ transitions. The energy gap between split $d$-orbitals corresponds to visible light:

\Delta E = \frac`\{hc}`{\lambda}

Spectrochemical series (increasing $\Delta$):

$$\mathrm{I}^- \lt \mathrm{Br}^- \lt \mathrm{Cl}^- \lt \mathrm{F}^- \lt \mathrm{H}_2\mathrm{O} \lt \mathrm{NH}_3 \lt \mathrm{en} \lt \mathrm{CN}^- \lt \mathrm{CO}$$

Catalytic Properties

Transition metals are effective catalysts because of variable oxidation states and the ability to form intermediate complexes.

Type	Example
Heterogeneous	Fe (Haber), V$_2$O$_5$ (Contact), Ni (hydrogenation)
Homogeneous	$\mathrm{Fe}^{2+}/\mathrm{Fe}^{3+}$ (Fenton)

Magnetic Properties

• Paramagnetic: unpaired $d$-electrons present (attracted to magnetic field).
• Diamagnetic: all $d$-electrons paired (weakly repelled).

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Practice Problems

Problem 1

Explain why the thermal decomposition temperature of Group 2 carbonates increases down the group.

Solution:

As the size of the $\mathrm{M}^{2+}$ ion increases down the group, its charge density decreases. The larger cation has lower polarising power, meaning it distorts the electron cloud of the $\mathrm{CO}_3^{2-}$ anion less. A less distorted carbonate ion is more stable and requires more thermal energy to decompose into $\mathrm{MO}$ and $\mathrm{CO}_2$.

Problem 2

Explain why $\mathrm{Al}_2\mathrm{O}_3$ is amphoteric but $\mathrm{MgO}$ is basic.

Solution:

$\mathrm{Al}^{3+}$ has a high charge density (small ion, $+3$ charge), which polarises the O--H bonds in water molecules coordinated to it, facilitating proton release. This gives aluminium oxide acidic character in addition to its basic character. $\mathrm{Mg}^{2+}$ has a lower charge density, so it acts only as a Lewis acid in accepting oxide ions but does not polarise water sufficiently to release protons. Therefore $\mathrm{MgO}$ is purely basic.

Problem 3

Predict and explain the trend in solubility of Group 2 sulfates down the group. Which Group 2 sulfate is the least soluble?

Solution:

Solubility of Group 2 sulfates decreases down the group. The hydration enthalpy (energy released when ions are hydrated) decreases more rapidly than the lattice energy as the cation size increases. Since the lattice energy does not decrease as fast, the enthalpy of solution becomes less favourable (less negative or more positive) down the group.

$\mathrm{BaSO}_4$ is the least soluble. This is exploited in barium meal X-ray procedures and in gravimetric determination of sulfate.

Problem 4

Explain why $\mathrm{TiCl}_4$ is a liquid at room temperature while $\mathrm{TiO}_2$ is a solid with a very high melting point ($1843\degree\mathrm{C}$).

Solution:

$\mathrm{TiCl}_4$ is a simple molecular (covalent) compound with weak London dispersion forces between molecules, so it is a liquid at room temperature (bp $136\degree\mathrm{C}$).

$\mathrm{TiO}_2$ has a giant ionic lattice structure. The strong electrostatic forces between $\mathrm{Ti}^{4+}$ and $\mathrm{O}^{2-}$ ions require a large amount of energy to overcome, resulting in a very high melting point. The high charges on both ions ($+4$ and $-2$) produce a particularly large lattice energy.

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Worked Examples

Worked Example: Determining if an Element is a Transition Metal

Is scandium a transition metal? Justify using the IB definition.

Solution

The IB defines a transition metal as an element with a partially filled $d$-subshell in the atom or any common oxidation state. Scandium has the electron configuration $[\mathrm{Ar}]\,3d^1\,4s^2$, so its atom has a partially filled $d$-subshell. However, its only common oxidation state is $\mathrm{Sc}^{3+}$ with configuration $[\mathrm{Ar}]\,3d^0$ — an empty $d$-subshell. Since no common ion of scandium has a partially filled $d$-subshell, scandium is not classified as a transition metal under the IB definition.

Worked Example: Halogen Displacement Prediction

Aqueous chlorine is added to a solution containing both potassium bromide and potassium iodide. Deduce what happens, and write the overall ionic equations.

Solution

Chlorine is more reactive than both bromine and iodine (higher position in Group 17), so it displaces both halide ions:

$$\mathrm{Cl}_2(aq) + 2\mathrm{KBr}(aq) \to 2\mathrm{KCl}(aq) + \mathrm{Br}_2(aq)$$$$\mathrm{Cl}_2(aq) + 2\mathrm{KI}(aq) \to 2\mathrm{KCl}(aq) + \mathrm{I}_2(aq)$$

In ionic form:

$$\mathrm{Cl}_2(aq) + 2\mathrm{Br}^-(aq) \to 2\mathrm{Cl}^-(aq) + \mathrm{Br}_2(aq)$$$$\mathrm{Cl}_2(aq) + 2\mathrm{I}^-(aq) \to 2\mathrm{Cl}^-(aq) + \mathrm{I}_2(aq)$$

Observations: the solution turns orange-brown due to dissolved $\mathrm{Br}_2$ and violet due to $\mathrm{I}_2$. If cyclohexane is added and shaken, $\mathrm{Br}_2$ appears orange in the organic layer and $\mathrm{I}_2$ appears violet.

Worked Example: Period 3 Oxide pH Prediction

Predict the approximate pH of a solution of $\mathrm{Na}_2\mathrm{O}$ dissolved in water, and write the equation for the reaction. Compare this with $\mathrm{P}_4\mathrm{O}_{10}$ dissolved in water.

Solution

$\mathrm{Na}_2\mathrm{O}$ is an ionic, basic oxide. It reacts with water to form sodium hydroxide:

$$\mathrm{Na}_2\mathrm{O}(s) + \mathrm{H}_2\mathrm{O}(l) \to 2\mathrm{NaOH}(aq)$$

Since $\mathrm{NaOH}$ is a strong base, the solution pH is well above 7 (typically 12--14 depending on concentration).

$\mathrm{P}_4\mathrm{O}_{10}$ is a covalent, acidic oxide. It reacts with water to form phosphoric acid:

$$\mathrm{P}_4\mathrm{O}_{10}(s) + 6\mathrm{H}_2\mathrm{O}(l) \to 4\mathrm{H}_3\mathrm{PO}_4(aq)$$

Phosphoric acid is a weak acid, giving a pH in the range 1--2 for a moderately concentrated solution. The trend from strongly basic ($\mathrm{Na}_2\mathrm{O}$) to acidic ($\mathrm{P}_4\mathrm{O}_{10}$) across Period 3 reflects the increasing electronegativity and decreasing ionic character of the oxides.

Worked Example: Thermal Decomposition Ranking

Arrange the following in order of increasing thermal decomposition temperature, and explain: $\mathrm{Na}_2\mathrm{CO}_3$, $\mathrm{MgCO}_3$, $\mathrm{CaCO}_3$.

Solution

Order: $\mathrm{Na}_2\mathrm{CO}_3$ $\lt$ $\mathrm{MgCO}_3$ $\lt$ $\mathrm{CaCO}_3$

The thermal stability of carbonates depends on the polarising power of the cation. $\mathrm{Na}^+$ is a large, singly charged ion with very low charge density — it barely distorts the $\mathrm{CO}_3^{2-}$ ion, so $\mathrm{Na}_2\mathrm{CO}_3$ does not decompose on heating (it melts at $851\degree\mathrm{C}$ without decomposition). $\mathrm{Mg}^{2+}$ has the highest charge density among these three (small radius, $+2$ charge), so it most effectively polarises the carbonate ion and destabilises it, giving the lowest decomposition temperature ($\approx 540\degree\mathrm{C}$). $\mathrm{Ca}^{2+}$ is larger than $\mathrm{Mg}^{2+}$, so it has lower polarising power, making $\mathrm{CaCO}_3$ more thermally stable ($\approx 840\degree\mathrm{C}$).

Worked Example: Complex Ion Formation and Colour

Explain why $\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}$ is blue, but anhydrous $\mathrm{CuSO}_4$ is white.

Solution

The blue colour of hydrated copper(II) sulfate arises from the $[\mathrm{Cu}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ complex ion. In the octahedral crystal field created by the six water ligands, the $d$-orbitals of the $\mathrm{Cu}^{2+}$ ion ($d^9$ configuration) split into two energy levels. When white light passes through the solution, photons of a specific wavelength (red-orange, $\approx 600$--$700\mathrm{ nm}$) are absorbed to promote electrons from the lower to the upper $d$-orbital set. The transmitted or reflected light is the complementary colour — blue.

In anhydrous $\mathrm{CuSO}_4$, no water ligands are present, so no crystal field splitting occurs and no visible light is absorbed by $d$-$d$ transitions. The compound appears white.

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Common Pitfalls

• Confusing van der Waals radius with covalent radius: Noble gas radii are van der Waals radii and are much larger than the covalent radii of halogens in the same period. Do not compare them directly on the same scale.

• Misidentifying transition metals: $\mathrm{Sc}$ and $\mathrm{Zn}$ are not transition metals under the IB definition. The criterion is a partially filled $d$-subshell in a common oxidation state, not merely being in the $d$-block.

• Assuming all Group 2 carbonates decompose easily: $\mathrm{Na}_2\mathrm{CO}_3$ and $\mathrm{K}_2\mathrm{CO}_3$ are Group 1 carbonates and are thermally stable — they do not decompose on heating. Group 2 carbonates do decompose, with decreasing ease down the group.

• Forgetting the oxide/peroxide/superoxide trend: Students often assume all Group 1 metals form normal oxides ($\mathrm{M}_2\mathrm{O}$). In reality, heavier alkali metals form peroxides ($\mathrm{M}_2\mathrm{O}_2$) and superoxides ($\mathrm{MO}_2$) in excess oxygen due to the decreasing charge density of the $\mathrm{M}^+$ ion.

• Reversing the spectrochemical series: The spectrochemical series ranks ligands by the magnitude of crystal field splitting ($\Delta$). $\mathrm{CN}^-$ produces a larger $\Delta$ than $\mathrm{H}_2\mathrm{O}$, not the other way around. A larger $\Delta$ means shorter-wavelength (higher-energy) light is absorbed.

• Saying transition metals are coloured: The metals themselves are not coloured; it is their complex ions that are coloured due to $d$-$d$ transitions. A $d^0$ or $d^{10}$ complex is colourless because there are no partially filled $d$-orbitals to allow such transitions.

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Exam-Style Problems

1. [Medium] Explain why the first ionization energy of aluminium ($577\mathrm{ kJ/mol}$) is lower than that of magnesium ($738\mathrm{ kJ/mol}$), even though aluminium has a greater nuclear charge.

2. [Medium] A student adds aqueous silver nitrate to three separate test tubes containing $\mathrm{NaCl}(aq)$, $\mathrm{NaBr}(aq)$, and $\mathrm{NaI}(aq)$. Describe the observations at each step, including what happens when dilute and concentrated ammonia are subsequently added.

3. [Hard] Write balanced equations for the reactions of sulfur dioxide with: (a) water, (b) sodium hydroxide solution (limited and excess), (c) oxygen in the presence of $\mathrm{V}_2\mathrm{O}_5$. State the role of $\mathrm{V}_2\mathrm{O}_5$ in reaction (c).

4. [Hard] The chloride of an unknown Period 3 element produces a white smoke when exposed to moist air and dissolves in water to give an acidic solution. Identify the element and justify your answer with reference to Period 3 trends in chlorides.

5. [Medium] Explain why the melting point of argon ($84\mathrm{ K}$) is significantly lower than that of chlorine ($172\mathrm{ K}$), despite both being simple molecular substances at low temperatures.

6. [Hard] A solution containing $\mathrm{Fe}^{3+}$ ions is pale yellow. When excess $\mathrm{KCN}$ is added, the solution becomes colourless. Account for both observations in terms of crystal field theory and the spectrochemical series.

7. [Medium] State and explain the trend in reactivity of Group 17 elements with hydrogen. Include balanced equations for the reactions of fluorine, chlorine, and bromine with hydrogen, noting any conditions required.

8. [Hard] EDTA ($\mathrm{C}_{10}\mathrm{H}_{16}\mathrm{N}_2\mathrm{O}_8$) forms a 1:1 complex with $\mathrm{Ca}^{2+}$. Calculate the mass of EDTA required to complex exactly $50.0\mathrm{ mg}$ of $\mathrm{Ca}^{2+}$ ions. ($M_r$ of EDTA $= 292\mathrm{ g/mol}$, $A_r$ of $\mathrm{Ca} = 40.1$)

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Worked Examples (Expanded)

Worked Example: Lattice Energy Calculation Using the Born-Haber Cycle

Calculate the lattice energy of sodium chloride using the Born-Haber cycle.

Quantity	Value ($\mathrm{kJ/mol}$)
$\Delta H_f^\circ(\mathrm{NaCl})$	$-411$
$\Delta H_\mathrm{atom}(\mathrm{Na})$	$+108$
$\Delta H_\mathrm{atom}(\mathrm{Cl}_2)$ (per mole of atoms)	$+122$
$IE_1(\mathrm{Na})$	$+496$
$EA_1(\mathrm{Cl})$	$-349$

Solution

The Born-Haber cycle relates the enthalpy of formation to the individual energy steps:

$\Delta H_f^\circ = \Delta H_\mathrm{atom}(\mathrm{Na}) + \Delta H_\mathrm{atom}(\mathrm{Cl}) + IE_1(\mathrm{Na}) + EA_1(\mathrm{Cl}) + U$

Solving for the lattice energy $U$:

$U = \Delta H_f^\circ - \Delta H_\mathrm{atom}(\mathrm{Na}) - \Delta H_\mathrm{atom}(\mathrm{Cl}) - IE_1(\mathrm{Na}) - EA_1(\mathrm{Cl})$

$U = -411 - 108 - 122 - 496 - (-349) = -788\;\mathrm{kJ/mol}$

The lattice energy of $\mathrm{NaCl}$ is $-788\;\mathrm{kJ/mol}$. The large negative value reflects the strong electrostatic attraction between $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ in the ionic lattice.

Worked Example: Predicting Ionisation Energy Anomalies

Explain why the first ionisation energy of oxygen ($1314\;\mathrm{kJ/mol}$) is lower than that of nitrogen ($1402\;\mathrm{kJ/mol}$), even though oxygen has a greater nuclear charge.

Solution

Nitrogen has the electron configuration $1s^2\;2s^2\;2p^3$ — the $2p$ subshell is exactly half-filled. Half-filled subshells have extra stability due to exchange energy (parallel spins in degenerate orbitals are quantum-mechanically favoured). Removing an electron from a half-filled $p$ subshell disrupts this symmetry, costing additional energy.

Oxygen has the configuration $1s^2\;2s^2\;2p^4$. The fourth $2p$ electron must pair with an existing electron in one of the $p$ orbitals. The paired electrons experience mutual electron-electron repulsion, which destabilises the atom slightly. Removing one electron from a paired orbital relieves this repulsion, making ionisation slightly easier (lower $IE_1$) than for nitrogen.

This is a specific example of a broader principle: half-filled and fully-filled subshells confer extra stability, producing local maxima in ionisation energy across a period.

Worked Example: Enthalpy of Solution and Hydration

The enthalpy of solution of $\mathrm{CaCl}_2$ is $-82.8\;\mathrm{kJ/mol}$. The lattice energy is $-2258\;\mathrm{kJ/mol}$. Calculate the total hydration enthalpy of $\mathrm{Ca}^{2+}$ and $2\mathrm{Cl}^-$. Given that the hydration enthalpy of $\mathrm{Cl}^-$ is $-363\;\mathrm{kJ/mol}$, determine the hydration enthalpy of $\mathrm{Ca}^{2+}$.

Solution

The enthalpy of solution is the sum of the endothermic lattice-breaking step and the exothermic hydration step:

$\Delta H_\mathrm{sol} = \Delta H_\mathrm{lattice} + \Delta H_\mathrm{hydration}$

$-82.8 = +2258 + \Delta H_\mathrm{hydration}$

$\Delta H_\mathrm{hydration} = -82.8 - 2258 = -2341\;\mathrm{kJ/mol}$

The total hydration enthalpy is $-2341\;\mathrm{kJ/mol}$, which is the sum of the hydration enthalpies of all ions:

$\Delta H_\mathrm{hydration} = \Delta H_\mathrm{hyd}(\mathrm{Ca}^{2+}) + 2 \times \Delta H_\mathrm{hyd}(\mathrm{Cl}^-)$

$-2341 = \Delta H_\mathrm{hyd}(\mathrm{Ca}^{2+}) + 2(-363)$

$\Delta H_\mathrm{hyd}(\mathrm{Ca}^{2+}) = -2341 + 726 = -1615\;\mathrm{kJ/mol}$

The hydration enthalpy of $\mathrm{Ca}^{2+}$ is $-1615\;\mathrm{kJ/mol}$. This large magnitude reflects the high charge density of the small, doubly charged cation and its strong interaction with water molecules.

Worked Example: Transition Metal Oxidation States in Redox Reactions

A solution of acidified potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) is used to oxidise $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$. The chromium is reduced from the $+6$ to the $+3$ oxidation state. (a) Write the balanced ionic equation. (b) Calculate the volume of $0.0200\;\mathrm{M}$ $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ solution required to oxidise $25.0\;\mathrm{mL}$ of $0.100\;\mathrm{M}$ $\mathrm{FeSO}_4$ solution.

Solution

(a) Balancing the equation:

Reduction half-reaction: $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Oxidation half-reaction: $\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-$

Balancing electrons ($\times 6$ for the iron half-reaction): $6\mathrm{Fe}^{2+} \to 6\mathrm{Fe}^{3+} + 6e^-$

Overall: $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{Fe}^{2+} \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 6\mathrm{Fe}^{3+}$

(b) Stoichiometric calculation:

$n(\mathrm{Fe}^{2+}) = 0.100 \times 0.0250 = 0.00250\;\mathrm{mol}$

From the equation, $6\;\mathrm{mol\;Fe}^{2+}$ react with $1\;\mathrm{mol\;Cr}_2\mathrm{O}_7^{2-}$:

$n(\mathrm{Cr}_2\mathrm{O}_7^{2-}) = \frac{0.00250}{6} = 4.17 \times 10^{-4}\;\mathrm{mol}$

$V(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7) = \frac{n}{c} = \frac{4.17 \times 10^{-4}}{0.0200} = 0.0208\;\mathrm{L} = 20.8\;\mathrm{mL}$

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Exam-Style Problems (Expanded)

Problem 9: Quantitative -- Comparing Lattice Energies

Arrange the following ionic compounds in order of increasing lattice energy (most negative first): $\mathrm{NaF}$, $\mathrm{MgO}$, $\mathrm{NaCl}$, $\mathrm{MgCl}_2$, $\mathrm{Al}_2\mathrm{O}_3$. Justify your ordering using Coulomb's law: $U \propto \frac{z_+ z_-}{r_+ + r_-}$.

Problem 10: Extended Response -- d-Block Contraction

The atomic radius of $\mathrm{Zr}$ ($160\;\mathrm{pm}$) is almost identical to that of $\mathrm{Hf}$ ($159\;\mathrm{pm}$), despite Hf being one period below Zr. Explain this observation using the concept of the $d$-block contraction (lanthanide contraction). Discuss the consequences for the chemical similarity of Group 4 elements and the difficulty of separating Zr and Hf in industrial processes.

Problem 11: Quantitative -- Halogen Displacement Equilibrium

The equilibrium constant for the reaction $\mathrm{Cl}_2(aq) + 2\mathrm{Br}^-(aq) \rightleftharpoons 2\mathrm{Cl}^-(aq) + \mathrm{Br}_2(aq)$ is $K = 4.0 \times 10^{11}$ at $298\;\mathrm{K}$. (a) Calculate $E_{\mathrm{cell}}^\circ$ for the reaction. (b) If $0.050\;\mathrm{mol}$ of $\mathrm{Cl}_2$ is bubbled into $1.0\;\mathrm{L}$ of $0.10\;\mathrm{M}$ $\mathrm{KBr}$, calculate the equilibrium concentration of $\mathrm{Br}_2$. ($E^\circ(\mathrm{Cl}_2/\mathrm{Cl}^-) = +1.36\;\mathrm{V}$, $E^\circ(\mathrm{Br}_2/\mathrm{Br}^-) = +1.09\;\mathrm{V}$)

Problem 12: Extended Response -- Crystal Field Theory and Colour

The complex ion $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ is pink and $[\mathrm{Co}(\mathrm{Cl}_4)]^{2-}$ is blue. (a) State the oxidation state of cobalt in each complex. (b) Use the spectrochemical series to explain the colour difference. (c) Calculate the approximate wavelength of light absorbed by $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ if it absorbs green light ($\lambda \approx 520\;\mathrm{nm}$) and state what colour is observed. (d) Explain why $[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ is colourless.

Problem 13: Quantitative -- Successive Ionisation Energies

The first five ionisation energies of an element ($\mathrm{kJ/mol}$) are: $578$, $1817$, $2745$, $11577$, $14842$. (a) Identify the group of the element. (b) Write the electron configuration of the element. (c) The element is in Period 3. Identify it and explain the large jump between $IE_3$ and $IE_4$.

Problem 14: Extended Response -- Amphoteric Hydroxides

Both $\mathrm{Al}(\mathrm{OH})_3$ and $\mathrm{Zn}(\mathrm{OH})_2$ are amphoteric. Write balanced equations for the reaction of each with excess $\mathrm{NaOH}(aq)$ and with excess $\mathrm{HCl}(aq)$. Explain the structural reason why these hydroxides are amphoteric while $\mathrm{Mg}(\mathrm{OH})_2$ is purely basic. Discuss the relevance of amphoterism to the extraction of aluminium from bauxite ore via the Bayer process.

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Common Pitfalls (Expanded)

• Assuming all $d$-block elements are transition metals: The IB definition requires a partially filled $d$-subshell in a common oxidation state. $\mathrm{Sc}^{3+}$ has $3d^0$ and $\mathrm{Zn}^{2+}$ has $3d^{10}$ — neither qualifies. Always check the common ion, not just the atom.

• Reversing the polarising power trend: Smaller, more highly charged cations have greater polarising power. $\mathrm{Be}^{2+}$ polarises more than $\mathrm{Ba}^{2+}$; $\mathrm{Al}^{3+}$ polarises more than $\mathrm{Na}^+$. This is why $\mathrm{BeCl}_2$ and $\mathrm{AlCl}_3$ have significant covalent character.

• Ignoring the disproportionation of chlorine in cold vs hot alkali: In cold dilute $\mathrm{NaOH}$, chlorine disproportionates to $\mathrm{Cl}^-$ and $\mathrm{ClO}^-$. In hot concentrated $\mathrm{NaOH}$, it disproportionates to $\mathrm{Cl}^-$ and $\mathrm{ClO}_3^-$. The product depends on temperature.

• Stating that all Group 2 hydroxides are strong bases: $\mathrm{Be}(\mathrm{OH})_2$ is amphoteric, not basic. The trend from basic to amphoteric applies from $\mathrm{Mg}$ upward, but $\mathrm{Be}$ is the exception due to its high charge density.

• Confusing the oxidation state of oxygen in peroxides and superoxides: In normal oxides ($\mathrm{O}^{2-}$), oxygen is $-2$. In peroxides ($\mathrm{O}_2^{2-}$), each oxygen is $-1$. In superoxides ($\mathrm{O}_2^-$), the average oxidation state is $-0.5$. This affects balancing redox equations involving these species.

• Writing $E_{\mathrm{cell}}^\circ = E_{\mathrm{anode}}^\circ - E_{\mathrm{cathode}}^\circ$: This is the single most common sign error in electrochemistry. The correct formula is always $E_{\mathrm{cell}}^\circ = E_{\mathrm{cathode}}^\circ - E_{\mathrm{anode}}^\circ$ (cathode minus anode). A negative result means the reaction is non-spontaneous under standard conditions.

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If You Get These Wrong, Revise:

• Electron configurations and ionization energy → Review ./atomic-theory
• Intermolecular forces and boiling point trends → Review ./states-of-matter
• Acid-base character of oxides → Review ./acids-bases-advanced
• Oxidation states and redox reactions → Review ./redox-advanced
• Bonding and structure (ionic vs covalent) → Review ./chemical-bonding-advanced