Chemical Reactions and Bonding

Higher Chemical Bonding

Types of Bonding

Ionic Bonding:

The electrostatic attraction between oppositely charged ions formed by electron transfer. Occurs between metals (Groups 1, 2) and non-metals (Groups 6, 7).

Properties of ionic compounds:

High melting and boiling points (strong electrostatic forces)
Conduct electricity when molten or dissolved (ions are free to move)
Brittle (repulsion between like charges when layers shift)
Soluble in polar solvents like water

Example: Formation of magnesium oxide.

$\mathrm{Mg \to \mathrm{Mg^{2+} + 2e^-$ $\mathrm{O + 2e^- \to \mathrm{O^{2-}$

$\mathrm{Mg(s) + \tfrac{1}{2}\mathrm{O_2\mathrm{(g) \to \mathrm{MgO(s)$

Worked Example 1: Draw a dot-and-cross diagram for calcium fluoride, $\mathrm{CaF_2$ .

Calcium ( $Z = 20$ ) has electron configuration $2, 8, 8, 2$ . It loses two electrons to form $\mathrm{Ca^{2+}$ ( $2, 8, 8$ ). Each fluorine ( $Z = 9$ ) has electron configuration $2, 7$ and gains One electron to form $\mathrm{F^-$ ( $2, 8$ ). The dot-and-cross diagram shows two electrons Transferred from calcium (crosses) to one electron each accepted by two fluorine atoms (dots). The Resulting $\mathrm{Ca^{2+}$ and two $\mathrm{F^-$ ions are held together by electrostatic attraction In a giant ionic lattice.

Covalent Bonding:

A shared pair of electrons between two atoms. Covalent bonds can be single, double, or triple.

Properties of covalent substances:

Simple molecular: low melting points, do not conduct electricity (e.g., $\mathrm{H_2\mathrm{O$ $\mathrm{CO_2$ )
Giant covalent (macromolecular): very high melting points (e.g., diamond, graphite, silicon dioxide)

Worked Example 2: Draw a dot-and-cross diagram for $\mathrm{H_2\mathrm{S$ .

Sulfur ( $Z = 16$ ) has electron configuration $2, 8, 6$ . It needs two electrons to complete its outer Shell. Each hydrogen ( $Z = 1$ ) contributes one electron. Two single covalent bonds form, each Consisting of one shared pair. The shape predicted by VSEPR is bent (two bonding pairs and two lone Pairs on sulfur).

Metallic Bonding:

Positive metal ions in a “sea” of delocalised electrons.

Properties of metals:

Good conductors of electricity (delocalised electrons)
Malleable and ductile (layers of ions can slide)
High melting points (strong metallic bonding)
Shiny (delocalised electrons absorb and re-emit light)

Comparison of Bonding Types

Property	Ionic	Covalent (simple molecular)	Covalent (giant)	Metallic
Melting/boiling point	High	Low	Very high	High (variable)
Electrical conductivity	Molten/dissolved only	None	Graphite only	Good
Solubility in water	Generally soluble	Generally insoluble	Insoluble	Insoluble
Hardness	Hard, brittle	Soft	Very hard	Malleable
Example	$\mathrm{NaCl$	$\mathrm{H_2\mathrm{O$	Diamond, $\mathrm{SiO_2$	$\mathrm{Fe$ , $\mathrm{Cu$

Electronegativity and Bond Polarity

Electronegativity is the ability of an atom to attract the bonding electrons towards itself in a Covalent bond.

Pauling Scale:

Fluorine is the most electronegative element (4.0)
Cesium and francium are the least electronegative (0.7)

Trends in electronegativity:

Increases across a period (nuclear charge increases, atomic radius decreases)
Decreases down a group (shielding increases, atomic radius increases)

Bond polarity:

If the electronegativity difference is 0-0.4: non-polar covalent
If 0.5-1.7: polar covalent
If greater than 1.7: ionic

Dipole moment: A polar molecule has an overall dipole moment, which is the vector sum of Individual bond dipoles.

Example: $\mathrm{H_2\mathrm{O$ is a polar molecule because the bond dipoles do not cancel (bent Shape). $\mathrm{CO_2$ is non-polar because the bond dipoles cancel (linear shape).

Worked Example 3: Determine whether $\mathrm{NH_3$ and $\mathrm{BF_3$ are polar or non-polar.

$\mathrm{NH_3$ : Nitrogen ( $\mathrm{EN = 3.04$ ) and hydrogen ( $\mathrm{EN = 2.20$ ) give $\Delta\mathrm{EN = 0.84$ So each N-H bond is polar. The molecule has trigonal pyramidal shape (three bonding pairs, one lone pair). The bond dipoles do not cancel, so $\mathrm{NH_3$ is polar.

$\mathrm{BF_3$ : Boron ( $\mathrm{EN = 2.04$ ) and fluorine ( $\mathrm{EN = 3.98$ ) give $\Delta\mathrm{EN = 1.94$ So each B-F bond is polar. The molecule is trigonal planar and the bond Dipoles cancel perfectly, so $\mathrm{BF_3$ is non-polar.

Molecular Shape (VSEPR Theory)

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular shapes based on electron Pair repulsion.

Electron Pairs	Shape	Bond Angle	Example
2	Linear	180°	$\mathrm{BeCl_2$
3	Trigonal planar	120°	$\mathrm{BF_3$
4	Tetrahedral	109.5°	$\mathrm{CH_4$
5	Trigonal bipyramidal	90°, 120°	$\mathrm{PCl_5$
6	Octahedral	90°	$\mathrm{SF_6$

Lone pairs occupy more space than bonding pairs, reducing bond angles:

Bonding/Lone Pairs	Shape	Bond Angle	Example
3 bonding, 1 lone	Trigonal pyramidal	107°	$\mathrm{NH_3$
2 bonding, 2 lone	Bent	104.5°	$\mathrm{H_2\mathrm{O$

Derivation of bond angles using VSEPR:

The electron pairs around a central atom repel each other to positions that maximise separation. For Four electron pairs (tetrahedral), the angle that maximises separation in three dimensions is $\arccos(-1/3) \approx 109.5^\circ$ . When lone pairs replace bonding pairs, the lone pair—bond pair Repulsion is greater than bond pair—bond pair repulsion, compressing the remaining bond angles. This explains why $\mathrm{NH_3$ has bond angles of $107^\circ$ and $\mathrm{H_2\mathrm{O$ has $104.5^\circ$ .

Worked Example 4: Predict the shape and bond angle of $\mathrm{SF_4$ .

Sulfur in $\mathrm{SF_4$ has 6 valence electrons. Four are used in S-F bonds, leaving one lone pair. Total electron domains = 5 (4 bonding + 1 lone pair). This corresponds to a trigonal bipyramidal Electron geometry. The lone pair occupies an equatorial position to minimise repulsion (two 90° Interactions rather than three). The molecular shape is see-saw with bond angles of Approximately $102^\circ$ (equatorial) and $173^\circ$ (axial-equatorial).

Intermolecular Forces

Van der Waals (London dispersion) forces:

Weakest intermolecular force
Caused by temporary dipoles due to electron movement
Increase with molecular size (number of electrons)
Present in all molecules

Derivation of London force dependence on molecular size:

Larger molecules have more electrons. The electron cloud is more diffuse and can be more Distorted (higher polarisability). A larger, more polarisable electron cloud produces a larger Instantaneous dipole, which induces a larger dipole in a neighbouring molecule. Therefore, the Resulting attractive force is stronger. This is why boiling points increase down Group 7 ( $\mathrm{F_2$ < $\mathrm{Cl_2$ < $\mathrm{Br_2$ < $\mathrm{I_2$ ).

Permanent dipole-dipole forces:

Between polar molecules
Stronger than London forces
Examples: $\mathrm{HCl$ , $\mathrm{SO_2$

Hydrogen bonding:

Strongest type of intermolecular force
Occurs when hydrogen is bonded to a highly electronegative atom (F, O, or N)
Responsible for the anomalously high boiling point of water

Example: Explain why $\mathrm{H_2\mathrm{O$ has a higher boiling point than $\mathrm{H_2\mathrm{S$ Despite $\mathrm{H_2\mathrm{S$ having a larger molar mass.

$\mathrm{H_2\mathrm{O$ can form hydrogen bonds between molecules (H bonded to O), which are much Stronger than the dipole-dipole forces in $\mathrm{H_2\mathrm{S$ . Although $\mathrm{H_2\mathrm{S$ has Stronger London forces due to more electrons, hydrogen bonding in water dominates.

Summary of Intermolecular Forces

Force Type	Strength Order	Present In	Dependence on molecular size
London dispersion	Weakest	All molecules	Increases with size
Permanent dipole-dipole	Moderate	Polar molecules	Moderate
Hydrogen bonding	Strongest	Molecules with H bonded to N, O, F	Depends on H-bond density

Bond Enthalpy

Bond enthalpy is the energy required to break one mole of a particular bond in the gaseous state (endothermic).

Average bond enthalpies are used because bond enthalpies vary slightly depending on the Molecular environment.

Bond	Enthalpy (kJ/mol)
C-C	348
C=C	612
C-H	412
O-H	463
C=O	743
H-H	436
O=O	496
N≡N	945
C≡C	839
C-O	358
C-Cl	328
C-N	305

Worked Example 5: Estimate the enthalpy change for the reaction: $\mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O$ .

Bonds broken: $4 \times \mathrm{C-H + 2 \times \mathrm{O=O = 4(412) + 2(496) = 1648 + 992 = 2640 \mathrm{ kJ/mol$ .

Bonds formed: $2 \times \mathrm{C=O + 4 \times \mathrm{O-H = 2(743) + 4(463) = 1486 + 1852 = 3338 \mathrm{ kJ/mol$ .

$\Delta H = 2640 - 3338 = -698 \mathrm{ kJ/mol$

The negative value confirms the reaction is exothermic.

Worked Example 6: Using bond enthalpies, estimate $\Delta H$ for the hydrogenation of ethene: $\mathrm{C_2\mathrm{H_4 + \mathrm{H_2 \to \mathrm{C_2\mathrm{H_6$ .

Bonds broken: $1 \times \mathrm{C=C + 4 \times \mathrm{C-H + 1 \times \mathrm{H-H$ $= 612 + 4(412) + 436 = 612 + 1648 + 436 = 2696 \mathrm{ kJ/mol$ .

Bonds formed: $1 \times \mathrm{C-C + 6 \times \mathrm{C-H$ $= 348 + 6(412) = 348 + 2472 = 2820 \mathrm{ kJ/mol$ .

$\Delta H = 2696 - 2820 = -124 \mathrm{ kJ/mol$

Proof that bond enthalpy gives an approximation:

Average bond enthalpies are mean values taken from many different compounds. The actual C-H bond Enthalpy in methane differs from that in ethane because the electronic environment is different. Therefore, using average values introduces systematic error. The true $\Delta H$ for the Hydrogenation of ethene is $-137 \mathrm{ kJ/mol$ Showing that the bond enthalpy estimate ( $-124 \mathrm{ kJ/mol$ ) is approximate.

Chemical Reactions

Balancing Equations

Worked Example 7: Balance the equation for the combustion of propane.

$\mathrm{C_3\mathrm{H_8 + 5\mathrm{O_2 \to 3\mathrm{CO_2 + 4\mathrm{H_2\mathrm{O$

Step-by-step: Balance C first (3 $\mathrm{CO_2$ ), then H (4 $\mathrm{H_2\mathrm{O$ ), then O (needs $6 + 4 = 10$ O atoms on left, so $5\mathrm{O_2$ ).

Worked Example 8: Balance: $\mathrm{Fe_2\mathrm{O_3 + \mathrm{CO \to \mathrm{Fe + \mathrm{CO_2$ .

Balance Fe: $2\mathrm{Fe$ on right, so $\mathrm{Fe_2\mathrm{O_3 + \mathrm{CO \to 2\mathrm{Fe + \mathrm{CO_2$ . Balance C: C is already balanced (1 on each side). Balance O: left has $3 + 1 = 4$ O atoms, right has $2$ . Need $2\mathrm{CO_2$ on Right, so $2\mathrm{CO$ on left.

$\mathrm{Fe_2\mathrm{O_3 + 3\mathrm{CO \to 2\mathrm{Fe + 3\mathrm{CO_2$

Ionic Equations

Only include species that actually change during the reaction (spectator ions are omitted).

Worked Example 9: Write the ionic equation for the reaction of silver nitrate with sodium Chloride.

Full equation: $\mathrm{Ag^+(aq) + \mathrm{NO_3^-(aq) + \mathrm{Na^+(aq) + \mathrm{Cl^-(aq) \to \mathrm{AgCl(s) + \mathrm{Na^+(aq) + \mathrm{NO_3^-(aq)$

$\mathrm{Na^+$ and $\mathrm{NO_3^-$ are spectator ions.

Net ionic equation: $\mathrm{Ag^+(aq) + \mathrm{Cl^-(aq) \to \mathrm{AgCl(s)$

Worked Example 10: Write the ionic equation for the reaction of zinc with dilute sulfuric acid.

$\mathrm{Zn(s) + 2\mathrm{H^+(aq) + \mathrm{SO_4^{2-}(aq) \to \mathrm{Zn^{2+}(aq) + \mathrm{SO_4^{2-}(aq) + \mathrm{H_2(g)$

$\mathrm{SO_4^{2-}$ is a spectator ion.

Net ionic equation: $\mathrm{Zn(s) + 2\mathrm{H^+(aq) \to \mathrm{Zn^{2+}(aq) + \mathrm{H_2(g)$

Oxidation and Reduction

Oxidation: Loss of electrons (increase in oxidation state).

Reduction: Gain of electrons (decrease in oxidation state).

OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Oxidation states rules:

Elements in their standard state: 0
Monatomic ions: equal to the charge
Oxygen: $-2$ (except in peroxides, $-1$ )
Hydrogen: $+1$ (except in metal hydrides, $-1$ )
Sum of oxidation states equals the overall charge

Worked Example 11: Determine the oxidation states in $\mathrm{KMnO_4$ .

$\mathrm{K = +1$ , $\mathrm{O = -2$ So $\mathrm{Mn + 1 + 4(-2) = 0$ Giving $\mathrm{Mn = +7$ .

Worked Example 12: Determine the oxidation states of each element in $\mathrm{H_2\mathrm{O_2$ and $\mathrm{Na_2\mathrm{Cr_2\mathrm{O_7$ .

$\mathrm{H_2\mathrm{O_2$ : $\mathrm{H = +1$ , $2(+1) + 2(\mathrm{O) = 0$ So $\mathrm{O = -1$ . This is a Peroxide where oxygen has an unusual oxidation state of $-1$ .

$\mathrm{Na_2\mathrm{Cr_2\mathrm{O_7$ : $\mathrm{Na = +1$ , $\mathrm{O = -2$ . $2(+1) + 2(\mathrm{Cr) + 7(-2) = 0$ So $2(\mathrm{Cr) = 12$ Giving $\mathrm{Cr = +6$ .

Redox Reactions

Worked Example 13: Balance the reaction of $\mathrm{MnO_4^-$ with $\mathrm{Fe^{2+}$ in acidic Solution.

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

Reduction: $\mathrm{MnO_4^- + 8\mathrm{H^+ + 5e^- \to \mathrm{Mn^{2+} + 4\mathrm{H_2\mathrm{O$

Multiply oxidation by 5:

$5\mathrm{Fe^{2+} + \mathrm{MnO_4^- + 8\mathrm{H^+ \to 5\mathrm{Fe^{3+} + \mathrm{Mn^{2+} + 4\mathrm{H_2\mathrm{O$

Worked Example 14: Balance $\mathrm{Cr_2\mathrm{O_7^{2-} + \mathrm{SO_2 \to \mathrm{Cr^{3+} + \mathrm{SO_4^{2-}$ in acidic solution.

Oxidation: $\mathrm{SO_2 + 2\mathrm{H_2\mathrm{O \to \mathrm{SO_4^{2-} + 4\mathrm{H^+ + 2e^-$

Reduction: $\mathrm{Cr_2\mathrm{O_7^{2-} + 14\mathrm{H^+ + 6e^- \to 2\mathrm{Cr^{3+} + 7\mathrm{H_2\mathrm{O$

Multiply oxidation by 3 to balance electrons:

$\mathrm{Cr_2\mathrm{O_7^{2-} + 3\mathrm{SO_2 + 2\mathrm{H^+ \to 2\mathrm{Cr^{3+} + 3\mathrm{SO_4^{2-} + \mathrm{H_2\mathrm{O$

Check: Cr: 2 left, 2 right. S: 3 left, 3 right. O: $7 + 6 = 13$ left, $12 + 1 = 13$ right. H: $2$ Left, $2$ right. Charge: $-2 + 0 + 2 = 0$ left, $6 + (-9) = -6 + 2(+2) = -2$ … Let us recheck.

Left charge: $-2 + 2 = 0$ . Right charge: $2(+3) + 3(-2) = 0$ . Balanced.

Summary Table: Oxidation States of Common Elements

Element	Common oxidation states	Notes
Hydrogen	$+1$ (most compounds), $-1$ (hydrides)
Oxygen	$-2$ , $-1$ (peroxides)	$+2$ in $\mathrm{OF_2$
Nitrogen	$-3$ to $+5$	Most variable after transition metals
Sulfur	$-2$ to $+6$
Chlorine	$-1$ to $+7$	$-1$ most common
Manganese	$+2, +4, +7$	$\mathrm{MnO_4^-$ has Mn at $+7$
Chromium	$+3, +6$	$\mathrm{Cr_2\mathrm{O_7^{2-}$ has Cr at $+6$

Common Pitfalls

Confusing ionic and covalent bonding: Ionic bonding involves electron transfer; covalent involves electron sharing.
VSEPR with lone pairs: Lone pairs repel more strongly than bonding pairs, reducing bond angles below the ideal values.
Bond polarity vs. Molecular polarity: A molecule can have polar bonds but be non-polar overall (e.g., $\mathrm{CCl_4$ , $\mathrm{CO_2$ ).
Using average bond enthalpies: These give approximate values only. They are less accurate when bonds are in different molecular environments.
Oxidation state of oxygen in peroxides: In $\mathrm{H_2\mathrm{O_2$ Oxygen has oxidation state $-1$ Not $-2$ .
Spectator ions in ionic equations: Always identify and remove spectator ions. Remember that state symbols are essential.
Hydrogen bonding criteria: H must be bonded directly to N, O, or F. H bonded to Cl is NOT hydrogen bonding, despite chlorine being electronegative.

Practice Questions

Draw dot-and-cross diagrams for (a) $\mathrm{MgCl_2$ and (b) $\mathrm{H_2\mathrm{S$ . State the type of bonding in each.
Explain why the boiling point of $\mathrm{NH_3$ (240 K) is higher than that of $\mathrm{PH_3$ (185 K), despite $\mathrm{PH_3$ having more electrons.
Predict the shapes and bond angles of $\mathrm{ClF_3$ and $\mathrm{XeF_4$ .
Using bond enthalpies, estimate $\Delta H$ for the hydrogenation of ethene: $\mathrm{C_2\mathrm{H_4 + \mathrm{H_2 \to \mathrm{C_2\mathrm{H_6$ .
Balance the following redox equation in acidic solution: $\mathrm{Cr_2\mathrm{O_7^{2-} + \mathrm{SO_2 \to \mathrm{Cr^{3+} + \mathrm{SO_4^{2-}$ .
Explain why graphite conducts electricity but diamond does not.
Determine the oxidation state of chromium in $\mathrm{Na_2\mathrm{Cr_2\mathrm{O_7$ .
Compare and contrast the properties of metallic and ionic bonding, explaining the differences in terms of structure.
Using bond enthalpy data, estimate $\Delta H$ for the combustion of ethyne: $\mathrm{C_2\mathrm{H_2 + \tfrac{5}{2}\mathrm{O_2 \to 2\mathrm{CO_2 + \mathrm{H_2\mathrm{O$ .
Explain why the boiling point increases in the order $\mathrm{CH_4 \lt \mathrm{SiH_4 \lt \mathrm{GeH_4 \lt \mathrm{SnH_4$ and why $\mathrm{NH_3$ does not follow this pattern.
Write ionic equations for the reactions of (a) barium chloride with sodium sulfate, and (b) magnesium with hydrochloric acid.
Balance the following in acidic solution: $\mathrm{MnO_4^- + \mathrm{C_2\mathrm{O_4^{2-} \to \mathrm{Mn^{2+} + \mathrm{CO_2$ .

Advanced Bonding Concepts

Ionic Lattice Energy

Lattice energy is the enthalpy change when one mole of an ionic compound is formed from its Gaseous ions. It is always exothermic (negative) and indicates the strength of ionic bonding.

Factors affecting lattice energy:

Ionic charge: Higher charges produce stronger attraction and more exothermic lattice energy. $\mathrm{MgO$ ( $\mathrm{Mg^{2+}$ , $\mathrm{O^{2-}$ ) has a much more exothermic lattice energy than $\mathrm{NaCl$ ( $\mathrm{Na^+$ , $\mathrm{Cl^-$ ).
Ionic radius: Smaller ions can approach more closely, increasing the attraction. $\mathrm{NaF$ has a more exothermic lattice energy than $\mathrm{NaCl$ .

Compound	Ionic charges	Sum of ionic radii (pm)	Lattice energy (kJ/mol)
NaCl	+1, -1	276	-787
MgO	+2, -2	210	-3791
NaF	+1, -1	231	-910
CaO	+2, -2	231	-3414

Born-Haber Cycles

A Born-Haber cycle is a thermochemical cycle that relates lattice energy to other measurable Enthalpy changes. It is an application of Hess’s Law to ionic compounds.

Worked Example 15: Construct a Born-Haber cycle for sodium chloride.

The cycle involves the following steps:

$\mathrm{Na(s) \to \mathrm{Na(g)$ — enthalpy of atomisation, $\Delta H_{\mathrm{at} = +108 \mathrm{ kJ/mol$
$\tfrac{1}{2}\mathrm{Cl_2\mathrm{(g) \to \mathrm{Cl(g)$ — enthalpy of atomisation, $\tfrac{1}{2}\Delta H_{\mathrm{at} = +122 \mathrm{ kJ/mol$
$\mathrm{Na(g) \to \mathrm{Na^+(g) + e^-$ — first ionisation energy, $\mathrm{IE_1 = +496 \mathrm{ kJ/mol$
$\mathrm{Cl(g) + e^- \to \mathrm{Cl^-(g)$ — first electron affinity, $\mathrm{EA_1 = -349 \mathrm{ kJ/mol$
$\mathrm{Na^+(g) + \mathrm{Cl^-(g) \to \mathrm{NaCl(s)$ — lattice energy, $\Delta H_{\mathrm{latt} = ?$

By Hess’s Law:

$\Delta H_f = \Delta H_{\mathrm{at}(\mathrm{Na) + \tfrac{1}{2}\Delta H_{\mathrm{at}(\mathrm{Cl) + \mathrm{IE_1 + \mathrm{EA_1 + \Delta H_{\mathrm{latt}$

$-411 = 108 + 122 + 496 + (-349) + \Delta H_{\mathrm{latt}$

$\Delta H_{\mathrm{latt} = -411 - 108 - 122 - 496 + 349 = -788 \mathrm{ kJ/mol$

Polarising Power and Covalent Character

Small, highly charged cations have high polarising power. Large, highly charged anions are Polarised. When polarisation is significant, the ionic bond acquires covalent character.

Fajans’ Rules:

A small cation has high polarising power (e.g., $\mathrm{Li^+$ vs. $\mathrm{Cs^+$ )
A large anion is more polarised (e.g., $\mathrm{I^-$ vs. $\mathrm{F^-$ )
A high charge on either ion increases polarisation

Consequences:

$\mathrm{AlCl_3$ has significant covalent character despite the electronegativity difference suggesting ionic bonding.
$\mathrm{SnCl_4$ is a covalent liquid at room temperature, whereas $\mathrm{SnCl_2$ is a solid ionic compound.

Metallic Bonding: Alloy Properties

Alloys are mixtures of metals (or a metal with a non-metal) that have been designed to have Improved properties.

Alloy type	Description	Effect on properties
Substitutional	Different-sized atoms replace host atoms	Distorts lattice, prevents slip planes, increases hardness
Interstitial	Small atoms fit in gaps between metal atoms	Blocks dislocation movement, increases hardness
Example: Steel	Iron with carbon (interstitial)	Much harder than pure iron
Example: Brass	Copper with zinc (substitutional)	Stronger than copper, retains malleability

Worked Example: Predicting Bond Type

Question: Predict the type of bonding in each of the following and justify your answer: (a) $\mathrm{KF$ (b) $\mathrm{CCl_4$ (c) $\mathrm{Cu$ (d) $\mathrm{AlCl_3$ .

(a) $\mathrm{KF$ : Potassium is a Group 1 metal and fluorine is a Group 17 non-metal. The Electronegativity difference is $3.98 - 0.82 = 3.16$ Which is greater than 1.7, so this is Predominantly ionic bonding.

(b) $\mathrm{CCl_4$ : Both carbon and chlorine are non-metals. The electronegativity difference is $3.16 - 2.55 = 0.61$ Which is in the polar covalent range. The molecule is tetrahedral and the bond Dipoles cancel, making the molecule non-polar overall. This is covalent bonding.

(c) $\mathrm{Cu$ : Copper is a metal. The bonding involves positive metal ions in a sea of Delocalised electrons. This is metallic bonding, accounting for copper’s conductivity, Malleability, and high melting point.

(d) $\mathrm{AlCl_3$ : Aluminium is a metal and chlorine is a non-metal. The electronegativity Difference is $3.16 - 1.61 = 1.55$ Suggesting polar covalent. $\mathrm{Al^{3+}$ is a small, highly Charged cation with high polarising power, and $\mathrm{Cl^-$ is a relatively large anion. According To Fajans’ Rules, this gives $\mathrm{AlCl_3$ significant covalent character. Indeed, $\mathrm{AlCl_3$ sublimes at $180^\circ\mathrm{C$ (low for an ionic compound) and forms dimer molecules $\mathrm{Al_2\mathrm{Cl_6$ in the gas phase.

Derivation: Why $\mathrm{H_2\mathrm{O$ Is Bent but $\mathrm{CO_2$ Is Linear

Both $\mathrm{H_2\mathrm{O$ and $\mathrm{CO_2$ contain three atoms, yet they have different shapes.

$\mathrm{CO_2$ : Carbon has 4 valence electrons. Each double bond counts as one electron domain. With two electron domains (two double bonds), the electron geometry is linear ( $180^\circ$ ). Since there Are no lone pairs, the molecular shape is also linear. The two C=O bond dipoles are equal in Magnitude and opposite in direction, so they cancel: $\mathrm{CO_2$ is non-polar.

$\mathrm{H_2\mathrm{O$ : Oxygen has 6 valence electrons. Two are used in O-H bonds, and four remain As two lone pairs. With four electron domains (two bonding pairs + two lone pairs), the electron Geometry is tetrahedral. The molecular shape (considering only atoms) is bent. The two O-H bond Dipoles do not cancel because they are not opposite each other: $\mathrm{H_2\mathrm{O$ is polar.

Practice Questions (Extended)

Using a Born-Haber cycle, calculate the lattice energy of $\mathrm{MgO$ given:

$\Delta H_{\mathrm{at}(\mathrm{Mg) = +148 \mathrm{ kJ/mol$
$\Delta H_{\mathrm{at}(\mathrm{O) = +248 \mathrm{ kJ/mol$
$\mathrm{IE_1(\mathrm{Mg) = +738 \mathrm{ kJ/mol$ $\mathrm{IE_2(\mathrm{Mg) = +1451 \mathrm{ kJ/mol$
$\mathrm{EA_1(\mathrm{O) = -141 \mathrm{ kJ/mol$ , $\mathrm{EA_2(\mathrm{O) = +798 \mathrm{ kJ/mol$
$\Delta H_f(\mathrm{MgO) = -602 \mathrm{ kJ/mol$

Explain why $\mathrm{BeCl_2$ is covalent while $\mathrm{CaCl_2$ is ionic, using Fajans’ Rules.
Using bond enthalpies, explain why the Haber process is exothermic: $\mathrm{N_2 + 3\mathrm{H_2 \to 2\mathrm{NH_3$ .
Predict and explain the trend in boiling points for the hydrogen halides $\mathrm{HF$ $\mathrm{HCl$$\mathrm{HBr$$\mathrm{HI$ .
Balance the following redox equation in basic solution: $\mathrm{MnO_4^- + \mathrm{Br^- \to \mathrm{MnO_2 + \mathrm{BrO_3^-$ .
Explain how the properties of steel differ from those of pure iron, in terms of metallic bonding structure.
A student states that “all Group 1 chlorides have ionic bonding.” Evaluate this statement.
Draw the dot-and-cross diagram for $\mathrm{CO_2$ and explain why it is a linear molecule.

Giant Covalent Structures in Detail

Diamond

Each carbon atom in diamond is covalently bonded to four other carbon atoms in a tetrahedral Arrangement, forming a rigid three-dimensional network. Every C-C bond is a strong sigma bond.

Properties explained:

Very high melting point ( $> 3500^\circ$ C): All bonds must be broken to melt diamond, requiring enormous energy.
Extremely hard: The three-dimensional network prevents any movement of atoms.
Does not conduct electricity: All four valence electrons of each carbon are localised in sigma bonds. There are no delocalised or free electrons.
Insoluble in all solvents: The covalent bonds are too strong to be broken by solvent interactions.

Graphite

Each carbon atom in graphite is covalently bonded to three other carbon atoms in a trigonal planar Arrangement, forming flat hexagonal layers. The fourth electron of each carbon is delocalised over The entire layer.

Properties explained:

High melting point: Strong covalent bonds within layers.
Soft and slippery: Weak van der Waals forces between layers allow them to slide over each other.
Conducts electricity along the planes: Delocalised electrons can move freely within each layer.
Insoluble: Strong covalent bonds within layers.

Silicon Dioxide ( $\mathrm{SiO_2$ )

Each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is bonded to two Silicon atoms, forming a giant covalent network similar to diamond.

Very high melting point ( $\sim 1700^\circ$ C)
Hard and insoluble
Does not conduct electricity

Ionic Structures

Sodium Chloride Structure

$\mathrm{NaCl$ crystallises in a face-centred cubic lattice. Each $\mathrm{Na^+$ ion is surrounded by Six $\mathrm{Cl^-$ ions (octahedral coordination) and vice versa.

Coordination number: 6:6

Caesium Chloride Structure

$\mathrm{CsCl$ has a body-centred cubic arrangement. Each $\mathrm{Cs^+$ ion is surrounded by eight $\mathrm{Cl^-$ ions.

Coordination number: 8:8

The larger size of $\mathrm{Cs^+$ allows it to accommodate more anions around it.

Comparison of Ionic Structures

Compound	Cation radius	Anion radius	Structure	Coordination number	Melting point
NaCl	102 pm	181 pm	FCC (rock salt)	6:6	801°C
CsCl	167 pm	181 pm	BCC	8:8	645°C
MgO	72 pm	140 pm	FCC (rock salt)	6:6	2852°C

Note: $\mathrm{MgO$ has a much higher melting point than $\mathrm{NaCl$ because the $\mathrm{Mg^{2+}$ And $\mathrm{O^{2-}$ ions have higher charges, producing stronger electrostatic attraction.

Advanced Redox: Disproportionation

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and Reduced.

Example: Chlorine with cold, dilute sodium hydroxide:

$\mathrm{Cl_2 + 2\mathrm{NaOH \to \mathrm{NaCl + \mathrm{NaClO + \mathrm{H_2\mathrm{O$

Oxidation states: $\mathrm{Cl_2$ (0) $\to$ $\mathrm{Cl^-$ in $\mathrm{NaCl$ (-1) and $\mathrm{Cl^+$ in $\mathrm{NaClO$ (+1). Chlorine is both reduced (0 to -1) and oxidised (0 to +1).

Example: Copper(I) oxide with dilute sulfuric acid:

$\mathrm{Cu_2\mathrm{O + \mathrm{H_2\mathrm{SO_4 \to \mathrm{Cu + \mathrm{CuSO_4 + \mathrm{H_2\mathrm{O$

Copper goes from +1 in $\mathrm{Cu_2\mathrm{O$ to 0 in $\mathrm{Cu$ (reduction) and to +2 in $\mathrm{CuSO_4$ (oxidation).

Worked Example: Multi-Step Redox Balancing

Question: Balance the following equation in acidic solution: $\mathrm{IO_3^- + \mathrm{I^- + \mathrm{H^+ \to \mathrm{I_2 + \mathrm{H_2\mathrm{O$

Step 1: Assign oxidation states. In $\mathrm{IO_3^-$ Iodine is $+5$ . In $\mathrm{I^-$ Iodine is $-1$ . In $\mathrm{I_2$ Iodine is 0.

$\mathrm{IO_3^-$ is reduced: $+5 \to 0$ (gain of 5 electrons) $\mathrm{I^-$ is oxidised: $-1 \to 0$ (loss of 1 electron)

Step 2: Write half-equations.

Reduction: $\mathrm{IO_3^- + 6\mathrm{H^+ + 5e^- \to \tfrac{1}{2}\mathrm{I_2 + 3\mathrm{H_2\mathrm{O$

Oxidation: $\mathrm{I^- \to \tfrac{1}{2}\mathrm{I_2 + e^-$

Step 3: Balance electrons. Multiply oxidation by 5:

$\mathrm{IO_3^- + 5\mathrm{I^- + 6\mathrm{H^+ \to 3\mathrm{I_2 + 3\mathrm{H_2\mathrm{O$

Check: I: $1 + 5 = 6$ left, $3 \times 2 = 6$ right. O: $3$ left, $3$ right. H: $6$ left, $6$ right. Charge: $-1 + 5(-1) + 6(+1) = 0$ left, $0$ right. Balanced.

Summary: Bonding and Properties Connection

Bonding type	Bond strength (kJ/mol)	Melting point	Conductivity	Typical examples
Ionic	600-4000	High	Molten/solution	$\mathrm{NaCl$ , $\mathrm{MgO$
Covalent (simple)	150-400	Low	None	$\mathrm{H_2\mathrm{O$ , $\mathrm{CO_2$
Covalent (giant)	350-400	Very high	Graphite only	Diamond, $\mathrm{SiO_2$
Metallic	100-500	Moderate-high	Good	$\mathrm{Fe$ , $\mathrm{Cu$
Hydrogen bonds	5-40	—	—	Between molecules of $\mathrm{H_2\mathrm{O$ , $\mathrm{NH_3$

Worked Examples

Example 1: Mole calculation

Calculate the number of moles in $12.0\,\text{g}$ of $\text{NaOH}$ ( $M_r = 40.0$ ).

Solution:

$n = \frac{m}{M_r} = \frac{12.0}{40.0} = 0.300\,\text{mol}$

Example 2: Reacting masses

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

What mass of $\text{CaCl}_2$ is produced from $10.0\,\text{g}$ of $\text{CaCO}_3$ ? ( $M_r[\text{CaCO}_3] = 100$ , $M_r[\text{CaCl}_2] = 111$ )

Solution:

$n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\,\text{mol}$

From the equation, ratio is $1:1$ , so $n(\text{CaCl}_2) = 0.100\,\text{mol}$ .

$m(\text{CaCl}_2) = 0.100 \times 111 = 11.1\,\text{g}$

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Chemical Reactions and Bonding

Chemical Reactions and Bonding

Higher Chemical Bonding

Types of Bonding

Comparison of Bonding Types

Electronegativity and Bond Polarity

Molecular Shape (VSEPR Theory)

Intermolecular Forces

Summary of Intermolecular Forces

Bond Enthalpy

Chemical Reactions

Balancing Equations

Ionic Equations

Oxidation and Reduction

Redox Reactions

Summary Table: Oxidation States of Common Elements

Common Pitfalls

Practice Questions

Advanced Bonding Concepts

Ionic Lattice Energy

Born-Haber Cycles

Polarising Power and Covalent Character

Metallic Bonding: Alloy Properties

Worked Example: Predicting Bond Type

Derivation: Why \mathrm{H_2\mathrm{O Is Bent but \mathrm{CO_2 Is Linear

Practice Questions (Extended)

Giant Covalent Structures in Detail

Diamond

Graphite

Silicon Dioxide (\mathrm{SiO_2)

Ionic Structures

Sodium Chloride Structure

Caesium Chloride Structure

Comparison of Ionic Structures

Advanced Redox: Disproportionation

Worked Example: Multi-Step Redox Balancing

Summary: Bonding and Properties Connection

Worked Examples

Derivation: Why $\mathrm{H_2\mathrm{O$ Is Bent but $\mathrm{CO_2$ Is Linear

Silicon Dioxide ( $\mathrm{SiO_2$ )