krit.club logo

Structure 2. Models of bonding and structure - From models to materials

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Metallic bonding is the electrostatic attraction between a lattice of positive metal ions (cations) and a 'sea' of delocalized valence electrons. The strength of the bond increases with the number of valence electrons and decreasing ionic radius, e.g., Mg2+Mg^{2+} has a higher melting point than Na+Na^{+} because of higher charge density.

•

Ionic bonding results from the electrostatic attraction between oppositely charged ions in a giant lattice. The strength of the ionic bond is determined by Coulomb's Law, where the force of attraction is proportional to the product of charges (q1q2q_1 q_2) and inversely proportional to the square of the distance between them (r2r^2).

•

Alloys are mixtures of metals with other elements. They are often harder than pure metals because atoms of different sizes disrupt the regular lattice layers, preventing them from sliding over each other easily. Examples include steel and brass.

•

Giant covalent structures like diamond, graphite, and SiO2SiO_2 involve atoms linked by covalent bonds in a continuous network. In diamond, each carbon is sp3sp^3 hybridized, forming a tetrahedral structure, while in graphite, carbon is sp2sp^2 hybridized, forming layers held by London dispersion forces.

•

Molecular materials are held together by intermolecular forces (IMFs) such as London dispersion forces, dipole-dipole attractions, and hydrogen bonding. Hydrogen bonding occurs when hydrogen is covalently bonded to highly electronegative elements (NN, OO, or FF).

•

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry based on the principle that electron domains (bonding and non-bonding pairs) around a central atom stay as far apart as possible to minimize repulsion.

•

Formal charge (FCFC) is used to determine the most stable Lewis structure among resonance contributors. The structure where the FCFC of each atom is closest to zero is generally preferred.

📐Formulae

FC=V−(N+12B)FC = V - (N + \frac{1}{2}B) (where VV is valence electrons, NN is non-bonding electrons, and BB is bonding electrons)

F=kq1q2r2F = k \frac{q_1 q_2}{r^2} (Coulomb's Law for ionic bond strength)

Bond Order=Total number of bonding pairsNumber of bonding regions\text{Bond Order} = \frac{\text{Total number of bonding pairs}}{\text{Number of bonding regions}}

💡Examples

Problem 1:

Explain why magnesium oxide (MgOMgO) has a significantly higher melting point than sodium chloride (NaClNaCl).

Solution:

Tm(MgO)≈2852∘CT_m(MgO) \approx 2852^{\circ}C whereas Tm(NaCl)≈801∘CT_m(NaCl) \approx 801^{\circ}C.

Explanation:

In MgOMgO, the ions are Mg2+Mg^{2+} and O2−O^{2-}, while in NaClNaCl they are Na+Na^{+} and Cl−Cl^{-}. According to the lattice enthalpy relationship U∝q1q2r2U \propto \frac{q_1 q_2}{r^2}, the product of charges for MgOMgO is 2×2=42 \times 2 = 4, while for NaClNaCl it is 1×1=11 \times 1 = 1. The higher charges and smaller ionic radii in MgOMgO result in much stronger electrostatic attractions, requiring more energy to break the lattice.

Problem 2:

Determine the molecular geometry and bond angle of the ammonia molecule (NH3NH_3).

Solution:

Geometry: Trigonal Pyramidal; Bond angle: ≈107.3∘\approx 107.3^{\circ}.

Explanation:

Nitrogen has 5 valence electrons. In NH3NH_3, there are 3 bonding pairs and 1 lone pair, making 4 electron domains (Tetrahedral electron domain geometry). Because the lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, the bond angle is reduced from the ideal tetrahedral 109.5∘109.5^{\circ} to approximately 107∘107^{\circ}.

Problem 3:

Calculate the formal charge of the central Carbon atom in Carbon Dioxide (CO2CO_2).

Solution:

FC=4−(0+12(8))=0FC = 4 - (0 + \frac{1}{2}(8)) = 0

Explanation:

Carbon is in group 14 and has V=4V = 4 valence electrons. In the Lewis structure O=C=OO=C=O, the central carbon has 0 non-bonding electrons (N=0N=0) and 4 bonding pairs (8 electrons, so B=8B=8). Using the formula FC=V−N−12BFC = V - N - \frac{1}{2}B, we get 4−0−4=04 - 0 - 4 = 0.