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Chemistry - Chemical Bonding (Ionic, Covalent, and Metallic)

Grade 10IB

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

🔑Concepts

Ionic Bonding: The electrostatic attraction between oppositely charged ions formed by the complete transfer of electrons from a metal to a non-metal, such as in NaClNaCl.

Covalent Bonding: The electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the atoms involved, typically occurring between non-metals like in H2OH_{2}O.

Metallic Bonding: The electrostatic attraction between a lattice of positive metal ions (cationscations) and a 'sea' of delocalized valence electrons.

The Octet Rule: Atoms tend to lose, gain, or share electrons to achieve a stable electron configuration with 88 electrons in their valence shell, similar to Noble Gases (Group18Group 18).

Electronegativity Difference (Δχ\Delta\chi): Used to predict bond type. If Δχ>1.8\Delta\chi > 1.8, the bond is generally ionic; if 0.5<Δχ<1.80.5 < \Delta\chi < 1.8, it is polar covalent; if Δχ<0.5\Delta\chi < 0.5, it is non-polar covalent.

Giant Covalent Structures: Substances like Diamond (CC), Graphite (CC), and Silicon Dioxide (SiO2SiO_{2}) which have high melting points due to a network of strong covalent bonds.

Properties of Ionic Compounds: High melting/boiling points due to strong lattice enthalpy, and the ability to conduct electricity only in molten or aqueous states (aqaq) where ions are mobile.

📐Formulae

F=kq1q2r2F = k \frac{q_{1}q_{2}}{r^2}

Δχ=χAχB\Delta\chi = |\chi_{A} - \chi_{B}|

Xn++YmXmYnX^{n+} + Y^{m-} \rightarrow X_{m}Y_{n}

FC=VNB2FC = V - N - \frac{B}{2}

💡Examples

Problem 1:

Determine the formula and bond type for a compound formed between Magnesium (MgMg) and Nitrogen (NN).

Solution:

Mg3N2Mg_{3}N_{2}, Ionic bond.

Explanation:

Magnesium is in Group 2 and loses 22 electrons to form Mg2+Mg^{2+}. Nitrogen is in Group 15 and gains 33 electrons to form N3N^{3-}. To balance the charges: 3×(+2)+2×(3)=03 \times (+2) + 2 \times (-3) = 0. Thus, the formula is Mg3N2Mg_{3}N_{2}.

Problem 2:

Explain why DiamondDiamond has a very high melting point while MethaneMethane (CH4CH_{4}) has a very low melting point, despite both having covalent bonds.

Solution:

Difference in structure: Giant Covalent vs. Simple Molecular.

Explanation:

DiamondDiamond is a giant covalent structure where every Carbon atom is bonded to four others by strong covalent bonds in a tetrahedral lattice, requiring massive energy to break. CH4CH_{4} consists of small molecules held together by weak intermolecular forces (London dispersion forces), which require very little energy to overcome.

Problem 3:

Calculate the formal charge (FCFC) of the Carbon atom in Carbon Dioxide (CO2CO_{2}) given the structure O=C=OO=C=O.

Solution:

FC=0FC = 0

Explanation:

For Carbon: Valence electrons (VV) = 44, Non-bonding electrons (NN) = 00, Bonding electrons (BB) = 88 (from two double bonds). Using the formula FC=VNB2FC = V - N - \frac{B}{2}, we get FC=4082=44=0FC = 4 - 0 - \frac{8}{2} = 4 - 4 = 0.