Chemical Bonding

Covalent Bonding

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to define covalent bonding and explain its formation.
  • By the end of this lesson students will be able to draw Lewis structures for simple covalent molecules and polyatomic ions.
  • By the end of this lesson students will be able to distinguish between polar and non-polar covalent bonds and molecules.
  • By the end of this lesson students will be able to predict molecular shapes using VSEPR theory (Higher Level).
  • By the end of this lesson students will be able to describe and explain the different types of intermolecular forces.

Key concepts

Covalent Bonding

Covalent bonding is a type of chemical bond formed when two non-metal atoms share one or more pairs of valence electrons to achieve a stable electron configuration, typically an octet (eight valence electrons). This sharing leads to a strong electrostatic attraction between the shared electrons and the positively charged nuclei of the bonded atoms.

Lewis Structures

Lewis structures (also known as Lewis dot diagrams or electron dot structures) are diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. They are used to visualise the distribution of valence electrons and predict the number of bonds an atom will form.

Electronegativity

Electronegativity is a measure of the relative attraction an atom has for a shared pair of electrons in a covalent bond. The Pauling scale is commonly used, with fluorine being the most electronegative element (4.0). Differences in electronegativity determine the polarity of a bond.

Polar Covalent Bond

A polar covalent bond forms when there is an unequal sharing of electrons between two atoms due to a significant difference in their electronegativities (typically between 0.4 and 1.7 on the Pauling scale). This unequal sharing results in partial positive (δ+) and partial negative (δ-) charges on the atoms.

Non-Polar Covalent Bond

A non-polar covalent bond forms when electrons are shared equally between two atoms. This occurs either when the two bonded atoms are identical (e.g., H₂ or Cl₂) or when their electronegativity difference is very small (typically less than 0.4).

Molecular Polarity

The overall polarity of a molecule depends on both the polarity of its individual bonds and its molecular geometry. Even if a molecule contains polar bonds, it can be non-polar overall if the bond dipoles cancel each other out due to the molecule's symmetrical shape (e.g., CO₂). If the bond dipoles do not cancel, the molecule is polar.

VSEPR Theory (Valence Shell Electron Pair Repulsion Theory) (Higher Level)

VSEPR theory is used to predict the three-dimensional geometry of molecules. It states that electron pairs (both bonding pairs and lone pairs) in the valence shell of a central atom repel each other and will arrange themselves as far apart as possible to minimise repulsion, thus determining the molecular shape.

Intermolecular Forces (IMFs)

Intermolecular forces are attractive forces that exist between molecules. They are much weaker than the intramolecular covalent bonds within molecules but significantly influence the physical properties of substances, such as boiling points, melting points, and solubility.

London Dispersion Forces (LDFs)

London dispersion forces are the weakest type of intermolecular force and are present in all molecules, whether polar or non-polar. They arise from temporary, instantaneous dipoles created by the momentary uneven distribution of electrons around an atom or molecule.

Dipole-Dipole Forces

Dipole-dipole forces are attractive forces that occur between the permanent dipoles of polar molecules. The partially positive end of one polar molecule is attracted to the partially negative end of an adjacent polar molecule.

Hydrogen Bonding

Hydrogen bonding is a particularly strong type of dipole-dipole interaction. It occurs when a hydrogen atom, covalently bonded to a highly electronegative atom (nitrogen, oxygen, or fluorine), is attracted to a lone pair of electrons on another highly electronegative atom (N, O, or F) in an adjacent molecule.

Key facts to remember

  • 1Covalent bonds involve the sharing of valence electrons between non-metal atoms.
  • 2Lewis structures are used to represent valence electrons and bonding in molecules.
  • 3The difference in electronegativity between bonded atoms determines bond polarity.
  • 4Molecular polarity depends on both bond polarity and the overall molecular shape (symmetry).
  • 5VSEPR theory (HL) states that electron pairs around a central atom repel each other and arrange to minimise repulsion, determining molecular geometry.
  • 6Intermolecular forces (IMFs) are attractive forces between molecules, much weaker than intramolecular covalent bonds.
  • 7London dispersion forces are present in all molecules and arise from temporary dipoles.
  • 8Hydrogen bonding is the strongest type of intermolecular force, occurring when H is bonded to N, O, or F.

Worked examples

Example 1

Draw the Lewis structure for carbon dioxide (CO₂).

I1. Count the total number of valence electrons: Carbon (Group 14) has 4 valence electrons. Oxygen (Group 16) has 6 valence electrons. Total = 4 + (2 × 6) = 16 valence electrons.
II2. Identify the central atom: Carbon is less electronegative than oxygen, so carbon is the central atom. Arrange the atoms as O-C-O.
III3. Draw single bonds between the central atom and the outer atoms: O-C-O. This uses 2 × 2 = 4 electrons. Remaining electrons = 16 - 4 = 12 electrons.
IV4. Distribute the remaining electrons to satisfy the octet rule for the outer atoms first: Each oxygen needs 6 more electrons to complete its octet. Place 6 electrons on each oxygen. This uses 2 × 6 = 12 electrons. Remaining electrons = 12 - 12 = 0 electrons.
V5. Check if the central atom has an octet: Carbon currently has only 4 electrons (from the two single bonds). It needs 4 more electrons.
VI6. Form multiple bonds by moving lone pairs from the outer atoms to shared positions until the central atom achieves an octet: Move one lone pair from each oxygen to form double bonds with carbon. This gives O=C=O.
VII7. Verify octets for all atoms: Each oxygen now has 2 bonding pairs (4 electrons) and 2 lone pairs (4 electrons), totalling 8 electrons. Carbon has 4 bonding pairs (8 electrons). All atoms have an octet.

Answer

The Lewis structure for CO₂ is: O=C=O, with two lone pairs of electrons on each oxygen atom.

Remember to always check that all atoms (except hydrogen) have an octet and that the total number of electrons used matches the initial count of valence electrons.

Example 2

Predict the molecular shape and overall polarity of ammonia (NH₃) using VSEPR theory.

I1. Draw the Lewis structure for NH₃: Nitrogen (Group 15) has 5 valence electrons. Hydrogen (Group 1) has 1 valence electron. Total = 5 + (3 × 1) = 8 valence electrons. Nitrogen is the central atom. Form three N-H single bonds (6 electrons used). The remaining 2 electrons form one lone pair on nitrogen. The Lewis structure shows a central nitrogen atom bonded to three hydrogen atoms and one lone pair.
II2. Determine the number of electron domains around the central atom: Nitrogen has 3 bonding pairs (to H atoms) and 1 lone pair. Total electron domains = 3 + 1 = 4.
III3. Determine the electron geometry: With 4 electron domains, the electron geometry is tetrahedral, as this arrangement minimises repulsion between the electron pairs.
IV4. Determine the molecular geometry (shape): The molecular geometry considers only the positions of the atoms. Since there is one lone pair, it distorts the tetrahedral electron geometry. The lone pair repels bonding pairs more strongly than bonding pairs repel each other. This results in a trigonal pyramidal shape.
V5. Assess bond polarity: Nitrogen is more electronegative than hydrogen (N: 3.0, H: 2.2). Therefore, each N-H bond is polar, with the nitrogen atom having a partial negative charge (δ-) and the hydrogen atoms having partial positive charges (δ+).
VI6. Determine overall molecular polarity: Due to the trigonal pyramidal shape, the bond dipoles do not cancel out. The lone pair on nitrogen also contributes to the overall dipole moment. The molecule has a net dipole moment, with the negative end towards the nitrogen atom and the positive end towards the hydrogen atoms. Therefore, NH₃ is a polar molecule.

Answer

Ammonia (NH₃) has a trigonal pyramidal molecular shape and is a polar molecule.

Lone pairs are crucial for determining molecular shape and often contribute to molecular polarity, even though they are not 'bonds'.

Example 3

Explain the significant difference in boiling points between methane (CH₄, boiling point -161.5 °C) and water (H₂O, boiling point 100 °C).

I1. Identify the type of bonding within each molecule: Both methane and water are covalent molecules, meaning atoms within each molecule are held together by strong covalent bonds.
II2. Determine the intermolecular forces (IMFs) present in methane (CH₄): Methane is a non-polar molecule (carbon and hydrogen have similar electronegativities, and the tetrahedral shape is symmetrical, causing bond dipoles to cancel). Therefore, the only intermolecular forces present between CH₄ molecules are weak London dispersion forces.
III3. Determine the intermolecular forces (IMFs) present in water (H₂O): Water is a polar molecule (oxygen is significantly more electronegative than hydrogen, and the bent shape means bond dipoles do not cancel). In addition to London dispersion forces, water molecules experience dipole-dipole forces. Crucially, because hydrogen is bonded directly to oxygen (a highly electronegative atom), water molecules can also form strong hydrogen bonds with each other.
IV4. Compare the strengths of the IMFs: Hydrogen bonds are significantly stronger than dipole-dipole forces, which are in turn stronger than London dispersion forces. Therefore, the intermolecular forces in water (primarily hydrogen bonding) are much stronger than the intermolecular forces in methane (only London dispersion forces).
V5. Relate IMF strength to boiling point: Boiling involves overcoming the intermolecular forces between molecules to allow them to escape into the gaseous phase. Since water has much stronger intermolecular forces (hydrogen bonding) compared to methane (only London dispersion forces), significantly more energy is required to separate water molecules. This results in water having a much higher boiling point than methane.

Answer

Methane (CH₄) is a non-polar molecule, so it only experiences weak London dispersion forces between its molecules. Water (H₂O) is a polar molecule and experiences London dispersion forces, dipole-dipole forces, and crucially, strong hydrogen bonding between its molecules. Hydrogen bonds are much stronger than London dispersion forces. Therefore, significantly more energy is required to overcome the strong hydrogen bonds in water to cause it to boil, compared to the weak London dispersion forces in methane, leading to water's much higher boiling point.

Always identify all types of IMFs present, but focus on the strongest type when comparing properties like boiling point.

Common mistakes

  • Confusing intramolecular covalent bonds (within a molecule) with intermolecular forces (between molecules).
  • Incorrectly counting valence electrons or failing to account for lone pairs when drawing Lewis structures.
  • Forgetting to consider lone pairs when determining molecular geometry using VSEPR theory.
  • Assuming a molecule with polar bonds is always polar; molecular symmetry can lead to overall non-polarity (e.g., CO₂).
  • Incorrectly identifying hydrogen bonding; it only occurs when hydrogen is directly bonded to nitrogen, oxygen, or fluorine.
  • Not explaining the link between the strength of intermolecular forces and physical properties like boiling point.

Exam tips

  • Always show all valence electrons (bonding and lone pairs) clearly in Lewis structures, especially for polyatomic ions where charge is important.
  • For VSEPR questions, systematically determine electron domains, electron geometry, and then molecular geometry. Clearly state the name of the shape.
  • When comparing physical properties (e.g., boiling points), explicitly state the types of intermolecular forces present in each substance and compare their relative strengths.
  • Remember that VSEPR theory applies to the electron pairs around the *central* atom, not necessarily the entire molecule.
  • For molecular polarity, consider both the polarity of individual bonds and how these bond dipoles add up or cancel out due to the molecule's 3D shape.

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