Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
A covalent bond is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the atoms involved.
The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration of eight valence electrons, similar to a noble gas. Exceptions include incomplete octets (e.g., , ) and expanded octets for elements in Period 3 and below (e.g., , ).
Bond polarity is determined by the difference in electronegativity () between atoms. A polar covalent bond occurs when , leading to a dipole moment represented by and .
Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry based on minimizing repulsion between electron domains. Total domains include both bonding pairs and lone pairs.
Resonance occurs when more than one valid Lewis structure can be drawn for a molecule (e.g., , , ). The actual structure is a resonance hybrid with delocalized electrons.
Sigma () bonds are formed by the head-on overlap of atomic orbitals along the internuclear axis. Pi () bonds are formed by the sideways overlap of -orbitals above and below the internuclear axis.
Hybridization involves the mixing of atomic orbitals (such as and ) to form new hybrid orbitals () that are degenerate (equal in energy).
Giant covalent structures like diamond, graphite, graphene, and have high melting points and varying electrical conductivities based on electron delocalization.
📐Formulae
(where is Formal Charge, is valence electrons, is non-bonding electrons, and is bonding electrons)
(Electronegativity difference)
💡Examples
Problem 1:
Predict the molecular geometry, electron domain geometry, and hybridization of the central atom in .
Solution:
Electron domain geometry: Trigonal bipyramidal. Molecular geometry: Seesaw. Hybridization: .
Explanation:
has 6 valence electrons. In , it forms 4 single bonds with and has 1 lone pair. This results in 5 electron domains. According to VSEPR, 5 domains lead to a trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimize repulsion, resulting in a 'seesaw' molecular shape.
Problem 2:
Calculate the formal charge of the central oxygen atom in the ozone () molecule for the resonance structure .
Solution:
Explanation:
The central oxygen atom () has one lone pair (2 electrons) and forms one double bond and one single bond (total 3 bonds, 6 bonding electrons). Using the formula , we get .
Problem 3:
Identify the number of and bonds in ethyne ().
Solution:
3 bonds and 2 bonds.
Explanation:
The structure of ethyne is . Each single bond () is a bond (2 total). The triple bond between carbons consists of 1 bond and 2 bonds. Summing these gives 3 and 2 bonds.