Valence Bond Theory primarily focuses on explaining the formation and properties of chemical bonds between atoms.

In Valence Bond Theory, the strength of a chemical bond is influenced by a combination of factors, including the number of valence electrons, the electronegativity of the atoms involved, and the distance between the atoms.

In a methane molecule, the carbon atom undergoes sp³ hybridization, resulting in four equivalent hybrid orbitals that form tetrahedral bonds with the four hydrogen atoms.

Ethylene (C₂H₄) contains a double bond between its two carbon atoms, whereas ethane (C₂H₆) has a single bond, acetylene (C₂H₂) has a triple bond, and propane (C₃H₈) has only single bonds.

Molecules with sp³ hybridization adopt a tetrahedral shape due to the four equivalent hybrid orbitals that are directed towards the corners of a tetrahedron.

Carbon dioxide (CO₂) has a linear shape because the two oxygen atoms are bonded to the carbon atom in a straight line.

In an ammonia molecule, the nitrogen atom undergoes sp³ hybridization, resulting in four equivalent hybrid orbitals that form tetrahedral bonds with the three hydrogen atoms and the lone pair of electrons.

Nitrogen trifluoride (NF₃) has a trigonal pyramidal shape due to the three fluorine atoms and the lone pair of electrons on the nitrogen atom.

In an ethene molecule, the carbon atoms undergo sp² hybridization, resulting in three equivalent hybrid orbitals that form trigonal bonds with the two hydrogen atoms and the double bond between the carbon atoms.

Water (H₂O) has a bent shape due to the two lone pairs of electrons on the oxygen atom, which repel the bonding pairs of electrons and cause the molecule to deviate from a linear geometry.

In a water molecule, the oxygen atom undergoes sp³ hybridization, resulting in four equivalent hybrid orbitals that form tetrahedral bonds with the two hydrogen atoms and the two lone pairs of electrons.

Methane (CH₄) has a tetrahedral shape because the four hydrogen atoms are bonded to the carbon atom in a tetrahedral arrangement.

In a pyridine molecule, the nitrogen atom undergoes sp² hybridization, resulting in three equivalent hybrid orbitals that form trigonal bonds with the two carbon atoms and the nitrogen atom in the ring, and one unhybridized p orbital that forms a lone pair of electrons.

Boron trifluoride (BF₃) has a trigonal planar shape due to the three fluorine atoms that are bonded to the boron atom in a trigonal arrangement.

In a benzene molecule, the carbon atoms undergo sp² hybridization, resulting in three equivalent hybrid orbitals that form trigonal bonds with the two neighboring carbon atoms and the hydrogen atom, and one unhybridized p orbital that participates in the formation of the delocalized π-electron system.