Calculating the Total Valence Electrons
Valence electrons are the electrons located in the outermost shell of an atom that participate in the formation of chemical bonds. These electrons dictate an atom’s chemical behavior. For the water molecule ($\text{H}_2\text{O}$), the total number of valence electrons must be calculated from its constituent atoms: one oxygen atom and two hydrogen atoms.
The oxygen atom belongs to Group 16 of the periodic table, contributing six valence electrons. Hydrogen is in Group 1, meaning each hydrogen atom contributes a single valence electron. To find the total number of electrons available for bonding, a simple summation is performed based on the number of each type of atom present.
The calculation is determined by multiplying the number of atoms by the valence electrons they contribute, then adding the results together. This involves taking the one oxygen atom (six valence electrons) and combining that with the two hydrogen atoms (one valence electron each). The total calculation is therefore $(1 \times 6) + (2 \times 1)$, which results in a sum of eight valence electrons for the entire $\text{H}_2\text{O}$ molecule.
Visualizing Electron Placement
The eight total valence electrons are arranged around the central oxygen atom to achieve a stable electronic configuration. This arrangement is represented by the molecule’s Lewis structure, which illustrates how these electrons are distributed as shared and unshared pairs. The oxygen atom acts as the central hub, while the two hydrogen atoms are positioned on the periphery.
Four of the eight valence electrons form two single covalent bonds, known as bonding pairs. Each hydrogen atom shares one pair of electrons with the oxygen atom, satisfying hydrogen’s need for a complete outer shell of two electrons.
The remaining four valence electrons are situated exclusively on the central oxygen atom and are known as lone pairs, forming two unshared pairs. These two lone pairs and the two bonding pairs constitute four distinct regions of electron density. This distribution ensures that the oxygen atom achieves a stable octet, meaning it is surrounded by eight electrons.
How Electron Arrangement Defines Water’s Properties
The arrangement of the two bonding pairs and the two lone pairs of valence electrons around the oxygen atom determines the water molecule’s shape and chemical properties. According to the Valence Shell Electron Pair Repulsion theory (VSEPR), all four electron density regions—shared and unshared pairs—repel each other. This forces the electron pairs into a roughly tetrahedral arrangement around the central oxygen atom.
The repulsion exerted by the unshared lone pairs is stronger than the repulsion between the shared bonding pairs. These lone pairs push the two hydrogen atoms closer together than they would be in a perfect tetrahedral shape. This results in the characteristic bent, or V-shaped, molecular geometry of the water molecule, rather than a linear structure.
This specific bent shape is responsible for water’s high polarity. Because the oxygen atom is more electronegative than the hydrogen atoms, it pulls the shared electrons closer to itself. Coupled with the molecule’s non-symmetrical, bent geometry, this unequal sharing of electrons creates a dipole moment. The oxygen side of the molecule develops a partial negative charge, while the hydrogen side acquires a partial positive charge, with the resulting $\text{H-O-H}$ bond angle measured at approximately $104.5^\circ$.