Physics » Condensed Matter Physics » Types of Molecular Bonds

# Bonding in Polyatomic Molecules

## Bonding in Polyatomic Molecules

A polyatomic molecule is a molecule made of more than two atoms. Examples range from a simple water molecule to a complex protein molecule. The structures of these molecules can often be understood in terms of covalent bonding and hybridization. Hybridization is a change in the energy structure of an atom in which mixed states (states that can be written as a linear superposition of others) participate in bonding.

To illustrate hybridization, consider the bonding in a simple water molecule, $${\text{H}}_{2}\text{O}\text{.}$$ The electron configuration of oxygen is $$1{s}^{2}2{s}^{2}2{p}^{4}.$$ The 1s and 2s electrons are in “closed shells” and do not participate in bonding. The remaining four electrons are the valence electrons. These electrons can fill six possible states ($$l=1$$, $$m=0$$, $$±1$$, plus spin up and down). The energies of these states are the same, so the oxygen atom can exploit any linear combination of these states in bonding with the hydrogen atoms. These linear combinations (which you learned about in the tutorial on atomic structure) are called atomic orbitals, and they are denoted by $${p}_{x},{p}_{y},$$ and $${p}_{z}.$$ The electron charge distributions for these orbitals are given in this figure.

Oxygen has four valence electrons. In the context of a water molecule, two valence electrons fill the $${p}_{z}$$ orbital and one electron fills each of the $${p}_{x}$$ and $${p}_{y}$$ orbitals. The $${p}_{x}$$ and $${p}_{y}$$ orbitals are used in bonding with hydrogen atoms to form $${\text{H}}_{2}\text{O}$$. Without repulsion of H atoms, the bond angle between hydrogen atoms would be 90 degrees.

The transformation of the electron wave functions of oxygen to $${p}_{x},{p}_{y},$$ and $${p}_{z}$$ orbitals in the presence of the hydrogen atoms is an example of hybridization. Two electrons are found in the $${p}_{z}$$ orbital with paired spins $$(↑↓)$$. One electron is found in each of the $${p}_{x}$$ and $${p}_{y}$$ orbitals, with unpaired spins. The latter orbitals participate in bonding with the hydrogen atoms. Based on this figure, we expect the bonding angle for $$\text{H—O—H}$$ to be $$90\text{°}$$. However, if we include the effects of repulsion between atoms, the bond angle is $$104.5\text{°}$$. The same arguments can be used to understand the tetrahedral shape of methane $$({\text{CH}}_{4})$$ and other molecules.

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