The valence electron configurations of several atoms are shown here. How many bonds can each atom make without hybridization? a. \(\mathrm{B} 2 \mathrm{~s}^{2} 2 p^{1}\) b. \(\mathrm{N} 2 s^{2} 2 p^{3}\) c. \(\mathrm{O} 2 \mathrm{~s}^{2} 2 p^{4}\)

Chemical bonding is a fundamental concept in chemistry that deals with the joining of atoms to form molecules. Bonds between atoms are created by the attraction between the positive nucleus of one atom and the negative electrons of another. The number of bonds an atom can form is often determined by its valence electrons, which are the electrons in the outermost shell. These valence electrons can either be shared with other atoms to form covalent bonds or transferred to achieve a more stable electron configuration, as in ionic bonding.

For example, in the exercise given, Boron can make one bond, Nitrogen can make three, and Oxygen can make two without hybridization, which means these are the direct ways these atoms can interact with others to form chemical bonds using their unpaired valence electrons.

Electron configuration refers to the arrangement of electrons in the orbitals of an atom. When writing electron configurations, we list the orbitals and the number of electrons they contain. Taking the valence electron configuration, which represents the electrons in the highest energy level, we can determine an element's reactivity and ability to bond with other atoms.

From our exercise, Boron's valence electron configuration is \(2s^2 2p^1\), indicating that it has three valence electrons that can engage in bonding. Similarly, Nitrogen \(2s^2 2p^3\) and Oxygen \(2s^2 2p^4\) have five and six valence electrons, respectively, portraying their potential to bond with other atoms.

Hybridization is a concept in molecular chemistry that describes the mixing of atomic orbitals to form new hybrid orbitals. These hybrid orbitals then combine to form bonds in molecules. The types of hybridization, such as sp, sp2, and sp3, correspond to different molecular shapes and bond angles.

An atom's ability to undergo hybridization is crucial for forming multiple bonds with specific geometries. In the exercise scenario, we observe that hybridization is not necessary for Boron, Nitrogen, or Oxygen to form the basic number of bonds as predicted by their unpaired valence electrons. However, in more complex molecules, hybridization is often essential for the atoms to achieve the necessary geometry for bonding.

Unpaired electrons play a significant role in the bonding capabilities of an atom. These are the electrons that are alone in an orbital and are not part of an electron pair. Atoms will often seek to either borrow or share these unpaired electrons with other atoms to achieve a full valence shell, leading to a more stable electron configuration.

Looking at our exercise, the number of bonds that each element can form is directly related to the number of unpaired electrons in its valence electron configuration. For instance, Boron with one unpaired electron can form one bond, while Nitrogen with three unpaired electrons can form three bonds. Oxygen, with two unpaired electrons, completes two bonds without the need for additional changes to its electron configuration.