Skip to main content
Chemistry LibreTexts

Valence Bond Theory and Hybrid Atomic Orbitals

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    However, the application of VSEPR theory can be expanded to complicated molecules such as

        H H         H   O
        | |         |  //
        |           |  \
        H           N   O-H
                   / \
                  H   H

    By applying the VSEPR theory, one deduces the following results:

    • \(\ce{H-C-C}\) bond angle = 109o
    • \(\ce{H-C=C}\) bond angle = 120o, geometry around \(\ce{C}\) trigonal planar
    • \(\ce{C=C=C}\) bond angle = 180o, in other words linear
    • \(\ce{H-N-C}\) bond angle = 109o, tetrahedral around \(\ce{N}\)
    • \(\ce{C-O-H}\) bond angle = 105 or 109o, 2 lone electron pairs around \(\ce{O}\)

    Confidence Building Questions

    1. In terms of valence bond theory, how is a chemical bond formed?

      Hint: A chemical bond is due to the overlap of atomic orbitals.

      Discussion -
      Molecular orbital theory considers the energy states of the molecule.

    2. When one s and two p atomic orbitals are used to generate hybrid orbitals, how many hybrid orbitals will be generated?

      Hint: Using three atomic orbitals generates three hybrid orbitals.

      Discussion -
      Number of orbitals does not change in hybridization of atomic orbitals.

    3. In the structures of \(\ce{SO2}\) and \(\ce{NO2}\), what are the values of the bond angles?

      Hint: The bond angles are expected to be less than 120 degrees.

      Discussion -
      Since the lone electron pair in \(\ce{:SO2}\) and lone electron in \(\ce{.NO2}\) take up more space, we expect the structure to distort leaving a smaller angle than 120 between the bonds.

    4. What is the geometrical shape of the molecule \(\ce{CH4}\), methane?

      Hint: Methane molecules are tetrahedral.

      Discussion -
      The 4 \(\ce{H}\) atoms form a tetrahedron, and methane has a tetrahedral shape.

    5. What do you expect the bond angles to be in the \(\ce{NH4+}\) ion?

      Hint: All bond angles are 109.5 degrees, the ideal value for a symmetric tetrahedral structure.

      Discussion -
      The structure of this ion is very similar to that of \(\ce{CH4}\).

    6. What hybrid orbitals does the \(\ce{C}\) atom use in the compound \(\ce{H-C\equiv C-H}\), in which the molecule is linear?

      Hint: The sp hybrid orbitals are used by the \(\ce{C}\) atom.

      Discussion -
      Sigma (s) bonds are due to sp hybrid orbitals, and 2 p orbitals are used for pi (p) bonds. The two sigma bonds for each \(\ce{C}\) are due to overlap of sp hybrid orbitals of each \(\ce{C}\) atom.

    7. What hybrid orbitals does \(\ce{C}\) use in the molecule:
      This is a trigonal planar molecule. It is called formaldehyde, a solvent for preserving biological samples. The compound has an unpleasant smell.
      Hint: The \(\ce{C}\) atom uses sp2 hybrid orbitals.

      Skill -
      The \(\ce{C}\) atom has 3 sigma (s) bonds by using three sp2 hybrid orbitals and a pi (p) bond, due to one 2p orbital.

    8. What is the shape of the molecule \(\ce{SF6}\)?

      Hint: Its shape is octahedral.

      Discussion -
      Since the \(\ce{S}\) atom uses d2sp3 hybrid orbitals, you expect the shape to be octahedral. The \(\ce{F}\) atoms form an octahedron around the sulfur.

    9. Phosphorus often forms a five coordinated compound \(\ce{PX5}\). What hybrid orbitals does \(\ce{P}\) use in these compounds?

      Hint: The \(\ce{P}\) atom uses dsp3 hybrid orbitals.

      Discussion -
      A total of 5 atomic orbitals are used in the hybridization: one 3d, one 3s and three 3p orbitals. The dsp3 hybrid orbitals of \(\ce{P}\) give rise to a trigonal bipyramidal coordination around the \(\ce{P}\) atom.

      The energy of d orbitals in \(\ce{N}\) is not compatible with 2s and 2p orbitals for hybridization. Thus, you seldom encounter a compound with formula \(\ce{NX5}\) with \(\ce{N}\) as the central atom.

    Contributors and Attributions

    Valence Bond Theory and Hybrid Atomic Orbitals is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?