Anonymous asked in 科學及數學化學 · 1 decade ago

hybirdization states of some compounds

How to determine the hybirdization state of a particular elements in a compound ?

Please give examples including carbon , oxygen and nitrogen. Thanks


I still cannot get the technique of finding the hybirdization states!

Update 2:

O=C=O in C = sp2??Why??

3 Answers

  • ?
    Lv 7
    1 decade ago
    Favorite Answer

    The very simple technique I use to work these out is:

    triple bond = sp hydridised

    double bone = sp2

    single bonds = sp3

    In a more physical chemical term, we need to think about the molecule as a whole. ie. not just the central atom but the substitutents attached to it. The reason why we have sp-x is because in a molecule, the s-orbital and the p-orbitals are 'mixed' when bonding (ie. sharng electrons). The 1s-orbital is always filled, and the rest of the electrons are spread out into the available 2s and 2p-orbitals, which are px, py, and pz (because 1s only holds 2 electrons, most molecules have more than 2 electrons)

    So, instead of: 1s, 2s, 2px, 2py, 2pz

    we'd have: 1s, 2s, 2spx, 2spy, 2spz <-- this is the case of sp3 hybrid, because 2s mixes with all the p-orbitals when forming a bond.

    In the case of sp2, the 2s orbital obly mixes wth two p-orbitals, and sp = 2s mixes with just one p-orbital

    To go further, it would require 'Molecular Orbital Theory' to explain everything, which includes concepts such as bonding and non-bonding orbitals.

    Hope that makes sense.

    2008-05-17 21:12:35 補充:

    Oops, examples:

    CH4 = sp3 -- 1s orbital filled, the rest spreads out into 2s, 2px, 2py and 2pz. When a bond is formed, the electrons are promoted to the higher orbital due to energy put into the system.

    2008-05-17 21:12:42 補充:

    Ethylene = sp2 -- same reason but only two p-orbitals are occupied.

    Hybridisation also considers the orientation the atom are most comfortable in, partly the reason for the different orbitals being mixed during bonding.

    Source(s): Me
  • 1 decade ago

    Steps to find the hybridization state of the central atom (for C, N and O) of a molecule :

    Step 1: Draw the structural formula (or 3-dimensional diagram) of the molecule.

    Step 2: Count the total number of single bonds, double bonds, triple bonds and lone pairs of the central atom. (Only the central atoms are hybridized.)

    Step 3: Determine the hybridization state of the central atom.

    If the total number is 2, the central atom is sp hybridized.

    If the total number is 3, the central atom is sp2 hybridized.

    If the total number is 4, the central atom is sp3 hybridized.


    Example 1: Methanl (HCHO)


    C is the central atom, which has 2 single bonds and 1 double bond, and the total number is 3. Therefore, the C atom in HCHO is sp2 hybridized.


    Example 2: Hydrogen cyanide (HCN)


    C is the central atom, which has 1 single bond and 1 triple bond, and the total number is 2. Therefore, the C atom in HCN is sp hybridized.


    Example 3: Ammonia (NH3)


    N is the central atom, which has 3 single bonds and 1 lone pair, and the total number is 4. Therefore, the N atom in NH3 is sp3 hybridized.


    Example 4: Water (H2O)

    Draw the structural formula of H2O. O is the central atom, which has 2 single bonds and 2 lone pairs, and the total number is also 4. Therefore, the O atom in H2O is sp3 hybridized.

  • 冷風
    Lv 5
    1 decade ago

    The direct method determining the hydridization state of a organic compound is distinguish the order of bondings involved in the compound.

    For example, like

    entrile group ─CN

    we know that the type of the bonding is carbon-nitrogen triple bond.


    the hydridization state of entrile group ─CN is 「sp hybridization」

    And other example is aldehyde group ─CHO

    the C-O bond is double bond, the C-H bond is single bond.

    Therefore, the hybridization state of aldehyde group is 「sp2 hybridization」.

    Source(s): Myself
Still have questions? Get your answers by asking now.