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Explain geometry of methane molecule on the basis of Hybridization. - Chemistry

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Explain geometry of methane molecule on the basis of Hybridization.

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Formation of methane (CH4) molecule on the basis of sp3 hybridization:

  1. Methane molecule (CH4) has one carbon atom and four hydrogen atoms.
  2. The ground state electronic configuration of C (Z = 6) is 1s2 2s2 \[\ce{2p^1_{{x}}}\] \[\ce{2p^1_{{y}}}\] \[\ce{2p^0_{{z}}}\].
    Electronic configuration of carbon:
    Ground state: 1s 2s 2p
    ↑↓ ↑↓  

    Excited state:

    1s 2s 2p
    sp3 Hybrid orbitals: ↑↓
    1s (four sp3 hybrid orbitals)
  3. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp3 hybridization.
  4. One electron from the 2s orbital of the carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py, and pz mix and recast to form four new sp3 hybrid orbitals having the same shape and equal energy. They are maximum apart and have tetrahedral geometry with an H–C–H bond angle of 109°28'. Each hybrid orbital contains one unpaired electron.
  5. Each of these sp3 hybrid orbitals with one electron overlaps axially with the 1s orbital of the hydrogen atom to form one C–H sigma bond. Thus, in CH4 molecule, there are four C–H bonds formed by the sp3–s overlap.
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Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 5 Chemical Bonding
Exercises | Q 3. (F) | Page 79
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