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Zorluk: KolayTetrahedral Carbon, Bonding, and Hybridization

What is the hybridization state and geometry of the central carbon atom in a molecule of methane (CH4CH_4)?

  1. sp3sp^3 hybridization with tetrahedral geometry and bond angles of 109.5∘109.5^\circCevap
  2. B
    sp2sp^2 hybridization with trigonal planar geometry and bond angles of 120∘120^\circ
  3. C
    spsp hybridization with linear geometry and bond angles of 180∘180^\circ
  4. D
    sp3sp^3 hybridization with square planar geometry and bond angles of 90∘90^\circ

Cevap

sp3sp^3 hybridization with tetrahedral geometry and bond angles of 109.5∘109.5^\circ
In methane (CH4CH_4), the central carbon atom forms four single covalent σ\sigma bonds with four hydrogen atoms. To achieve equivalent bonding, one 2s2s and three 2p2p orbitals hybridize to form four sp3sp^3 hybrid orbitals directed toward the corners of a regular tetrahedron, giving bond angles of 109.5∘109.5^\circ.

Adım Adım Çözüm

1
Determine the valence electron configuration and bonding of the central carbon atom.
Carbon has 4 valence electrons and forms 4 single σ\sigma (sigma) bonds with four hydrogen atoms in methane (CH4CH_4).
The number of single bonds and lone pairs determines the steric number of the central atom.
2
Determine the hybridization state based on the steric number.
Steric number = 4 (four σ\sigma bonds, zero lone pairs), requiring four equivalent hybrid orbitals formed by combining one 2s2s and three 2p2p atomic orbitals (sp3sp^3).
Mixing one ss orbital and three pp orbitals yields four equivalent sp3sp^3 hybrid orbitals.
3
Determine the spatial geometry and ideal bond angle.
According to VSEPR theory, four bonding pairs position themselves as far apart as possible in 3D space, forming a tetrahedral geometry with bond angles of 109.5∘109.5^\circ.
The tetrahedral arrangement minimizes electrostatic repulsion among the four bonding electron pairs.

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Tetrahedral Carbon, Bonding, and Hybridization
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Tetrahedral Carbon, Bonding, and Hybridization

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