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Annette [7]
4 years ago
15

The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is

Chemistry
1 answer:
WINSTONCH [101]4 years ago
3 0

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

  • sp^{3}, four electron domains, as in ethane \text{C}_2\text{H}_6
  • sp^{2}, three electron domains, as in ethene \text{C}_2\text{H}_4
  • sp, two electron domains, as in ethyne \text{C}_2\text{H}_2

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

  • Carbon atoms with a sp^{3} hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately 109.5\textdegree{}
  • Carbon atoms with a sp^{2} hybridization would demonstrate a triangular planar configuration with a bond angle of 120\textdegree{}
  • Carbon atoms with a sp hybridization would demonstrate a linear configuration with a bond angle of 180\textdegree{}

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

  • sp^{3} in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
  • sp^{2} in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.
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For the given chemical reaction:

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To calculate the E^o_{cell} of the reaction, we use the equation:

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\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

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