Question:

The valence electron MO configuration of \( {C}_2 \) (atomic number of C = 6) molecule is:

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Remember that \( \pi \)-orbitals are filled before \( \pi^* \)-orbitals in molecules where the number of electrons is less than or equal to 8.
Updated On: Mar 10, 2025
  • \( (\sigma 2s)^3 (\sigma^* 2s)^3 (\pi 2p)^2 \)
  • \( (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 \)
  • \( (\sigma 2s)^2 (\sigma^* 2s)^3 (\pi 2p)^3 \)
  • \( (\sigma 2s)^2 (\sigma^* 2s)^4 (\pi 2p)^2 \)
  • \( (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^5 \)
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The Correct Option is B

Solution and Explanation

The molecular orbital (MO) configuration for \( {C}_2 \) can be determined by filling the molecular orbitals starting from the lowest energy. The energy levels for the \( 2s \) and \( 2p \) orbitals will follow this order: \[ \sigma 2s, \sigma^* 2s, \pi 2p, \pi^* 2p \] For \( {C}_2 \), each carbon atom contributes 4 electrons, giving a total of 8 electrons. According to the MO theory for molecules with even numbers of electrons, the configuration is: \[ (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 \]
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