| 5:01:58 PM PDT - Mon, Mar 10th 2014 |
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Hi Sanjib,
Your occupation numbers are correct for carbon and nitrogen atoms, but things are different when you switch to molecular orbitals in the CN radical. The simplest explanation is probably through orbital hybridization. A quick internet search or chemistry textbook would give more details, but in short:
You can ignore the core electrons (the 1s electrons) from each atom and focus on the valence electrons. There are 4 valence electrons in carbon and 5 valence electrons for nitrogen. Since there are only two atoms, the molecule will be linear with sp hybridized orbitals mixing contributions from the 2s orbitals and one of the 2p orbitals on each atom. You will have two sp hybrid orbitals and two remaining 2p orbitals. One pair of sp orbitals (one from each atom) will overlap to form a sigma bond with one electron from each atom. Both pairs of 2p orbitals will overlap to form two pi bonds (with each bond having one electron from each of the atoms). The sp orbital left on nitrogen will be doubly occupied (lone pair) because nitrogen will be surrounded by 5 valence electrons (one from each bond and both from the lone pair) to maintain a neutral charge. The final electron will be unpaired in the remaining sp orbital on carbon.
The final result is a triple bond between the two atoms, a lone pair on the nitrogen, and a single unpaired electron on carbon. 2S+1 = 2
Cheers,
Matt
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