Calculate the best geometry for the allyl radical using the 6-31G basis set and

Question

Calculate the best geometry for the allyl radical using the 6-31G basis set and HF method (IQmol program). Note you will need to specify that this molecule has a “multiplicity” of 2; this will run the calculation as the radical. I suggest putting the hydrogens on yourself so that you get a planar molecule: C3H5.
a. What are the bond lengths for each of the carbon-carbon bonds?

b. Look at the molecular orbitals, which orbitals (give number) correspond to the pi bonding, non-bonding and pi* anti-bonding orbitals. Draw a sketch of what you see.

Explanation / Answer

a) The carbon carbon bonds in the allyl radical is in the range of 1.391 Angstroms using 6-31G basis and HF level of theory.

b) The pi bonding is the HOMO (1b1) and the non bonding (1a2) has the one unpaired electron housed in it. The LUMO (2b1) is pi-antibonding. The allyl radical is at the C2v symmetry which is also the minima in the potential energy surface.

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