Question

What is the most likely spin multiplicity in the ground electronic state of an alkane, and of a radical such as ·CH3? Explain how you deduce this.

Answer 1

most organic molecules in the ground states are closed shells, and thus, have a multiplicity of one and no charge. Since multiplicity only means the number of ways that an electron can orient itself to a magnetic field, no unpaired electrons means it can only be in one orientation. 

In most cases the multiplicity can be set according to the following rules:
 All electrons in pairs --> singlet (multiplicity = 1)
 One unpaired electron (free radical) --> doublet (multiplicity = 2)
 Two unpaired electrons (diradical) -->
 and unpaired electrons have opposite spin --> singlet (multiplicity = 1)
 or
 and unpaired electrons have same spin --> triplet (multiplicity = 3)
 Multiplicity is the number of degenerate spin eigenfunctions that can be attached
 to the spatial part of the wavefunction.

First of all, for closed shell molecules--including most of the isolable
 organic molecules--the multiplicity is 1, hence for CH4 it will be 1
 Essentially "closed shell" means that the molecule follows the octet rule.
 Molecular fragments such as radicals and carbenes may have multiplicity >1.
 Organic radicals can be assumed to have multiplicity 2 (with very rare
 exceptions).  Carbenes can be either 1 or 3 (singlet or triplet), as can
 biradicals such as tetramethylene and "trimethylenemethane"

For CH3• the multiplicity will be 2