The PF4 radical, as an example of a radical containing both fluorine and second-row atoms, is treated using simple quantum-chemical methods. It is possible to estimate the structure from only the isotropic hyperfine splittings. In the present work, the molecular structure of the PF4 radical is assumed to belong the symmetry [formula omitted] with all the P-F internuclear distances equal to 1.57 Å. The calculations are performed by using the self-consistent field atomic orbitals. The bond angles are obtained from both the electron pair bond and the extended Hückel molecular orbital treatment by comparing the calculated hyperfine splittings with the observed values and are in fair agreement with each other. The most probable values are [formula omitted] and [formula omitted], which are fairly close to those of the SF4 molecule. The isotropic hyperfine coupling constant of the two fluorine nuclei making the larger bond angle, ∠F1PF1′, is larger than that of the other pair making the smaller angle, ∠F2PF2′. © 1969, American Institute of Physics. All rights reserved.