Explicit calculations are made to examine whether the tetrahedral model of CH_{4} is really the most stable in the Hund‐Mulliken and Slater‐Pauling theories. The secular determinant involved in the H‐M procedure is too complicated (degree 8) to solve generally, but three methods of approximation are given applicable in three different limiting cases, in all of which the tetrahedral model proves to be that of least energy quite irrespective of repulsions between the H atoms. The fact that two of the methods do not assume the carbon *s‐p* separation to be small shows that *s‐p* hybridization is not a necessary condition for tetrahedral valence bonds. Calculations are also given which show that in the Heitler‐London‐Pauling‐Slater method the regular tetrahedron is superior to other models of somewhat lower symmetry. It is further calculated according to both the H‐M and H‐L‐P‐S schemes that in compounds of the form CH_{2}X_{2}, CHX_{3}, and CH_{3}X, the most stable models are tetrahedra of less symmetry than the regular tetrahedron; i.e., models with the valence angles somewhat distorted from 109.5° unless the C – H and C — X bonds are of equal intensity. The predicted directions of the deviations from 109.5° are in agreement with x‐ray diffraction data for CH_{2}Cl_{2} and CHCl_{3}. The conclusion is rather prevalent in the literature that with *s‐p* hybridization and electron pairing two bond axes tend to set themselves at 109.5°. This is shown incorrect; instead the angle can be anything between 90° and 180° depending on the relative intensities of the *s* and *p* bonds. If the *s* bonding power is not negligible, the angle between an NH axis and the pyramidal axis in NH_{3} should be somewhat greater than the value 54.7° which is obtained if the three NH axes are orthogonal and which is characteristic of pure *p*‐bonding. Actually Dennison and Uhlenbeck find 68°, and Lueg and Hedfeld 73°, from band spectra. It is shown that CH_{4}^{+} should be a flattened rather than regular tetrahedron, conceivably even being plane. Also CH_{3} should be a flatter pyramid than NH_{3}.

## REFERENCES

*r*in (4), use must be made of the fact that if $\psi \pi ,$ $\psi \pi \u2032$ denote π wave functions of C with different phases,

*Y*which interests us when $WC\u2212WH\u226bQ$ is $Z\u2212X$ where

*Z*is the spur of the complete matrix (4) and

*X*is our previous root sum. The value of

*Z*is independent of the angles, and hence

*Y*has an extremum when

*X*does. When $WC\u2212WH\u226bQ$ the tetrahedral arrangements gives a minimum of $Z\u2212X$ and a maximum of

*X*rather than the reverse as previously. This is because sign changes in τ and in certain of the integrals (3), occasioned by the electronic drift being from C to H rather than from H to C, make the expressions β, γ defined in (19, 20) become negative.

*ms*and $mp\pi .$ Such a choice of the axis of $mp\sigma $ is clearly demanded for overlapping favorable to bonding. With these assumptions, only the $2p\sigma $ electron of the halide is involved in the secular problem (4), and the integrals (3) obviously are to be calculated for an $mp\sigma $ rather than 1

*s*state of X. Also no harm is done if the letter H similarly symbolize some atom other than hydrogen and not necessarily in an

*S*state. Thus the conclusions of our calculations also apply to compounds such as $CCl4,$ $CCl2Br2,$ $CCl3Br,$ as well as $CH4,$ $CH2X2,$ $CH3X.$

*Valence and the Structure of Atoms and Molecules*, Chem. Cat. Co., 1923.

*d*states are admitted to the linear combinations, but less general in that all the functions are supposed to have the same hybridization ratio. Eqs. (33) and (34) each involve two such ratios. Four would be required for CWXYZ. Presumably the number of different ratios required is the same as the number of different kinds of atoms attached by C. If this is the case then all the C wave functions of $CH4$ have the same ratio and the tetrahedral model follows uniquely simply from the requirement of orthogonality. Nevertheless it seems of some interest for us to verify that introduction of two unequal ratios in place of one raises the value of (35) in $CH4.$

*K*is our

*J*.

*s‐p*hybridization ratios for double bonds will not be the same as for $CH4.$ We suggest this combination of (42) with (34) because it gives three directed coplanar valences,

*viz.*, those corresponding to $\psi aC,$ $\psi bC,$ $\psi d\u2032.$ The C–C bond is here to be located along the

*z*axis, so as to bisect X–C–X. This subject of double bonds will be discussed more fully in a paper by W. G. Penney.

*s‐p*separation in the same fashion as in connection with (41). The addition of $\u22122Nss$ and the omission of $f(\omega \u2032)$ are consequences of the fact that in (38) those terms containing $N\sigma \sigma ,$ $N\sigma s,$ $N\pi \pi $ which arise from interaction of the third and fourth electrons of C with $X2,$ are proportional to $f(\omega \u2032),$ while the corresponding terms containing $Nss$ are proportional to $1\u2212f(\omega \u2032).$ Such terms must be dropped and new terms added because of the substitution of new wave functions in place of the third and fourth of (34).

*s*states.