# Bonding in Sulphur dioxide [duplicate]

In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 82, under section 3.8 - Types of Covalent Bonds (Sigma and Pi Bonds) it is given (for Sulphur dioxide molecule):

The $$\pi$$ bonds do not alter the shape, but merely shorten the bond lengths. The bond angle is reduced from ideal $$120°$$ to $$119°30'$$ because of the repulsion by the lone pair of electrons. Problems arise when we examine exactly which atomic orbitals are involved in $$\pi$$ overlap. If the $$\sigma$$ bonding occurs in the $$xy$$ plane then $$\pi$$ overlap can occur between the $$3p_z$$ orbital on sulphur and the $$2p_z$$ orbital on one oxygen atom to give one $$\pi$$ bond. The second $$\pi$$ bond involves a d orbital. Though the $$3d_{z^2}$$ orbital on sulphur is in the correct orientation for $$\pi$$ overlap with the $$2p_z$$ orbital on the other oxygen atom the symmetry of the $$3d_{z^2}$$ orbital is wrong (both lobes have a + sign); whilst for a p orbital one lobe is + and the other -. Thus overlap of these orbital does not result in bonding. The $$3d_{xy}$$ orbital in sulphur is in the correct orientation and has the correct symmetry to overlap with the $$2p_z$$ orbital on the second oxygen atom, and could give the second $$\pi$$ bond.

I am unable to understand the part which has been typed in boldface.

We know $$d_{xy}$$ orbital lies in the $$xy$$ plane. We also know that $$p_z$$ orbital lies along the z-axis. Then how can these two orbital participate in $$\pi$$ bonding?

The following image is my attempt to show bonding in sulphur dioxide. The oxygen atom under consideration is the second one marked with a $$\bigstar$$.

In the following image the lone pair present in the s orbital is not shown for the sake of simplicity. There will be some distortion in the bonds and sulphur dioxide would be bent and will not be linear. But here it is shown to be linear.

Clearly, for the second oxygen atom, the overlapping of atomic orbitals is not possible due to out of phase approach. I feel the formation of $$\pi$$ bond between the sulphur atom and the second oxygen atom must take place between $$2p_x$$ orbital of the second oxygen atom and $$3d_{xy}$$ orbital of the sulphur atom. Kindly tell whether I am correct or not.

Kindly clarify my doubt.

• This won't help solving your problem, but it is important: Sulfur does not use d orbitals for bonding. The statement in the book is old, it is no longer considered correct. – Martin - マーチン Aug 25 '19 at 11:04