I guess that an alternative way of asking this question is what is the
hybridization of H2SO4?
A better way to ask this might be: "What is the hybridization of the sulfur atom in Hydrogen Sulfate?"
The sulfur atom has a bond with four other atoms in this molecule. Because that the electrons try to stay as far away from each other as possible according to Valence Shell Electron Pair Repulsion Theory (VSEPR), the other atoms will seperate as far as possible, which results in bonds that are 109.5 degrees apart from one another. If this were a square planar arrangement, the bonds would only be 90 degrees apart, which is not as satisfactory as the tetrahedral arrangement would provide.
According to my book, if the number of effective pairs is 4, the arrangement is tetrahedral, and therefore the hybridization required is $sp^3$
Zumdahl, Zumdahl, Chemistry: An Atoms First Approach
This certainly answers the question, but why?
It has to do with molecular orbitals, and how they hold only a certain number of electrons, and they prefer to spread themselves out evenly because in the quantum world, electrons tend to prefer the lowest possible energy state. Once the sigma bond is occupied, which is the first orbital referred to as the s orbital, the electrons end up occupying the p orbitals, and when electrons are shared through these orbitals they are called pi bonds, they actually form a whole new orbital, a molecular orbital, which allows the sharing of electrons between two atoms. Let me refer you to this website to help visually show you what is happening.