Only valence shell electrons participate in bonding then how is Cr in H2CrO4 forming 6 covalent bonds although it has only 1 valence electron?
The electronic configuration of $\ce{Cr}$ is: $[\ce{Ar}]4s^2 3d^4$. Because of very little energy difference between 4s- and 3d- orbitals, it can lose 6 of it's valence electrons and thus form 6 covalent bonds.