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 structure of $\ce{H2CrO4}$ is as follows:



Shell configuration of chromium is 2,8,13,1 i.e. it has only 1 valence electron and only valence electron participate in bonding but Chromium in the above structure is forming 6 covalent bonds i.e. it is sharing 6 electrons. How's that possible ? How can it share 6 electrons when it only has 1 valence electron.

The same is the case for $\ce{Mn}$ in $\ce{H2MnO4}$ where it has only 2 valence electrons but is forming 6 covalent bonds and for $\ce{Mo}$ in $\ce{H2MoO4}$ where it has only 1 valence electron but forming 6 covalent bonds.
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.