# 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 definition of valence electrons is a lot more blurry than it seems. – Ivan Neretin Feb 17 at 11:29
• And covalent bonding capacity is based on valence orbitals (which can also get blurry), not valence electrons. Boron forms compounds where four valence orbitals are involved in covalent bonds even though it brings in only three valence electrons. – Oscar Lanzi Feb 17 at 14:55

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.