Chlorine can be $sp^3d^3$ hybridized.
If so, it can form $\ce{ClH_7}$ and then chlorine, being more electronegative, will have (-7) as its oxidation number. But we know that the oxidation number of $\ce{Cl}$ lies between (-1) and (+7).
Thoughts?
Chlorine can be $sp^3d^3$ hybridized.
If so, it can form $\ce{ClH_7}$ and then chlorine, being more electronegative, will have (-7) as its oxidation number. But we know that the oxidation number of $\ce{Cl}$ lies between (-1) and (+7).
Thoughts?
Your question is based on a skewed premise, as gsurfer04 already pointed out. Rather than invoking d-orbitals, which are usually rather far away in energy from the corresponding p-orbitals (in fact, usually higher in energy than the next shell’s s-orbital, so for chlorine 3p → 4s → 3d), these hypervalent halogen compounds are considered to derive from multi-centre bonds.
For details in the bonding mechanism, I am going to forward you to this excellent answer.
I haven’t seen any multi-hydrogen chloride yet. In fact, when discussing interhalogens like $\ce{ClF3}$, our professor in the introductory chemistry course went as far as saying even $\ce{H3Cl}$ is impossible. I’ve forgotten the exact reasoning (and unfortunately it’s not explicitly stated in the web material any more) but I believe it to be along the lines of hydrogen’s s-orbitals not being able to form the outside parts of a four-electron three-centre bond.
First of all, $d$ orbitals play no part in hypervalency.
If you look at existing hypervalent molecules, you'll see that the central atom is always less electronegative. As H is less electronegative than $\ce{Cl}$, $\ce{ClH7}$ will be very unstable.