2006-06-29 05:15:29 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
If you consider the H-H system at 90% confidence.
2006-06-29 05:23:12 · update #1
***If you're still there. Check the site below. It explains how the electron in the H 1s orbital is found mostly between .26 and 1.0 angstroms, with the Bohr radius of .526 Angstroms (the accepted distance of the electron from the nucleus in H atom)
For H2 the bondlength is given as .76 Angstroms.
Same order of magniture (e.g. angstroms) but somewhat smaller since you overlap atomic orbitals to make a bond (molecular orbital).
That is for the valence, or outermost orbitals. The inner ones are much smaller as they closer to the nucleus.
There are also unoccupied orbitals (except for excited or Rydberg states) that can be very large...3-5 times the lowest ground state orbital size.
All of this depends on your definition of orbital size too. This is usually a limit on the volume defined by the confidence (e.g. 95%) of finding an electron in that volume.
2006-06-29 05:20:53 · answer #1 · answered by Iridium190 5 · 0⤊ 0⤋
orbital dimensions aren't measured. you can get bond length through crystal measurements. there might be a relationship, but orbitals can't be measured only estimated
2006-06-29 05:20:01 · answer #2 · answered by shiara_blade 6 · 0⤊ 0⤋
I think it has something to do with the affinity quotient towards the nucleus.
2006-07-11 17:05:19 · answer #3 · answered by thewordofgodisjesus 5 · 0⤊ 0⤋