2007-12-03 06:44:25 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Helium cannot combine with Helium, or any other element, since it is a noble gas.
(This might have changed since I last read about it, as new science has been combining some noble gases).
2007-12-03 06:48:07 · answer #1 · answered by Feeling Mutual 7 · 1⤊ 0⤋
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RE:
explain why helium is monatomic but hydrogen exists as diatomic molecules?
2015-08-07 01:45:05 · answer #2 · answered by Anonymous · 0⤊ 0⤋
All atoms seek stability. Hydrogen has one quantum shell. This shell can hold 2 electrons but hydrogen has only 1 so it forms covalent bonds with other hydrogen atoms to obtain a stable confirguration. Helium, on the other hand, belongs to group 8 and so is a noble gas and all noble gases are unreactive. Also, Helium contains 1 quantum shell and two electrons so it has a stable configuration.
2007-12-03 06:56:15 · answer #3 · answered by silent R 1 · 0⤊ 0⤋
Is Helium Diatomic
2016-12-16 08:05:16 · answer #4 · answered by jaffe 4 · 0⤊ 0⤋
Hydrogen is 1s^1, which is not a full shell. If two of them share electrons, each has a share in two electrons, which is a full shell. Helium is1s^2, which is a full shell, so He is stable as is without forming diatomic molecules.
2007-12-03 06:48:54 · answer #5 · answered by steve_geo1 7 · 3⤊ 0⤋
Is Hydrogen Diatomic
2016-10-02 00:20:03 · answer #6 · answered by caitlyn 4 · 0⤊ 0⤋
Try to build a simple molecular orbital plot for the elements of the first row and you will see that He2 cannot exist because bond order is zero because the same number of electrons are present in bondig- and anti-bonding MO.
2007-12-03 06:55:50 · answer #7 · answered by LU 6 · 2⤊ 0⤋
Helium has a full s shell.
2007-12-03 06:48:19 · answer #8 · answered by Anonymous · 1⤊ 0⤋
Hydrogen atoms get lonely.
2007-12-03 06:47:22 · answer #9 · answered by Reverend Black Grape 6 · 3⤊ 0⤋
Why didn't you tell me you wanted it the other way before I went through all that work?!
2007-12-03 06:48:03 · answer #10 · answered by LovesAssFlakes 1 · 1⤊ 0⤋