We know that an Oxygen atom has two lone pairs. Why doesn't oxygen then form coordinate covalent bonds using those since it has nothing to lose?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Of course it does, why anyone would think otherwise? $\endgroup$– MithoronJan 2, 2020 at 15:25
-
$\begingroup$ Take a deep breath, and think about how that oxygen reaches your brain bound to the iron atoms of heme in the hemoglobin in the red blood cells. $\endgroup$– Karsten ♦Jan 2, 2020 at 17:45
-
$\begingroup$ chemistry.stackexchange.com/questions/67220/… $\endgroup$– MithoronJan 2, 2020 at 22:10
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Oxygen very much does form bonds in which both electrons come from the oxygen atom. Examples include:
1. The $\ce{H3O^+}$ ion at the center of the solvated proton in aqueous acids, also available as salts of some of the strongest acids such as $\ce{(H3O)(ClO4)}$
2. Carbon monoxide, with its triple rather than double bond.
3. Ozone, in which the oxygen atoms form a bent chain and one of the three bonds must have both electrons coming from the middle atom. The resulting charge separation makes ozone one of the few one-element substances with a nonzero dipole moment.
answered Jan 2, 2020 at 13:33
$\begingroup$
$\endgroup$
The number of complex compounds that are surrounded by water molecules is quite big , a few examples of water being a ligand include [Cu(H2O)4]2+ or [Co(H2O)6]2+. Anyway oxygen is indeed less eager to donate its electrons because its electronegativity is considerably high, nitrogen for example works better as a ligand (actually the first complex compounds to be observed and classified were with NH3 as ligands).But the examples where O donates his electrons can be found quite readily.
