# Bond angles in a oxonium cation

What are the bond angles in protonated alcohol? The bonds in the carbon chain are presumably unaffected by the $\ce{-O^{+}H2}$ group and so in their usual tetrahedral arrangement; what are the angles $\ce{ROH}$ and $\ce{HOH}$?

• In $\ce{NH3}$ there are 4 substituents around the nitrogen atom (3 hydrogens plus 1 lone pair); what are the approximate angles? Analogous to methane, also with 4 substituents, we'd estimate ~109°. In $\ce{H2O}$ there are again 4 substituents (2 substituents and 2 lone pairs); what are the approximate bond angle here (see this earlier answer for more detail)? ...continued – ron Feb 22 '18 at 1:42
• continued... In $\ce{R-O^{+}H2}$ there are again 4 substituents (1 alkyl group, 2 hydrogens and 1 lone pair) around the oxygen atom. Based on the previous examples, what would you estimate the bond angles to be in this case? – ron Feb 22 '18 at 1:43
• Thank you for the edits and answer. I understand the principles behind VSEPR, but not the mathematics. How would I calculate an exact angle? Presumably it, like ammonium, is a little less than the standard tetrahedral 109.47. – speedstyle Feb 22 '18 at 12:43
• You can search Chem SE for "Coulson's Theorem" which relates angle and hybridization. If you know one, then you can calculate the other. – ron Feb 22 '18 at 14:35