Take acetylene, for example.

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It is defined that the two outer bonds are at 180 degrees.

How is it possible to theoretically define all $\ce{C-H}$ bonds in acetylene as being 180 degrees?

  • $\begingroup$ With a protractor? — Kidding aside, we have no idea what's your background, what sort of explanation you want (theoretical? experimental?), and at what level. Thus this question can hardly be answered in its present form. $\endgroup$
    – F'x
    Dec 9, 2013 at 15:53
  • $\begingroup$ Theoretical, I mean, why can't those outer bonds be, say, 60 degrees? $\endgroup$ Dec 9, 2013 at 15:54
  • $\begingroup$ Again: what's your level of understanding of molecules and their shapes? $\endgroup$
    – F'x
    Dec 9, 2013 at 15:56
  • $\begingroup$ Not a whole lot. I have a working knowledge of the octet rule and the purpose of bonds, but the measure of those bonds is what I do not understand $\endgroup$ Dec 9, 2013 at 15:58
  • 2
    $\begingroup$ I think this question is clear, and shouldn't be on hold. There is a simple and definite answer: acetylene is perfectly linear when you 'freeze' its vibrations, because there is nothing pushing H's to bend. Look at its charge density at uam.es/departamentos/ciencias/qfa/DAM/acetylene.html $\endgroup$
    – sencer
    Dec 10, 2013 at 2:44

1 Answer 1


If we look at the molecule we see it is symmetrical. The forces acting upon the 2 hydrogen atoms are identical. As you know, when hydrogen forms a covalent bond with carbon, it's electron is drawn more strongly by the carbon than by the hydrogen, which leaves the hydrogen with a net positive charge (carbon is more electronegative than hydrogen).

Also, the 3 covalent bonds between the carbons are kind of bowed outward in the middle. The 3 bonds between the carbons constitute the most negative part of the molecule.

Naturally, the positive and negative charged moieties are drawn toward each other as much as possible. But, given that the force of attraction is equal on all sides of the hydrogen, it will be aligned equidistant on all sides to the negative triple bond.

I don't know if I am painting a good word picture here, but the forces exerted on the hydrogens place them exactly on a straight line, i.e. their bonds are 180 degrees.

Of course there are other forces in play, namely the forces of neighboring molecules. However, they will also align themselves as much as possible so as to minimize the repulsion of having hydrogen atoms close to one another and to maximize proximity of hydrogen atoms to the triple bonds of neighboring molecules. This arrangement will be preserved as much as possible as the molecules move about. So, the net forces upon the hydrogen-carbon bonds will always force those bonds to remain 180 degrees.

  • $\begingroup$ How would this explanation work for a water molecule that has a 120 degree bond angle? $\endgroup$
    – Michiel
    Dec 10, 2013 at 21:24
  • 1
    $\begingroup$ In addition to the two hydrogens, the oxygen atom has 2 pairs of unbonded electrons (lone pairs) that point towards the remaining two vertices of a tetrahedron with the oxygen in the center. Because the lone pairs are less localized than the electrons in the oxygen-hydrogen bond, they take up a little more room, causing the H-O-H angle to deviate away from the ideal 109.5 degrees. $\endgroup$
    – craigim
    Dec 13, 2013 at 1:52

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