# Determine the bond angle in a compound [closed]

How can we find the bond angle between 3 atoms in a compound? Please specify an equation which can be used for all compounds.

## closed as too broad by Geoff Hutchison, Klaus-Dieter Warzecha, DavePhD, jerepierre, Loong♦Nov 25 '14 at 16:16

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• There is no such equation. The bond angle can be estimated by some simple rules, such as VSEPR theory. – Geoff Hutchison Nov 25 '14 at 14:45
• If you know atomic coordinates (say, from a single-crystal diffraction experiment), then the task is basically the same as finding an angle between two vectors in 3D space. – andselisk Oct 30 at 10:49

In carbon compounds Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.

$$1+\lambda_{i} \lambda_{j} \cos(\theta_{ij})=0$$

where $$\ce{\lambda_{i}}$$ represents the hybridization index of the $$\ce{C-i}$$ bond (the hybridization index is the square root of the bond hybridization) and $$\ce{\theta_{ij}}$$ represents the $$\ce{i-C-j}$$ bond angle. Let's look at some examples.

• In the case of methane, each $$\ce{C-H}$$ bond is $$\ce{sp^3}$$ hybridized and the hybridization index of each $$\ce{C-H}$$ bond is $$\sqrt3$$. Using Coulson's theorem we find that the $$\ce{H-C-H}$$ bond angle is 109.5 degrees.
• In the case of the methyl carbocation ($$\ce{CH3^{+}}$$) the $$\ce{C-H}$$ bonds are $$\ce{sp^2}$$ hybridized and the hybridization index of each $$\ce{C-H}$$ bond is $$\sqrt2$$. In this case, using Coulson's theorem we find that the $$\ce{H-C-H}$$ bond angle is 120 degrees.
• Finally, in the case of acetylene the $$\ce{C-H}$$ bond is $$\ce{sp}$$ hybridized and the hybridization index of each $$\ce{C-H}$$ bond is $$\sqrt1$$. In this case, using Coulson's theorem we find that the $$\ce{C-C-H}$$ bond angle is 180 degrees.

Conversely, if the bond angle is known, the hybridization can be determined.