# 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.

• There is no such equation. The bond angle can be estimated by some simple rules, such as VSEPR theory. 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. Oct 30 '19 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.