Q. how to calculate bond angle

Definition. The bond angle at atom B for atoms A–B–C is the geometric angle between the two bond vectors from B to A and from B to C. We will compute that angle using vector algebra and also show an equivalent formula using interatomic distances. The procedure works in two or three dimensions.

Step 1. Write the Cartesian coordinates of the three atoms. Let the position vectors be

\[ \mathbf{r}_A = (x_A,\; y_A,\; z_A), \qquad \mathbf{r}_B = (x_B,\; y_B,\; z_B), \qquad \mathbf{r}_C = (x_C,\; y_C,\; z_C). \]

Step 2. Form the two bond vectors that meet at B. These are the vectors from B toward A and from B toward C. Compute

\[ \mathbf{BA} = \mathbf{r}_A – \mathbf{r}_B = (x_A – x_B,\; y_A – y_B,\; z_A – z_B), \]

\[ \mathbf{BC} = \mathbf{r}_C – \mathbf{r}_B = (x_C – x_B,\; y_C – y_B,\; z_C – z_B). \]

Step 3. Compute the dot product of the two vectors. The dot product is

\[ \mathbf{BA} \cdot \mathbf{BC} = (x_A – x_B)(x_C – x_B) + (y_A – y_B)(y_C – y_B) + (z_A – z_B)(z_C – z_B). \]

Step 4. Compute the magnitudes (Euclidean norms) of the two vectors. Use

\[ \lVert \mathbf{BA} \rVert = \sqrt{(x_A – x_B)^2 + (y_A – y_B)^2 + (z_A – z_B)^2}, \]

\[ \lVert \mathbf{BC} \rVert = \sqrt{(x_C – x_B)^2 + (y_C – y_B)^2 + (z_C – z_B)^2}. \]

Step 5. Use the dot product formula to get the cosine of the bond angle \( \theta \) at B. The cosine is

\[ \cos \theta = \dfrac{\mathbf{BA} \cdot \mathbf{BC}}{\lVert \mathbf{BA} \rVert \; \lVert \mathbf{BC} \rVert}. \]

Step 6. Numerically protect against floating point rounding. If the computed right hand side is slightly greater than 1 or slightly less than -1 because of rounding, clamp it into the interval \([-1,\;1]\). That is, set the value to 1 if it is > 1, and to -1 if it is < -1.

Step 7. Compute the bond angle. Take the inverse cosine to obtain the angle in radians, then convert to degrees if desired:

\[ \theta = \arccos\!\left(\dfrac{\mathbf{BA} \cdot \mathbf{BC}}{\lVert \mathbf{BA} \rVert \; \lVert \mathbf{BC} \rVert}\right). \]

If you want degrees, compute

\[ \theta_{\text{deg}} = \theta \times \dfrac{180}{\pi}. \]

Alternate method using distances (law of cosines). If you already have only the three pairwise distances \(d_{AB},\; d_{BC},\; d_{AC}\), then apply the law of cosines to the triangle ABC to get the angle at B:

\[ \cos \theta = \dfrac{d_{AB}^2 + d_{BC}^2 – d_{AC}^2}{2\, d_{AB}\, d_{BC}}, \]

and then

\[ \theta = \arccos\!\left(\dfrac{d_{AB}^2 + d_{BC}^2 – d_{AC}^2}{2\, d_{AB}\, d_{BC}}\right). \]

Worked numerical example. Let A = (1, 0, 0), B = (0, 0, 0), C = (0, 1, 0). Compute the bond angle at B step by step.

Compute vectors:

\[ \mathbf{BA} = (1 – 0,\; 0 – 0,\; 0 – 0) = (1,\;0,\;0), \]

\[ \mathbf{BC} = (0 – 0,\; 1 – 0,\; 0 – 0) = (0,\;1,\;0). \]

Dot product:

\[ \mathbf{BA} \cdot \mathbf{BC} = 1\cdot 0 + 0\cdot 1 + 0\cdot 0 = 0. \]

Magnitudes:

\[ \lVert \mathbf{BA} \rVert = \sqrt{1^2 + 0^2 + 0^2} = 1, \]

\[ \lVert \mathbf{BC} \rVert = \sqrt{0^2 + 1^2 + 0^2} = 1. \]

Cosine of the angle:

\[ \cos \theta = \dfrac{0}{1\cdot 1} = 0. \]

Angle:

\[ \theta = \arccos(0) = \dfrac{\pi}{2} \text{ radians} = 90^\circ. \]

Summary. To calculate a bond angle A–B–C, form the vectors from B to A and from B to C, compute their dot product and magnitudes, find the cosine by dividing dot product by the product of magnitudes, clamp to \([-1,\;1]\) if necessary, and take the inverse cosine. Alternatively, use the law of cosines with pairwise distances.