Question
A central atom with one lone pair and three single bonds would usually have a tetrahedral shape. Suppose a compound has extra pressure from its lone pairs and has a bond angle of 107 degrees. What is the difference between the bond angle predicted by the geometry and the real bond angle for this compound?(1 point)
Responses
2.5 degrees
2 degrees
1.7 degrees
0 degrees
Responses
2.5 degrees
2 degrees
1.7 degrees
0 degrees
Answers
GPT-4o mini
In a tetrahedral structure, the ideal bond angle is 109.5 degrees. However, with one lone pair, the actual bond angle is typically slightly less due to the repulsion exerted by the lone pair, which pushes the bonding pairs closer together. In this case, you mentioned that the bond angle is 107 degrees.
To find the difference between the predicted bond angle (109.5 degrees) and the actual bond angle (107 degrees), we can calculate:
\[
\text{Difference} = 109.5^\circ - 107^\circ = 2.5^\circ
\]
Therefore, the difference between the bond angle predicted by the geometry and the real bond angle for this compound is **2.5 degrees**.
To find the difference between the predicted bond angle (109.5 degrees) and the actual bond angle (107 degrees), we can calculate:
\[
\text{Difference} = 109.5^\circ - 107^\circ = 2.5^\circ
\]
Therefore, the difference between the bond angle predicted by the geometry and the real bond angle for this compound is **2.5 degrees**.