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

1.7 degrees
1.7 degrees

2 degrees
2 degrees

0 degrees
0 degrees

2.5 degrees

1 answer

In a tetrahedral molecular geometry, the ideal bond angle is 109.5 degrees. If a compound has one lone pair and three single bonds, it adopts a trigonal pyramidal shape, which experiences some repulsion from the lone pair, resulting in a bond angle less than 109.5 degrees.

In your scenario, the real bond angle is 107 degrees. To find the difference between the predicted bond angle (109.5 degrees) and the actual bond angle (107 degrees), you can subtract:

109.5 degrees - 107 degrees = 2.5 degrees

Therefore, the difference between the bond angle predicted by the geometry and the real bond angle for this compound is 2.5 degrees.