Saved Bookmarks
| 1. |
The bond angle between two hybrid orgials is105^(@) . Calculate the percentege of s-character of the hybrid orbital .Given that cos75^(@) = 0 .2588. |
|
Answer» Solution : ` cos alpha = - (1)/(m) ` where m = hybridisation index ` therefore - (1)/(m) cos 105^(@) = cos (180 - 75 ) = - cos ^(@) = - 0.2588 orm = (1)/(0.2588) = 3.86` % s-character ` = (1)/(1+m) xx10- = (1)/(1+3.86) XX100 = 20.58%` Alternatively , s-character decreases as the bond angle decreases. For example , `{:(" Hybrid orbital" ""sp^(3)"" sp^(2) ""sp),("s-character " ""25% ""33.3% ""50%),(" Bond angle """ 109.5^(@) "" 120^(@) ""180^(@)):}` THUS, when bond angle decreases below ` 109.5^(2)`, the s-character will dcrease accordingly. Although the decrease is not linear , even then approsimate value can be calculated as follows : Decrease in angle= `120 - 109.5^(@) = 10.5^(@)` Decrease in s-character =` 33.3 - 25 = 83 ` ACTUAL decrease in bond angle=` 109.5^(@) - 105^(@) = 4.5^(@)` ` therefore ` EXPECTED decrease in s-character `= (8.3)/(10.5) xx4.5 = 3.56%` Thus, s-character should decrease by about 3.56% HENCE, s-character= ` 25 - 3.56 = 21 .44%` |
|