Bond angle between two hybrid orbitals is `105^@` Percentage of s-orbital character of hybrid orbital is between

Explanation

` cos alpha = - (1)/(m) ` where m = hybridisation index
` therefore - (1)/(m) cos 105^(@) = cos (180 - 75 ) = - cos ^(@) = - 0.2588 or m = (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%`