The bond angle between two hybrid orbital is 60°.The percentage s character of hybrid orbital is
Answers
Answer:
50%
Explanation:
How?
COS60=1÷2
Then,
(1÷2×100)=50
e.g. For x = 180
Cos 180 => -1 = s/s-1 => -s+1 = s => s=1/2. Now % s character = (1/2 × 100)% = 50%. So it contributes 50% s orbital from 100% to form new hybrid orbital. And other 50% should be contributed by p orbital. Therefore s orbital contributes 50% and p orbital contributes 50% so new hybridization will be SP.
For x = 120
Cos 120 => -1/2 = s/s-1 => -s+1 = 2s => s=1/3. Now % s character = (1/3 × 100)% = 33.3%. So it contributes 33.3% s orbital from 100% to form new hybrid orbital. And other 66.7% should be contributed by p orbital. Therefore s orbital contributes 33.3% and p orbital contribute s66.7% so new hybridization will be SP2.
And if we put x = 109.5, then the hybridization is SP3.
So, order of bond angle as, SP>SP2>SP ,180>120>109.5
Where SP has more s character and bond angle than SP2, and SP2 have more s character and bond angle than SP3.
So, S Character ∝ Bond Angle.