Chemistry, asked by xyz1234561, 6 months ago

The bond angle between two hybrid orbital is 60°.The percentage s character of hybrid orbital is​

Answers

Answered by DadaGare
1

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.

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