Chemistry, asked by tanmaytak289, 10 months ago

What is the % ionic character in HI? Given electro negativity of H=2 and I=2.5?

Answers

Answered by abhiraiwla1

2

Answer:

8.8 %

Explanation:

% ionic = 16(electronegativity diff)-3.5(electronegativity diff)²

it will give 8.875 %

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