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  4. If the electronegativity difference between two atoms A and B is 2.0, then the percentage of co-valent character in the molecule is

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Q. If the electronegativity difference between two atoms A and B is 2.0, then the percentage of co-valent character in the molecule is

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Solution:

According to Hannay and Smith equation $ \therefore $ % ionic character $ =16({{x}_{A}}-{{x}_{B}})+3.5{{({{x}_{A}}-{{x}_{B}})}^{2}} $ where, $ {{x}_{A}} $ and $ {{x}_{B}} $ are the electronegativitys of the atoms A and B respectively. $ \therefore $ % ionic character $ =16(2)+3.5{{(2)}^{2}} $ $ =32+14=46% $ c