The electronegativities of elements A and B are 1.2 and 3.4 units respectively. The type of bond connects A and B in compound AB is:

covalent

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ionic

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coordinate covalent

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polar covalent

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Solution

The correct option is D ionic
$%$ ionic character $= [1 - e^{\frac{- (X_{A} - X_{B})^{2}}{4}}] \times 100$
$X_{A} = 1.2, X_{B} = 3.4$
$%$ ionic character $= [1 - e^{\frac{- (1.2 - 3.4)^{2}}{4}}] \times 100$
$= 71 %$

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