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Question

Two elements A & B of 3rd and 4th periods are such that B. E. of A-A, B-B & A-B are 81 eV, 63 eV and 76 eV respectively. The electronegativity of B is 2.4. What will be the electronegativity of 'A' approximately?

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Solution

We know,χAχB = Ed(AB)Ed(AA)+Ed(BB)2)12Where, χA, χB = electronegativity of A and B respectively χA2.4 = (7681+632)12χA = 4.4

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