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Scales of Electronegativity

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Two elements A & B of $3^{r d}$ and $4^{t h}$ periods are such that B. E. of A-A, B-B & A-B are $81 e V$ , $63 e V$ and $76 e V$ respectively. The electronegativity of B is 2.4. What will be the electronegativity of 'A' approximately?

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Solution

$W e k n o w, χ_{A} - χ_{B} = ⎛ ⎝ E_{d (A - B)} - \frac{E_{d} (A - A) + E_{d} (B - B)}{2})^{\frac{1}{2}} W h e r e, χ_{A,} χ_{B} = e l e c t r o n e g a t i v i t y o f A a n d B r e s p e c t i v e l y \Rightarrow χ_{A} - 2.4 = {(76 - \frac{81 + 63}{2})}^{\frac{1}{2}} \Rightarrow χ_{A} = 4.4$

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Similar questions

Q. Two elements A & B of

3^{r d}

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respectively. The electronegativity of B is 2.4. What will be the electronegativity of 'A' approximately?

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(A) Bridge elements (a) Elements of group IA-VIIA

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(E) Transuranic elements (e) Present in 4th, 5th and 6th periods

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