The values of EN of A and B are 1.2 and 4.0 respectively. $%$ ionic character in bond A - B is

$50 %$

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$43 %$

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$55.3 %$

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$72.24 %$

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Solution

The correct option is D
$72.24 %$

$16 [E_{A} - E_{B}] + 3.5 [E_{A} - E_{B}]^{2} = %$ Ionic character

$16 [4.0 - 4. q] + 3.5 (4 - 1.2)^{2} = %$ Ionic character

$% I o n i c c h a r a c t e r = 72.24 %$

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Similar questions

The values of EN of A and B are 1.2 and 4.0 respectively. $%$ ionic character in bond A - B is

A

B

1.20

4.0

A - B

= 18 (X_{B} - X_{A})^{1.4}

X_{B}

X_{A}

X_{B} > X_{A} .

X_{A} = 1.84 X_{B} = 3.48,

(1.64)^{7 / 5} = 2.0)

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