Question

Calculate the percentage of ionic character of a bond having 1.275 A it's length and 1.03 d it's dipole moment
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Answer 1

Explanation:

l hope this answer is right

Answer 2

$we \: know \: , μ = q + 1</p><p></p><p>$

$= 1.6 \times {10}^{ - 29} c \: \times 1.275( {10}^{ - 10})m$

$= 2.04 \times {10}^{ - 29} cm$

$actual \: , μ \: of \: HCL \: = 1.03d </p><p></p><p>$

$= 3.36 \times {10}^{ - 30} cm$

$= 0.34 \times {10}^{ - 29} cm$

$so \: percentage \: of \: ionic \: \\ charecter$

$= \frac{actual μ}{calculated μ} \times 100$

$= \frac{0.34 \times {10}^{ - 29} }{2.04 \times {10}^{ - 29} } \times 100$

$= 16.667\%(approx)$