Given - Distance and bond length
Find - Percentage ionic character
Solution - Percentage ionic character is 76.9%.
Percentage ionic character can be calculated by the formula - observed dipole moment/theoretical dipole moment*100
Now calculating theoretical dipole moment by the formula - charge * bond length.
Bond length in centimetres = 1.6*10-⁸ centimetres
Theoretical dipole moment = 4.8*10-¹⁰*1.6*10-⁸
Theoretical dipole moment = 6.24*10-¹⁸ esu cm
Now, 10-¹⁸ esu cm is 1 Debye or 1 D
So, 6.24*10-¹⁸ esu cm will be 6.24 D.
Now, calculating percentage ionic character -
Percentage ionic character = 4.8/6.24*100
Percentage ionic character = 76.9%
Thus, percentage ionic character is 76.9%.
Author:
gretamnjt
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4Given :
Dipole moment of A-B molecule = 4.8 D
Bond Length = 167 pm
To Find :
The percentage ionic character of A-B bond
Solution :
Theoretical dipole moment (μ) = Charge (q) × Distance (d)
= 1.6 × 10⁻¹⁹ × 167 × 10⁻¹²
= 267.2 × 10⁻³¹ Cm
Given, Observed Dipole moment = 4.8 D
Taking 1 D = 3.33564 × 10⁻³⁰ Cm
So, Observed Dipole moment = 4.8 × 3.33564 × 10⁻³⁰ Cm
Percentage ionic character = [tex]\frac{Observed Dipole Moment}{Theoretical Dipole Moment} * 100%[/tex]
= [tex]\frac{4.8 * 3.33564 * 10^-^3^0}{267.2 * 10^-^3^1} *100 %[/tex]
= 60%
Therefore, the percentage ionic character of A-B bond is 60%
Hence, option (d) is correct.
Author:
esthero13m
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