A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 A0 . Find the percentage ionic character in the molecule.

μtheo=4.8×10−10×1.6×10−8=7.68×10−18esu.cm=7.68 D.% ionic character =3.27.68×100=41.6%

A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 A0 . Find the percentage ionic character in the molecule.

μtheo=4.8×10−10×1.6×10−8=7.68×10−18esu.cm=7.68 D.% ionic character =3.27.68×100=41.6%

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A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 $\overset{0}{A}$ . Find the percentage ionic character in the molecule.

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71.2%

41.6%

39.6%

31%

answer is B.

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Detailed Solution

$μ_{theo} = 4.8 \times 10^{- 10} \times 1.6 \times 10^{- 8} = 7.68 \times 10^{- 18} esu.cm = 7.68 D$ .
% ionic character $= \frac{3.2}{7.68} \times 100 = 41.6 %$

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A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 A0 . Find the percentage ionic character in the molecule.

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A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 $\overset{0}{A}$ . Find the percentage ionic character in the molecule.