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Question

The observed dipole moment for a molecule AB is 1.45 D and its bond is 1.654 $$\mathring { A } $$. Calculate the percentage of ionic character in the bond.


Solution

$$\%\quad of\quad IC=\cfrac { { \mu  }_{ Ovserved } }{ { \mu  }_{ ionic } } \times 100\\ { \mu  }_{ Ovserved }=1.45D=1.45\times { 10 }^{ -30 }cm\\ { \mu  }_{ ionic }=1.654\times { 10 }^{ -10 }m=1.654\times { 10 }^{ -8 }cm\\ { \mu  }_{ ionic }c=q\times d\\ =1.9\times { 10 }^{ 19 }\times 1.65\times { 10 }^{ -8 }\\ =3.1426\times { 10 }^{ -27 }cm\\ %\quad IC=\cfrac { 1.45\times { 10 }^{ -30 } }{ 3.1426\times { 10 }^{ -27 } } \times 100\\ =0.46\times { 10 }^{ -1 }\\ =0.046%v$$

Chemistry

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