Dipole moment

Dipole moment

The degree of polarity is expressed in terms of dipole moment, i.e., the product of charge and distance between the tow charged ends, i.e.,

$\text{Dipole moment}(\mu) = \text{Charge} (e) \times \text{Distance} (d)$.

Since charge and distance are of the order of $4.8 \times 10^{-10} e.s.u. \text{and} 10^{-8}cm$ respectively hence $\mu \text{is of the order} 10^{-18} e.s.u. cm \text{or Debye D I.E.,} D = 10^{-18}e.s.u. cm$. In S.I. units e and d are of the order of $10^{-19} \text{and} 10^{-10}m$ respectively. Thus $\mu \text{is of the order } 10^{-19} \times 10^{-10}m C \text{or} 10^{-29}mC$(meter coulomb).

Dipole moment of non-polar diatomic molecule is zero. But in the case of tri-atomic, tetra-atomic etc. molecules, it is not so.

In such cases the value of $\mu$ also depends upon the geometry of the molecule e.g. the value of $\mu$for $CO_2$ and $CCl_4$ is zero although both have polar bonds. This is due to the fact that $CO_2$ is a linear molecule and $CCl_4$ is regular tetrahedral in shape hence the resultant $\mu$ is zero. The molecule such as $H_2O, H_2S, SO_2, NH_3$ etc. hae certain dipole mo ent because these molecules have bent structures. . (Un-symmetrical structure).

Similarly, trans-isomers have zero dipole moment while cis-isomers have certain dipole moment. The value of $\mu$ also depends upon the orientation in disubstituted benzene e.g. p-disubstituted benzene having same group has dipole moment zero while tha of o- and m –derivatives have certain values.

D.M. Electronegativity difference

$D.M. \propto \dfrac{1}{\Theta} \\ D.M. \propto \text{No. of lone pair of electrons}$.

All molecules with normal shapes like Linear, Trigonal, Tetrahedral etc. have zero dipole moment. But molecules with abnormal shapes due to $e^- e^-$ repulsion have dipole moment.

For ex. $\to$ Pyramidal, V-shape (Angular) T-shape, Sea-Saw shape etc.

Ex. $NH_3, H_2O, I_3^-, XeF+_2, H_2S, PH_3, SO_2, CH_3Cl, PCl_3$ etc.

Orders:

$\to H - F > H_2O > NH_3 > SO_2 \\ \to \text{C is-alkene} > \text{trans-alkene mostly zero due to symmetry}$.

Exception: trans-2 pentene has some value of dipole moment, i.e.,

Such trans alkene has some D.M. but less than cis-form.

With the knowledge of dipole moment % ionic character of the bond may be calculated as:

%$\text{ionic character} = \dfrac{\text{Observed dipole moment} \times 100}{\text{Dipole moment for 100 \% ionic bond, i.e.,} q \times r}$

Example1. In the HCl molecule, $d_{HCl } \text{is} 0.127nm$. The dipole moment of gaseous HCl is $0.347 \times 10^{-29}mC$. Calculate \% ionic character of the bond.

Solution: for 100% ionic bond,

$\mu = e \times d \\ = 1.6 \times 10^{-19}C \times 1.27 \times 10^{-10}m \\ 2.032 \times 10^{-29}mC$

%ionic character

% ionic character = $\dfrac{\text{Observed dipole moment} \times 100}{\text{Dipole moment for 100 \% ionic bond}} \\ \dfrac{0.347 \times 10^{-29} mC \times 100}{2.032 \times 10^{-29}mC} = 17.07 \%$

Pauling gave a relationship between the % ionic character of the bond and the difference of electronegativities for diatomic molecule as:

% ionic character = $(0.16 \Delta + 0.035 \Delta^2) \times 100$ where $\Delta$ is the difference between the electronegativities of the two atoms.

Electronegativity $\to$ 0.1 0.2 0.61 1.1 1.9 2.3 3.0 difference

% ionic character $\to$ 0.5 1.0 10 25 50 75 90

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