Question

The dipole moment of KCl is $3.336 \times 10^{-29}$ cm which indicates that it is highly polar molecule. The inter-atomic distance between $K^{+}$ and $Cl^{-}$ in this molecule is $2.6 \times 10^{-10}$ m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl.

Answer 1

The formula of dipole moment is as follows.

               ${ \mu  }_{ B }\quad =\quad \delta \quad \times \quad d$

From the given dipole moment of $KCl = 3.336\quad \times \quad { 10 }^{ -29 }$

Substitute the values

               $3.336\quad \times \quad { 10 }^{ -29 }\quad =\quad \delta \quad \times \quad d$

               $\delta \quad =\quad \frac { 3.36\quad \times \quad { 10 }^{ -29 } }{ 2.6\quad \times \quad { 10 }^{ -10 } } \quad =\quad 1.283\quad \times \quad { 10 }^{ -19 }\quad coulomb$

               $1.602\quad \times \quad { 10 }^{ -19 } charge on each, % ionic character = 100$

               $1.283\quad \times \quad { 10 }^{ -19 } charge on each, % ionic character =\quad \frac { 1.283\quad \times \quad { 10 }^{ -19 } }{ 1.602\quad \times \quad { 10 }^{ -19 } } \quad \times \quad 100\quad =\quad 80.09%$

If one unit charge then $\delta \quad =\quad 1.602\quad \times \quad { 10 }^{ -19 }\quad C$

$\mu \quad =\quad 1.602\quad \times \quad { 10 }^{ -19 }\quad \times \quad 2.6\quad \times \quad { 10 }^{ -10 }\quad =\quad 4.1652\quad \times \quad 10^{ -29 }\quad C.m$

Answer 2

The formula of dipole moment is as follows.

From the given dipole moment of KCl = 3.336\quad \times \quad { 10 }^{ -29 }KCl=3.336×10

−29

Substitute the values

3.336\quad \times \quad { 10 }^{ -29 }\quad =\quad \delta \quad \times \quad d3.336×10

−29

=δ×d

\delta \quad =\quad \frac { 3.36\quad \times \quad { 10 }^{ -29 } }{ 2.6\quad \times \quad { 10 }^{ -10 } } \quad =\quad 1.283\quad \times \quad { 10 }^{ -19 }\quad coulombδ=

2.6×10

−10

3.36×10

−29

=1.283×10

−19

coulomb

If one unit charge then \delta \quad =\quad 1.602\quad \times \quad { 10 }^{ -19 }\quad Cδ=1.602×10

−19

C

\mu \quad =\quad 1.602\quad \times \quad { 10 }^{ -19 }\quad \times \quad 2.6\quad \times \quad { 10 }^{ -10 }\quad =\quad 4.1652\quad \times \quad 10^{ -29 }\quad C.mμ=1.602×10

−19

×2.6×10

−10

=4.1652×10

−29

C.m