# How you can explain that C Cl_4 has no dipole moment?

Sep 19, 2017

Well, the molecular dipole moment....

#### Explanation:

....is the vector sum of the individual bond dipoles. There is some difference in negativity between carbon and chlorine, enuff to give a ""^(+delta)C-Cl^(delta-) dipole, but when these dipoles are summed together in a vector fashion, the resultant is ZERO...... On the other hand, chloroform, ${\text{HCCl}}_{3}$, which is more of less identical to carbon tet. geometrically, is a polar molecule, because the vector sum of the individual bond dipoles DOES NOT sum to ZERO.

Mar 5, 2018

See below...

#### Explanation: Let's have a vector approach.

• The angle between two $\text{Cl}$ atom is always color(red)(109^@28'.
• So, we have to take the cos component of the three $\text{Cl}$ atoms showing in the pic downside.
• I mark the upper one as $\left(1\right)$, left one as $\left(2\right)$, bottom one is $\left(3\right)$ and the right one is $\left(4\right)$.

For $\left(2\right) , \left(3\right) , \left(4\right)$ no. chlorine,the resultant dipole moment is

mu_((2-3-4))=3xxmu_1 cdot cos(180^@-109^@28')=3xxmu_1xx0.333=3xxmu_1xx1/3=mu_1" "[${\mu}_{1}$ is generated dipole moment for each $\text{Cl}$ atom.]

Now, the upper $\text{Cl}$ atom's dipole moment and the resultant dipole moment cancel out each other.

Hence,$\textcolor{red}{{\mu}_{{\text{CCl}}_{4}}} = {\mu}_{1} - {\mu}_{1} = 0$

This is the answer proved by vector.