How can I tell if a molecule is polar or not?


> 
> Consider $\ce{PCl3}$, $\ce{CH2Cl2}$, $\ce{HCN}$, and $\ce{SiO2}$. Which of these molecules are polar substances?
> 
> 
> 


I appear to be lacking on the concept of "polar substance". I was just studying about Lewis structures and this question popped up in the book, but I'm not quite sure how is this related.


I think that a chemical bond is polar when one atom is more electronegative than another. But my basic understanding tells me that a chemical bond is non-polar only when the two atoms are the same (like $\ce{F-F}$ or $\ce{O-O}$). But all the above options have different atoms, so the answer would be "none", which is clearly wrong because that's not one of the options.


How can I tell if a molecule is polar or not?




**Every sufficiently asymmetric molecule will be polar, but some more than others**


The polarity of molecules is related to the polarity of bonds within the molecule, but just having polar bonds is not enough to create a polar molecule.
 Consider, for example, CCl4 and CHCl3. Carbon tetrachloride has 4 fairly polar bonds but they form a regular tetrahedron and the polarity of the individual bonds cancel each other out to leave a molecule which has no net dipole moment and is non-polar. But the chloroform molecule (which swaps a hydrogen for a chlorine) is fairly polar (~1.1 Debye–the unit used to measure dipole moments–usually abbreviated to D) because the symmetry is lower and the three C-Cl bonds now all point away from the C-H bond and their dipoles don't cancel out.


Similarly carbon dioxide is a linear molecule so the dipoles of the (fairly polar) C=O bonds point in exactly the opposite direction and so cancel out. Water, however is bent so the dipoles of the polar O-H bonds do not cancel out leaving a polar molecule with a significant dipole moment (~1.8 D).


Saturated hydrocarbons are usually thought to be non-polar for most meaningful purposes. But those that lack symmetry still have measurable (but tiny) dipole moments. Propane $\ce {CH3CH2CH3} $ and isobutane $\ce {CH3CH(CH3)CH3} $, for example, both have small but measurable dipole moments (0.081 D and 0.132 D) so are, strictly, polar. In practice, though, these molecules have such low polarity that the effects of their net dipole are dominated by other non-polarity related intermolecular forces and can be though of as non-polar.


But the critical message is that the most fundamental feature that determines polarity is *symmetry* or the lack of it. 


Also note that we are talking about discrete molecules here. One of your initial examples, $\ce {SiO2}$, is a network solid not a discrete molecule and the ideas won't apply.


 1

*Definition of dipole moment* $$\vec{\mu}=q\times\vec{r}$$ 


where $q$ is the charge of the two atoms involved, $\vec{r}$ the distance between them.


* It is a vector parallel to $\vec{r}$.
* It origins in a separation of two equal in magnitud and opposite
charges,
* Usually: bigger electronegativity difference ($\chi$) between atoms implies bigger charge ($q$) and so bigger $\mu$. [Source.](https://www.chemguide.co.uk/atoms/bonding/electroneg.html)


But both $q$ and $r$ determine the $\mu$-value.


When there are many atoms, and many bonds ($b$) net dipole is obtained by summing up:


$$\mu\_{net}=\sum\_b\vec{\mu\_b}=\sum\_b\,q\_b\times\vec{r\_b}$$


Importantly, to sum up vectors we use bond angles between them. In consequence, molecular geometry is needed.


*Molecule geometry* 


Lewis structures and $\texttt{VSEPR}$ are necessary to predict molecular geometries, and molecular geometry will indicate the angles between dipoles.


One could summarize the previous lines as: $\texttt{VSEPR}$ $\rightarrow$ molecular geometry $\rightarrow$ net dipole moment. 


Let's write two examples:


* Carbon dioxide


[![co2](https://i.stack.imgur.com/rXGad.png)](https://i.stack.imgur.com/rXGad.png)


We can see dipoles are on the same line ($x$), and opposite direction. The sum is zero: $\vec{\mu}\_{net}=\mu\_x - \mu\_x = 0$.


* Water


[![water](https://i.stack.imgur.com/xJgpC.png)](https://i.stack.imgur.com/xJgpC.png)


In the $\ce{H2O}$ example, there is a net dipole. If angles are measured from $x$ axis, which bisects the bond angle, and contributions are added it yields:


$\vec{\mu}\_{net}=2\,q\, r\times cos(52.25)\vec{i}$


Only components on $x$ axis survive. For a detailed explanation see [this video.](https://www.youtube.com/watch?v=tnT5hNMSm8A)


Next step could be $\ce{SO2}$, which has a net dipole smaller than water. Same analysis gives zero dipole moment for $\ce{BF3}$. 




---


*Applications*


See [this post](https://chemistry.stackexchange.com/questions/65744/how-does-one-apply-the-knowledge-of-the-direction-of-a-molecular-dipole-moment/74000#74000) for simple applications of this knowledge in reality.


\*Molecules were drawn with `chemfig` package.


 0

To find answer to the question that you encountered you can follow these steps


Step 1 : 1st you have to understand what is hybridization and how to find hybridization of molecules , also each hybridization has different structure .


Step 2 : Once you have determined the structure successfully you can use the formula for calculating dipole moment 


μ = q\*d and this formula shows dipole moment for a single bond .


A more generalised form :


[![enter image description here](https://i.stack.imgur.com/Bvois.jpg)](https://i.stack.imgur.com/Bvois.jpg)


Here are some examples which can help :


[![enter image description here](https://i.stack.imgur.com/C8122.jpg)](https://i.stack.imgur.com/C8122.jpg)


For multiple bonds use concept of vector addition/subtraction.


Remember that polar compounds means the compounds that have a net dipole moment , but Polar Solvent means that the compound has high dielectric constant .
[![enter image description here](https://i.stack.imgur.com/GAvPP.jpg)](https://i.stack.imgur.com/GAvPP.jpg)


[![enter image description here](https://i.stack.imgur.com/hmXwk.jpg)](https://i.stack.imgur.com/hmXwk.jpg)


 -1