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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?

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    $\begingroup$ Try drawing the three-dimensional structures of each molecule, including lone pairs. $\endgroup$ – Shafter Jul 16 '15 at 8:10
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    $\begingroup$ Bond polarity and molecular polarity are different (though related) concepts. In short the molecular dipole moment is the vector sum of the individual bonds dipole moments, so you have to know the geometry of the molecule to check is it polar or not. See e.g. here. $\endgroup$ – Wildcat Jul 16 '15 at 8:12
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    $\begingroup$ Strongly related: Why is tetrafluoromethane non-polar and fluoroform polar? $\endgroup$ – bon Jul 16 '15 at 8:57
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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.

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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.

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

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

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.

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 for simple applications of this knowledge in reality.

*Molecules were drawn with chemfig package.

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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

Here are some examples which can help :

enter image description here

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

enter image description here

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    $\begingroup$ Hybridisation is not the cause for a molecular structure, it is, - if at all -, the result of it. Better yet, it is a mathematical model to explain orbitals within a molecule. Step 1 therefore does not work. I know that it has often been taught that way, but unfortunately this is wrong. $\endgroup$ – Martin - マーチン Oct 8 '18 at 7:24
  • $\begingroup$ @Martin-マーチン Yes your are correct Hybridization is a hypothetical concept , but while solving problems. ,like mentioned , it is the easiest way to find whether a molecule has dipole moment or not , Its simply not possible to remember all structures and until the structure is known , we will not be able to find the net dipole moment . $\endgroup$ – Chloritone_360 Oct 8 '18 at 8:06
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    $\begingroup$ It is wrong to use hybridisation to determine the structure; even if you think it works and is easy. It fails for hyper-coordinate compounds like $\ce{SF6}$, and everything (else) that involves predominantly ionic bonds. Simpler models to estimate structure exist, are equally (un)reliable, and don't rely on wrong theory. Also note that the model you are proposing fails for $\ce{CO}$. $\endgroup$ – Martin - マーチン Oct 8 '18 at 8:14
  • $\begingroup$ @Martin-マーチン Yes There are many exception I cant diagree . But This is the way I would solve the question that is mentioned. $\endgroup$ – Chloritone_360 Oct 8 '18 at 8:23
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    $\begingroup$ I see there is no reason arguing with you; this discussion is already too long. I just wanted to tell you - as a courtesy - why I down-voted your answer. Regarding Drago's rule, please see What is Drago's rule? Does it really exist? Unfortunately the lone pair repulsion is also a post rationalisation of the failing VSEPR model. The bond angle in $\ce{PCl3}$ is about 100°; which means there is some hybridisation, but far from sp³, too. Anyway, I respectfully bow out of this discussion. $\endgroup$ – Martin - マーチン Oct 8 '18 at 8:56

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