Why is CH4 Nonpolar but CH2Cl2 is polar?

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Solution 1

There are two reasons why dichloromethane is polar and the tetrahedral shape is only one

The reasons why any molecule is polar–which is usual chemistry talk means that the molecule has a dipole moment–is that the individual dipole moments of its constituent bonds don't balance out. That means that we need to know both the polarity of the constituent bonds and the 3D shape of the molecule.

All the bonds in methane are the same (and are fairly non-polar to start with). So, even if methane were flat with bonds forming a square with carbon at the centre, it would be non-polar. It isn't flat: it is tetrahedral but, since all the bonds are the same, even a tiny dipole on each would still cancel out. carbon tetrachloride has 4 polar C-Cl bonds but is still non-polar as those dipoles cancel out exactly.

Dichloromethane, were it a flat molecule, could be either polar or non-polar (the dipoles of C-Cl bonds on opposite sides would exactly cancel out, but on the same side would not, obviously). But it isn't that shape, it is tetrahedral. But there is no way to organise two C-Cl bonds in a tetrahedral structure so their dipoles balance out. Build a 3D model if this isn't obvious.

So you need to know two facts to judge molecular polarity: the bonds making up the molecule and the shape of the molecule. Methane isn't non-polar because of its tetrahedral shape: it is non-polar because all the bonds are the same. Dichloromethane is polar because it has different polarity bonds and its shape cannot arrange those bond dipoles to cancel out.

Solution 2

Note that polarity can be considered for the whole molecule, functional group or particular bonds.E.g. $\ce{CO2}$ has zero permanent dipole moment, as bond dipoles cancel each other. But the bonds themselves are polar enough due their dipole moment, being able to involve the molecule in polar intermolecular interactions.

Bonds of both $\ce{CH4}$ and $\ce{CH2Cl2}$ have small dipole moments ( $\ce{C-Cl}$ bigger than $\ce{C-H}$ ). But $\ce{CH2Cl2}$ have nonzero dipole moment as a molecule, having its bond dipoles oriented asymetrically. $\ce{CH4}$ has dipole zero moment due molecule symmetry. Even though, $\ce{CH2Cl2}$ is rather somewhat polar than polar ( like e.g. methanol or water)

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Updated on January 30, 2023

Comments

  • Epistrophe
    Epistrophe 9 months

    I keep reading that the reason why $\ce{CH2Cl2}$ is polar because due to its tetrahedral shape, the dipoles can not cancel each other out but doesn't $\ce{CH4}$ also have tetrahedral shape too? I assumed, since the reason for $\ce{CH2Cl2}$ being polar is apparently due to the 109.5° angles from the tetrahedral shape, that this means that the surrounding atoms in all tetrahedral-shaped compounds are not placed directly opposite of each other and that is why the dipoles can't cancel each other out. Am I understanding this wrong? Or if this is the case, then shouldn't the dipoles of the $\ce{H}$ atoms in $\ce{CH4}$ not be able to cancel each other out either because they aren't directly opposite of each other? Is it different because all the surrounding atoms in this case would have the same dipole?

  • matt_black
    matt_black over 3 years
    IT is not correct to describe carbon dioxide as "polar" in almost any sense in chemistry. Its use as an example here is both wrong and deeply misleading.
  • Poutnik
    Poutnik over 3 years
    OK, it may be better to describe it as non polar with polar bonds :-)