Chemistry Figuring molecule shapes and polarity

AI Thread Summary
To determine if a molecule has a trigonal planar shape, it must have three bonding pairs of electrons and no lone pairs, following the AX3 notation, as seen in BF3. Polarity is influenced not only by electronegativity differences but also by molecular geometry, which can lead to nonpolar molecules despite polar bonds, as illustrated by CO and CO2. Lone pairs can significantly affect molecular geometry, exemplified by XeF4, which is octahedral rather than tetrahedral due to its electron configuration and lone pairs. Additionally, cis- molecules are generally polar, while trans- molecules tend to be nonpolar due to their symmetrical arrangements. Understanding these concepts is crucial for mastering molecular shapes and polarity for exams.
touma
Messages
18
Reaction score
0
I am having a lot of trouble understanding the following:

1) How do I determine if the shape is a trogonal planar? Is it because it would have more bonds than the tetrahedral?

2)Is figuring the polarity of a molecule only based on the electronegativity?

Please help! I have a big test tomarrow and I need ot understand this material!
 
Physics news on Phys.org
Seriously, it would be amazing if someone could help!
 
1) How do I determine if the shape is a trogonal planar? Is it because it would have more bonds than the tetrahedral?
Trigonal planar structures follow the form of AX3, they have 3 bonding pairs of electrons and no non-bondin pairs. An example would be BF3.

2)Is figuring the polarity of a molecule only based on the electronegativity?
No, it is also based on geometry. There, are instances when the atoms in molecules have large enough electronegativity differences to make them polar, but due to their geometry, it all cancels out and it is non polar. A good example of this would be Carbon Monoxide and Carbon Dioxide. CO is polar due to the electronegativity differece between C and O, but in CO2, since the Oxygen atoms are located opposite each other, the polarities cancel out and it is nonpolar.
 
Great! That was all I needed! Thanks!
 
Just to add on, non-bonding or lone pairs can also affect the geometry of the molecule. Take for instance XeF4. While one would initially think it would be tetrahedral, but when one counts up the electrons (36 in this case) and places them on Xenon (it can expand its octet, since it is period 3 or beyond). In actuality, this molecule is octahedral. The Fluorine atoms take on equatorial positions and the two lone pairs take up the axial positions. The bond angles for this one, since it is octahedral, are 90 degrees instead of 109.5 for tetrahedral.

In terms of AXn notation, XeF4 is AX4, but the hybridization is sp3d2 due to the lone pairs.

Also, just in case you didn't know, cis- molecules tend to be polar compared to trans- molecules. If you draw out a molecule that has a cis- or trans- distinction, you'll notice that there is a net dipole moment on the cis- molecule, as the placement of the atoms in the trans- molecule cancel any charge out.

Good luck for your exam. Hope that helps :)
 
Last edited:
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

Similar threads

Replies
6
Views
5K
Replies
34
Views
4K
Replies
5
Views
2K
Replies
3
Views
1K
Replies
2
Views
3K
Replies
2
Views
2K
Replies
6
Views
7K
Replies
2
Views
1K
Back
Top