Bond angle in methane? Have been doing for 1 hr

Click For Summary
SUMMARY

The bond angle in methane (CH4) is derived from the geometry of a regular tetrahedron, where each hydrogen atom occupies a corner and the carbon atom is at the center. The angle between two C-H bonds can be calculated using vector mathematics, specifically the formula cosθ = (AxBx + AyBy + AzBz) / (AB), where A and B are the vectors representing the C-H bonds. The magnitudes of these vectors are determined to be √3, leading to the conclusion that the H-C-H bond angle is approximately 109.5 degrees, characteristic of tetrahedral molecular geometry.

PREREQUISITES
  • Understanding of vector mathematics and dot product
  • Familiarity with tetrahedral geometry
  • Basic knowledge of molecular structure and bond angles
  • Ability to visualize three-dimensional shapes, particularly cubes and tetrahedra
NEXT STEPS
  • Study vector operations in three-dimensional space
  • Learn about molecular geometry and hybridization in chemistry
  • Explore the relationship between bond angles and molecular shape
  • Investigate the properties of tetrahedra and their applications in chemistry
USEFUL FOR

Chemistry students, molecular geometry enthusiasts, and anyone studying the structural properties of molecules will benefit from this discussion.

elpermic
Messages
29
Reaction score
0
Bond angle in methane? Have been doing for 1 hr!

Homework Statement


In the methane molecule, CH4, each hydrogen atom is at a corner of a regular tetrahedron with the carbon atom at the center. In coordinates where one of the C-H bonds is in the direction of i + j + k, an adjacent C-H bond is in the i - j - k direction. Calculate the angle between these two bonds?



Homework Equations


I know this is a pyramid.



The Attempt at a Solution

 
Physics news on Phys.org


If A and B are vectors, then the angle between them is given by
cosθ= A.B/A*B
 


Would that mean that cos theta is equal to the (Ax+Bx) + (Ay+By) divided by the magnitudes of A and B? I think that each side of the pyramid is 1. Is this right?
 


elpermic said:
Would that mean that cos theta is equal to the (Ax+Bx) + (Ay+By) divided by the magnitudes of A and B? I think that each side of the pyramid is 1. Is this right?
No.
cosθ = (AxBx + AyBy + AzBz)/AB, where A and B are the magnitudes of the vectors.
 


So how would I use the method of components to find Ax and Bx, and Ay and By?? I never learned how to do Az and Bz either. I only have one fact that I think, each side is 1.
 


You have expressed bonds is in the direction of i + j + k, an adjacent C-H bond is in the i - j - k direction.
Here coefficient of i, j and k are Ax, Ay and Az. i.e. 1, 1, 1.
Similarly find Bx, By and Bz. Magnitudes of A and B are sqrt(3)
Now find the angle using the formula.
 


It might help to know that a tetrahedron can be embedded in a cube. I think it's far more intuitive to obtain the H-C-H angle by examining the cube.

anim5in1.gif


umm. that wasn't supposed to be spinning, but it's the best I could find.
 
Last edited by a moderator:


I have been trying it out and I still cannot do this
 


Consider if the cube is one unit on a side. The blue line at the top of the cube connects two hydrogen atoms. The length of the line is sqrt(2).

The center of this line lies at the center of a cube face. The center of the cube face is 1/2 unit from the center of the cube. The carbon atom is at the center of the cube.

Draw a picture. Does the line from the center of the cube intersect the blue line at a right angle?
 

Similar threads

Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 4 ·
Replies
4
Views
5K