In the methane molecule, each carbon atom is surrounded by four hydrogen atoms, which form a tetrahedron, with the carbon atom at its center. The bond angle theta is the angle formed by the H-C-H combination. Find the tetrahedron angle theta. Take the C atom to be at (1/2, 1/2, 1/2) and the hydrogen atoms at (1,0,0), (0,1,0), and (0,0,1).

Question

In the methane molecule, each carbon atom is surrounded by four hydrogen atoms, which form a tetrahedron, with the carbon atom at its center. The bond angle theta is the angle formed by the H-C-H combination. Find the tetrahedron angle theta. Take the C atom to be at (1/2, 1/2, 1/2) and the hydrogen atoms at (1,0,0), (0,1,0), and (0,0,1).

Answer 1

As usual, google is a good place to start:

https://www.google.com/search?client=firefox-b-1-d&q=methane+geometry

Answer 2

To find the tetrahedron angle theta, we can use the concept of vector dot product. The dot product of two vectors is given by the formula:

A · B = |A| |B| cos(θ)

Where A and B are vectors, θ is the angle between them, and |A| and |B| are their magnitudes.

In this case, we will consider the positions of the carbon atom and the three hydrogen atoms as vectors. Let's define the three vectors:

Vector CA = Position of Carbon atom - Position of Atom A
Vector CB = Position of Carbon atom - Position of Atom B
Vector CC = Position of Carbon atom - Position of Atom C

Now, let's calculate the dot product between these vectors:

CA · CB = |CA| |CB| cos(θ1)
CA · CC = |CA| |CC| cos(θ2)
CB · CC = |CB| |CC| cos(θ3)

The magnitudes of the vectors can be calculated using the distance formula, which is the square root of the sum of the squares of the differences in coordinates:

|CA| = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
|CB| = sqrt((x3 - x1)^2 + (y3 - y1)^2 + (z3 - z1)^2)
|CC| = sqrt((x4 - x1)^2 + (y4 - y1)^2 + (z4 - z1)^2)

Now, let's substitute the given values into the formulas:

Position of Carbon atom = (1/2, 1/2, 1/2)
Position of Atom A = (1, 0, 0)
Position of Atom B = (0, 1, 0)
Position of Atom C = (0, 0, 1)

|CA| = sqrt((1/2 - 1)^2 + (1/2 - 0)^2 + (1/2 - 0)^2)
|CB| = sqrt((1/2 - 0)^2 + (1/2 - 1)^2 + (1/2 - 0)^2)
|CC| = sqrt((1/2 - 0)^2 + (1/2 - 0)^2 + (1/2 - 1)^2)

Now, calculate these magnitudes:

|CA| = sqrt((1/2 - 1)^2 + (1/2 - 0)^2 + (1/2 - 0)^2)
= sqrt(1/4 + 1/4 + 1/4)
= sqrt(3/4)
= sqrt(3) / 2

|CB| = sqrt((1/2 - 0)^2 + (1/2 - 1)^2 + (1/2 - 0)^2)
= sqrt(1/4 + 1/4 + 1/4)
= sqrt(3/4)
= sqrt(3) / 2

|CC| = sqrt((1/2 - 0)^2 + (1/2 - 0)^2 + (1/2 - 1)^2)
= sqrt(1/4 + 1/4 + 1/4)
= sqrt(3/4)
= sqrt(3) / 2

The magnitude of the vector CA, CB, and CC is sqrt(3) / 2 for all three vectors.

Now, calculate the dot products:

CA · CB = |CA| |CB| cos(θ1)
CC · CB = |CC| |CB| cos(θ2)
CA · CC = |CA| |CC| cos(θ3)

Using the formula for dot product, we rewrite those equations:

CA · CB = (sqrt(3) / 2) * (sqrt(3) / 2) * cos(θ1)
CC · CB = (sqrt(3) / 2) * (sqrt(3) / 2) * cos(θ2)
CA · CC = (sqrt(3) / 2) * (sqrt(3) / 2) * cos(θ3)

Simplifying the equation:

CA · CB = 3/4 cos(θ1)
CC · CB = 3/4 cos(θ2)
CA · CC = 3/4 cos(θ3)

Now, substitute the positions into the equations:

CA · CB = (1/2 * 1/2) + (1/2 * 1/2) + (1/2 * 0) = 1/2
CC · CB = (1/2 * 0) + (1/2 * 1/2) + (1/2 * 1/2) = 1/2
CA · CC = (1/2 * 1/2) + (1/2 * 1/2) + (1/2 * 0) = 1/2

Now, solve for the angles:

1/2 = (3/4) cos(θ1)
1/2 = (3/4) cos(θ2)
1/2 = (3/4) cos(θ3)

Simplifying further:

cos(θ1) = (2/3)
cos(θ2) = (2/3)
cos(θ3) = (2/3)

Finally, to find the tetrahedron angle theta, we take the inverse cosine (arccos) of the values we obtained:

θ1 = arccos(2/3)
θ2 = arccos(2/3)
θ3 = arccos(2/3)

So, the tetrahedron angle theta in the methane molecule is approximately 48.19 degrees.