How Do You Calculate the Dipole Moment of CO2 with a 120 Degree Bond Angle?

[vaazu]

Homework Statement

r-bond length (in anströms)
q-charge (in debyes)
CO2 r(CO)=1.1867 q(C)=1.1352 q(O)=-0.5676
angle between them is 120 deg

Homework Equations

The Attempt at a Solution

I know I have to make vectors and then add them, but I don't know how to make the vectors. I know the angle in CO2 is 180, just need to find the dipole moment if the angle was 120.

 

[chemisttree]

Try http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/vectors/u3l1b.html" site.

 

[Blueshift5]

To calculate the dipole moment of a molecule, we need to find the vector sum of the individual bond dipole moments. The bond dipole moment is calculated by multiplying the bond length (r) by the charge (q) on each atom. In this case, we have two bond dipole moments, one for the C-O bond and one for the O-O bond.

For the C-O bond, the bond length (r) is given as 1.1867 angstroms and the charge (q) is 1.1352 debyes. The bond dipole moment for this bond can be calculated as:

μ(C-O) = r(C-O) * q(C) = 1.1867 * 1.1352 = 1.3473 debyes

For the O-O bond, the bond length (r) is the same as the C-O bond, but the charge (q) is different. The bond dipole moment for this bond can be calculated as:

μ(O-O) = r(O-O) * q(O) = 1.1867 * (-0.5676) = -0.6737 debyes

To find the total dipole moment of the molecule, we need to add these two bond dipole moments together. Since the angle between the two bond dipole moments is 120 degrees, we can use the formula:

μ = √(μ(C-O)^2 + μ(O-O)^2 + 2μ(C-O)μ(O-O)cosθ)

Where θ is the angle between the two bond dipole moments, which is 120 degrees in this case.

Plugging in the values, we get:

μ = √((1.3473)^2 + (-0.6737)^2 + 2(1.3473)(-0.6737)cos120)

μ = √(1.8175 + 0.4544 - 1.3969) = 0.9148 debyes

Therefore, the dipole moment of CO2 with a bond angle of 120 degrees is 0.9148 debyes.