In chemistry, a molecule is considered non-polar when its positive charge center and negative charge — Physical Chemistry Chemistry Question
Polar and non-polar molecules
In chemistry, a molecule is considered non-polar when its positive charge center and negative charge center coincide, i.e. the charge distribution is symmetrical in the molecule.
On the other hand, when a molecule has two distinct centers for positive and negative charges, it is considered polar.
This charge distribution property is measured by a quantity called the dipole moment which is defined as the magnitude of the charge q and the distance l between the charges:
µ = q l
The dipole moment is a vector pointing from the positive charge center to the negative one.
The dipole moment is often expressed in debyes (D). The relationship between debyes (D) and coulomb meters (C·m) in SI units is as follows: 1 D = 3.33 × 10–30 C·m.
I. The dipole moment is closely related to the molecular geometry. In order to calculate the net dipole moment µ of multi-atomic molecules, we can add the dipole moment vectors for individual bonds. In this case, an individual bond is considered to have its own dipole moment called the bond moment. For a non-linear molecule with three atoms, ABC, the net dipole moment µ can be calculated by adding vectors in which µ1 and µ2 are the bond moments for AB and AC bonds, and α is the bond angle.
1.1 Determine the general equation for calculating the net dipole moment.
Model Answer
I.
1.1 The net dipole moment µ is calculated as follows:
µ^2 = µ1^2 + µ2^2 + 2µ1µ2cosα (1)
II. The directions of the individual bond moments should be considered. The molecule of CO2 is linear.
1.2 Calculate the net dipole moment µ of the molecule.
Model Answer
II.
1.2 The geometry of CO2: O C O
Because two bond moments of µCO have opposite directions and cancel each other out, the net dipole moment for CO2 is zero. Therefore:
µCO2 = 0
1.3 A non-linear molecule of A2B such as H2S has the net dipole moment µ ≠ 0. Determine µ for H2S if µSH = 2.61 ⋅ 10–30 C·m and the bond angle α = 92.0 °.
Model Answer
1.3 The geometry of H2S:
From the general equation (1),
µ^2_H2S = µ^2_HS + µ^2_HS + 2µ_HSµ_HScosα = 2µ^2_HS(1+cosα) = 4µ^2_HScos^2(α/2)
µ_H2S = 2µ_HScos(α/2)
Therefore, µ_H2S = 2 × (2.61 ⋅ 10^-30 / 3.33 ⋅ 10^-30) × cos(92/2) = 1.09 D
III. The bond angle HCH in the formaldehyde molecule is determined experimentally to be approximately 120o; the bond moments for C–H and C–O bonds are µC-H = 0.4 D and µC=O = 2.3 D, respectively.
1.4 Determine the orbital hybridization of C and O atoms, and plot the overlaps of orbitals in the formaldehyde molecule.
Model Answer
III.
1.4
1.5 Calculate the net dipole moment (µ) of the formaldehyde (D), given the order of the electronegativity as O > C > H. (Hints: Electronegativity is the ability of an atom in a molecule to attract shared electrons to itself).
Model Answer
1.5 Since χO > χC > χH, µC-H has the direction showed in the above plot, and
µ_HCH = 2µ_C-H cos(120/2) = 2 × 0.4 × 0.5 = 0.4 D
µ_C=O also has the direction toward the O atom. Therefore, the net dipole moment µ of the molecule is
µ = µ_HCH + µ_C=O = 0.4 + 2.3 = 2.7 D
IV. The dipole moments of water and dimethylether in gaseous state are determined as 1.84 D, and 1.29 D, respectively. The bond angle formed by two bond moments of O–H in the water molecule is 105o. The bond angle formed by two bond moments of O–C in the ether molecule is 110o.
1.6 Estimate the bond angle formed by the bond moments of O–H and C–O in the methanol molecule, given that the dipole moment of methanol molecule is 1.69 D. Assume that individual bond moments are unchanged in different molecules.
Model Answer
IV.
1.6 We can plot the geometry of the three molecules involved in this problem in the following scheme:
- The dipole moment µ is a vector which can be calculated by adding individual bond moments µ1 and µ2
µ^2 = µ1^2 + µ2^2 + 2µ1µ2cosα (1)
α is the angle formed by the individual bond moments.
- The dipole moment µ in the water molecule with the bond angle α formed by the two bond moments of O-H can be calculated as follows.
From equation (1) we have
µ_H2O = [4µ^2_OH cos^2(α/2)]^1/2 = 2µ_OHcos(α/2)
Given α = 105o, the bond moment µOH in water can be calculated:
1.84 = 2µ_OHcos(105/2) → µ_OH = 1.51 D
Similarly, we can calculate the bond moment for O-CH3 in dimethylether:
1.29 = 2µ_OCH3cos(110/2) → µ_OCH3 = 1.12 D
- In methanol, the individual bond moments are given as µ1 = µ_OH and µ2 = µ_OCH3 as in water and dimethylether. The bond angle α is formed by the two individual bond moments.
From equation (1), cosα is:
cosα = (µ^2 – µ1^2 – µ2^2) / (2µ1µ2) = (1.69^2 – 1.51^2 – 1.12^2) / (2 × 1.51 × 1.12) = -0.198
→ α = 101o 57'
Therefore, the bond angle C – O – H in methanol is of 101o 57'