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IB DP Chemistry – S2.2.6 Molecular polarity- Study Notes

IB DP Chemistry - S2.2.6 Molecular polarity- Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – S2.2.6 Molecular polarity- Study Notes – New Syllabus

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Structure 2.2.6 — Molecular Polarity and Net Dipole Moment

Structure 2.2.6 — Molecular Polarity and Net Dipole Moment

Molecular Polarity

A molecule is said to be polar if it has a net dipole moment — that is, an overall separation of positive and negative charges across the molecule.

This depends on two key factors:

  • Bond Polarity – due to electronegativity differences
  • Molecular Geometry – determines whether bond dipoles add up or cancel out

Step-by-Step to Determine Molecular Polarity

Step 1:

  • Determine if each bond is polar
  • → Use electronegativity values to decide if bonds have dipoles.
  • → Represent bond dipoles with arrows: \( \delta^- \rightarrow \delta^+ \)

Step 2:

  • Consider the geometry of the molecule
  • → Use VSEPR to predict the shape.

Step 3:

  • Check vector addition of dipoles
  • → If dipoles cancel out: molecule is non-polar
  • → If dipoles do not cancel: molecule is polar

Examples of Electronegativity Differences:

  • H (2.1), C (2.5), N (3.0), O (3.5), F (4.0), Cl (3.0)

Representing Dipoles

  • Partial Charges: \( \delta^+ \), \( \delta^- \)
  • Vectors: Arrows pointing from positive to negative ends (longer arrow = stronger dipole)

Examples of Polar and Non-Polar Molecules:

   Water, \( \text{H}_2\text{O} \):

Polar bonds + bent shape → polar molecule

Carbon dioxide, \( \text{CO}_2 \):

Polar bonds but linear shape → non-polar (dipoles cancel)

Ammonia, \( \text{NH}_3 \):

Polar bonds + trigonal pyramidal → polar molecule

Boron trifluoride, \( \text{BF}_3 \):

Polar bonds but trigonal planar → non-polar (dipoles cancel)

Example

Determine whether the molecule \( \text{CHCl}_3 \) (chloroform) is polar or non-polar.

▶️Answer/Explanation
  • Step 1: Bond Polarity – C–Cl and C–H bonds are polar due to electronegativity differences (Cl = 3.0, C = 2.5, H = 2.1).
  • Step 2: Molecular Geometry – The molecule is tetrahedral (4 bonding pairs around central carbon).
  • Step 3: Dipole Analysis – 3 Cl atoms are more electronegative than H, creating an asymmetrical electron distribution.

 

Conclusion: Dipoles do not cancel → net dipole → molecule is polar.

Example 

Is \( \text{CO}_2 \) (carbon dioxide) polar or non-polar?

▶️Answer/Explanation
  • Step 1: Bond Polarity – C=O bonds are polar (O = 3.5, C = 2.5, so \( \Delta \chi = 1.0 \)).
  • Step 2: Molecular Geometry – Linear molecule (2 bonding regions around carbon).
  • Step 3: Dipole Analysis – The two dipoles are equal in magnitude and directly opposite in direction.

Conclusion: Dipoles cancel → no net dipole → molecule is non-polar.

Example

Predict the polarity of \( \text{SO}_2 \ ) (sulfur dioxide).

▶️Answer/Explanation
  • Step 1: Bond Polarity – S=O bonds are polar (S = 2.5, O = 3.5).
  • Step 2: Molecular Geometry – Bent shape (due to lone pair on sulfur), VSEPR predicts bond angle around 119°.
  • Step 3: Dipole Analysis – The dipoles do not cancel due to the bent geometry.

Conclusion: Dipoles reinforce each other → molecule is polar.

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