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Bond vectors add — geometry decides polarity — interactive Chemistry simulation for IIT-JEE.

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Concept

Each polar bond is a vector (points − → + by chemistry convention here: toward the more electronegative atom). The molecule's dipole is the vector sum — so geometry rules polarity: bent H₂O (1.85 D) is polar while linear CO₂ (0 D) is not, despite C=O being MORE polar than O–H.

Key formula

μres=μ12+μ22+2μ1μ2cos⁡θ,μ=q×d  (1 D=3.34×10−30 C m)\mu_{res} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}, \qquad \mu = q \times d \;(1\,D = 3.34\times10^{-30}\,C\,m)

Derivation

Standard vector addition with the included angle θ. Equal bonds: μ_res = 2μcos(θ/2).

θ = 180° → zero (CO₂, BeCl₂); θ = 104.5° → strong resultant (H₂O). Symmetric shapes (tetrahedral CCl₄, trigonal BF₃, square-planar XeF₄) cancel completely regardless of bond polarity.

Scenarios to explore

  • Dipole Moments — Bond vectors add — geometry decides molecular polarity.

Real-world applications

  • Microwave heating needs polar molecules (water's μ).
  • Miscibility: like dissolves like — polarity matching.
  • Distinguishing cis (polar) from trans (often μ = 0) alkenes experimentally.

JEE exam tips

  • Dipole order o-dichlorobenzene > m- > p- (0) follows 2μcos(θ/2) with θ = 60°, 120°, 180°.
  • % ionic character ≈ μ_observed/μ_ionic × 100 (HCl ≈ 17%).
  • trans-2-butene μ ≈ 0; cis > 0 — geometry fingerprint.

Common mistakes

  • Polar bonds ⇒ polar molecule (false for symmetric shapes).
  • Ignoring lone pairs: NH₃ (1.47 D) vs NF₃ (0.23 D) — lone pair adds in NH₃, opposes in NF₃.
  • Adding magnitudes instead of vectors.

Exam traps to avoid

  • NH₃ > NF₃ dipole despite N–F being more polar — the lone-pair direction question.
  • CO₂ has polar bonds AND zero dipole; O₃ has 'non-polar' O–O bonds AND non-zero dipole (formal charges).