Q2(a). Dipole moments of NOF and NO₂F; resonance in NO₂F
Of the two compounds NOF and NO₂F, one has a resultant dipole moment of μ=1.81 D, and the other, μ=0.47 D. Which dipole moment do you predict for each compound? Explain.
Also write resonance structures of NO₂F and identify the dominant structure.
Answer in brief
NOF: 1.81 D; NO₂F: 0.47 D. In bent NOF the bond moments and the N lone pair reinforce; in planar NO₂F the two N–O moments oppose the N–F moment. The dominant forms of NO₂F are the two equivalent structures F–N⁺(=O)–O⁻.
Key points
- O=N–F, N has one lone pair and two bonding regions, so the molecule is bent, like NO₂⁻.
- N has three bonding regions and no lone pair: trigonal planar, isoelectronic with NO₃⁻ and BF₃-shaped, with one O replaced by F.
- the N=O and N–F bond moments point towards O and F, both on the same side of N. Their vector sum is large and the lone pair adds its own moment; nothing cancels.
- the two N–O bond moments lie at about 120° and their resultant points along the C₂ axis away from F; the N–F moment points along the same axis towards F. The two nearly cancel, leaving a small net moment.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Shapes (VSEPR)
- NOF (18 valence electrons): O=N–F, N has one lone pair and two bonding regions, so the molecule is bent, like NO₂⁻.
- NO₂F (24 valence electrons): N has three bonding regions and no lone pair: trigonal planar, isoelectronic with NO₃⁻ and BF₃-shaped, with one O replaced by F.
Assigning the moments
- NOF, 1.81 D: the N=O and N–F bond moments point towards O and F, both on the same side of N. Their vector sum is large and the lone pair adds its own moment; nothing cancels.
- NO₂F, 0.47 D: the two N–O bond moments lie at about 120° and their resultant points along the C₂ axis away from F; the N–F moment points along the same axis towards F. The two nearly cancel, leaving a small net moment.
- The cancellation is incomplete because N–O and N–F bonds differ in polarity and length; it would be exact only if all three bonds were alike, as in BF₃ or NO₃⁻. The planar symmetry, not the size of any one bond moment, decides the low value.
Resonance structures of NO₂F
COUNT24 valence electronsRULESoctets, fewest charges, - on ODOMINANTI and II (equal)
- I and II: F–N single, one N=O and one N–O⁻; formal charges N +1, single-bonded O -1, F 0. Every atom has an octet and the negative charge sits on the electronegative O. These two equivalent forms are dominant; their average gives two equal N–O bonds (order 1.5).
- III: F=N⁺ with both O single-bonded: three formal charges and a positive charge on F, the most electronegative element. Minor.
- IV: no-bond ionic form F⁻ [O=N=O]⁺ (nitronium fluoride). It contributes a little and explains why the N–F bond is long and weak, and why NO₂F acts as a fluoride donor towards Lewis acids, giving NO₂⁺ salts.
Application. the same vector-sum argument fixes the shape of a molecule from its dipole moment: μ=0 for BF₃, CO₂ and SF₆ shows symmetric shapes, while a non-zero μ for SO₂, H₂O and NOF shows bent ones.
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All parts of Question 2
- 2(a) Dipole moments of NOF and NO₂F; resonance in NO₂F20 marks
- 2(b) Expansion of an ideal gas against a constant pressure of 1 atm20 marks
- 2(c) Preparation of XeO₃; reactions of HXeO₄⁻ and Ba₂XeO₆10 marks