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Concept 02 / 07Chemical Bonding and Molecular Structure

VSEPR Theory and the Shapes of Molecules

Valence shell electron pair repulsion: counting steric number, the order of repulsions lp-lp > lp-bp > bp-bp, electron geometry versus molecular shape for 2 to 6 electron domains, effects of lone pairs and electronegativity on bond angles, and standard examples from BeCl2 to IF7.

Figure 1
The valence shell electron pair repulsion (VSEPR) theory predicts molecular shapes from a simple idea: electron pairs around a central atom repel each other and arrange themselves as far apart as possible. Procedure. 1. Find the number of electron domains (steric number) around the central atom: SN=\text{SN} = bonded atoms ++ lone pairs. A quick formula: lone pairs =12(V−valency used±charge)= \frac{1}{2}(V - \text{valency used} \pm \text{charge}), where VV is the central atom's valence electrons. Equivalently, SN=12(V+monovalent atoms−charge for cations+charge for anions)\text{SN} = \frac{1}{2}(V + \text{monovalent atoms} - \text{charge for cations} + \text{charge for anions}). A multiple bond counts as one domain. 2. The domains adopt the electron geometry for that number: 22 linear (180∘180^{\circ}), 33 trigonal planar (120∘120^{\circ}), 44 tetrahedral (109.5∘109.5^{\circ}), 55 trigonal bipyramidal (90∘90^{\circ} and 120∘120^{\circ}), 66 octahedral (90∘90^{\circ}), 77 pentagonal bipyramidal. 3. The molecular shape describes only the positions of atoms, ignoring lone pairs. Repulsion order: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. Lone pairs are held by only one nucleus and spread out more, so they compress bond angles: CH4\mathrm{CH_{4}} 109.5∘109.5^{\circ}, NH3\mathrm{NH_{3}} 107∘107^{\circ} (one lone pair), H2O\mathrm{H_{2}O} 104.5∘104.5^{\circ} (two lone pairs). Standard shapes (see figure): SN 2: BeCl2\mathrm{BeCl_{2}}, CO2\mathrm{CO_{2}} linear. SN 3: BF3\mathrm{BF_{3}} trigonal planar; SO2\mathrm{SO_{2}}, O3\mathrm{O_{3}} bent (one lone pair). SN 4: CH4\mathrm{CH_{4}} tetrahedral; NH3\mathrm{NH_{3}} trigonal pyramidal (1 lp); H2O\mathrm{H_{2}O} bent (2 lp). SN 5: PCl5\mathrm{PCl_{5}} trigonal bipyramidal; SF4\mathrm{SF_{4}} see-saw (1 lp); ClF3\mathrm{ClF_{3}} T-shaped (2 lp); XeF2\mathrm{XeF_{2}}, I3−\mathrm{I_{3}^{-}} linear (3 lp). Lone pairs occupy equatorial positions, where they have fewer 90∘90^{\circ} repulsions. SN 6: SF6\mathrm{SF_{6}} octahedral; BrF5\mathrm{BrF_{5}} square pyramidal (1 lp); XeF4\mathrm{XeF_{4}} square planar (2 lp, trans to each other). SN 7: IF7\mathrm{IF_{7}} pentagonal bipyramidal. In the trigonal bipyramid, axial bonds are longer than equatorial ones in PCl5\mathrm{PCl_{5}}, because axial bonds suffer three 90∘90^{\circ} repulsions while equatorial bonds suffer two. Effect of electronegativity on bond angles. With the same central atom, more electronegative surrounding atoms pull bonding pairs away from the centre, reducing bond-pair repulsion and the bond angle: NH3\mathrm{NH_{3}} (107∘107^{\circ}) > NF3\mathrm{NF_{3}} (102∘102^{\circ}). With the same surrounding atoms, a more electronegative central atom pulls bonding pairs in, increasing repulsion: NH3\mathrm{NH_{3}} (107∘107^{\circ}) > PH3\mathrm{PH_{3}} (93.5∘93.5^{\circ}) > AsH3\mathrm{AsH_{3}} > SbH3\mathrm{SbH_{3}}. (For PH3\mathrm{PH_{3}} and heavier hydrides, the near-90∘90^{\circ} angles also indicate little hybridisation.) Multiple bonds, with more electron density, repel more than single bonds: in HCHO\mathrm{HCHO} the H-C-H angle (116∘116^{\circ}) is less than 120∘120^{\circ}. Key Takeaways: - Steric number == bonded atoms ++ lone pairs; multiple bonds count once. - lp-lp > lp-bp > bp-bp repulsion; lone pairs compress bond angles. - SN 5: lone pairs go equatorial (see-saw, T-shape, linear). SN 6: two lone pairs go trans (square planar). - NH3>NF3\mathrm{NH_{3}} > \mathrm{NF_{3}} and NH3>PH3\mathrm{NH_{3}} > \mathrm{PH_{3}} in bond angle. - XeF2\mathrm{XeF_{2}} linear, XeF4\mathrm{XeF_{4}} square planar, XeF6\mathrm{XeF_{6}} distorted octahedral. - Pitfall: naming the electron geometry (tetrahedral) as the shape of H2O\mathrm{H_{2}O} (bent). Brain-Teaser 1: Predict the shapes of ClF3\mathrm{ClF_{3}}, SF4\mathrm{SF_{4}} and I3−\mathrm{I_{3}^{-}}. Solution: ClF3\mathrm{ClF_{3}}: Cl has 77 valence electrons, uses 33 in bonds, leaving 44 = 22 lone pairs; SN 55 with 22 lp: T-shaped. SF4\mathrm{SF_{4}}: S has 66, uses 44, 11 lp; SN 55: see-saw. I3−\mathrm{I_{3}^{-}}: central I has 7+1=87 + 1 = 8, uses 22, leaving 33 lp; SN 55 with 33 equatorial lone pairs: linear. The same electron geometry gives three different shapes as lone pairs replace atoms. Brain-Teaser 2: Arrange H2O\mathrm{H_{2}O}, H2S\mathrm{H_{2}S}, NH3\mathrm{NH_{3}}, CH4\mathrm{CH_{4}} in order of bond angle. Solution: CH4\mathrm{CH_{4}} (109.5∘109.5^{\circ}) > NH3\mathrm{NH_{3}} (107∘107^{\circ}) > H2O\mathrm{H_{2}O} (104.5∘104.5^{\circ}) > H2S\mathrm{H_{2}S} (92∘92^{\circ}). Lone pairs reduce the angle from methane to water; the less electronegative, larger sulfur holds its bonding pairs farther out, so bond-pair repulsion is weak and the angle is close to 90∘90^{\circ}. Brain-Teaser 3: Why is XeF4\mathrm{XeF_{4}} square planar and not tetrahedral? Solution: Xe has 88 valence electrons; four bond to F, leaving 44 electrons = 22 lone pairs. SN =6= 6, so the electron geometry is octahedral. The two lone pairs take opposite (trans) positions to minimise lp-lp repulsion (180∘180^{\circ} apart instead of 90∘90^{\circ}), leaving the four F atoms in a square plane. A tetrahedral shape would require SN 44, which does not match the electron count.

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VSEPR Theory and the Shapes of Molecules — DPP

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