Electron domains · Electron geometry · Molecular geometry & bond angles · 3-D redraw · Bond polarity · Molecular polarity · The 9/30 worksheet — 59 cards
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Valence Shell Electron Pair Repulsion.
Electron pairs around the central atom repel each other, so they spread out as far apart as possible. That arrangement sets the shape of the molecule (NCERT §4.4).
A region of electron density around the central atom. Each bonded atom counts as 1 (single, double or triple bond), and each lone pair on the central atom counts as 1.
One. NCERT: a multiple bond is treated like a single "super pair".
No. Only the lone pairs on the central atom affect the shape.
2: two double bonds, and each counts once.
4: 3 bonded H atoms + 1 lone pair.
3: 2 bonded O atoms + 1 lone pair.
No. VSEPR works on any one resonance structure, and they all give the same number of domains (NCERT §4.4). HCO₂⁻ and NO₂⁻ each have 2 resonance forms and 1 shape.
lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
A bonding pair is held between two nuclei. A lone pair belongs only to the central atom, so it spreads out and takes more space.
Linear, 180°.
Trigonal planar, 120° (360° ÷ 3, all in one plane).
Tetrahedral, 109.5°.
In 3-D, four pairs can spread farther than a flat "+". Pointing them to alternate corners of a cube gives cos θ = (1 − 1 − 1)/3 = −1/3, so θ = 109.5°.
Electron geometry includes every domain, lone pairs too. Molecular geometry describes only where the atoms are.
When the central atom has no lone pairs.
Trigonal bipyramidal (90° and 120°) and octahedral (90°), as in PCl₅ and SF₆. They need an expanded octet.
Linear, 180°: CO₂, HCN.
Trigonal planar, 120°: BF₃, SO₃, HCO₂⁻.
Bent, slightly less than 120°: SO₂ (119.5°), NO₂⁻, NOCl.
Tetrahedral, 109.5°: CH₄, CCl₄, CF₂Cl₂.
Trigonal pyramidal, ≈107° (less than 109.5°): NH₃, PBr₃.
Bent, ≈104.5° (less than 109.5°): H₂O.
The lone pair on N repels more strongly than the bonding pairs, so it squeezes the H–N–H angle below 109.5° (NCERT Table 4.8).
Water has two lone pairs, and lone pair–lone pair repulsion is the strongest, so the bonds are squeezed further.
3 domains with 1 lone pair (less than 120°, e.g. SO₂), and 4 domains with 2 lone pairs (≈104.5°, e.g. H₂O). Write the electron geometry too, so it is clear which one you mean.
A = the central atom, X₃ = 3 bonded atoms, E = 1 lone pair. That is trigonal pyramidal, e.g. NH₃ (NCERT Table 4.7).
The bond lies in the plane of the page.
The bond comes out of the page, toward you.
The bond goes back, away from you.
2 plain lines in the page, 1 wedge and 1 dash. In a tetrahedron every corner is equivalent, so it does not matter which atom goes where.
The lone pair on top of P; below it, 1 plain line, 1 wedge and 1 dash to the three Br atoms.
Linear, trigonal planar and bent: all their atoms lie in one plane.
When the two atoms have different electronegativities. The more electronegative atom pulls the shared pair closer and becomes δ−; the other becomes δ+.
ΔEN less than 0.5: nonpolar covalent · 0.5 to 1.7: polar covalent · more than 1.7: mostly ionic.
ΔEN = 2.5 − 2.1 = 0.4, so it is nonpolar.
ΔEN = 2.8 − 2.1 = 0.7, so it is polar, with Br δ−.
ΔEN = 4.0 − 2.5 = 1.5, so it is polar.
ΔEN = 3.0 − 3.0 = 0, so it is nonpolar (with NCERT values).
µ = Q × r: charge times the distance between the centres of + and − charge. Its unit is the debye: 1 D = 3.33564 × 10⁻³⁰ C m. It is a vector (NCERT §4.3.6).
As a crossed arrow: the cross (+) at the δ+ end and the arrowhead at the δ− end. It points toward the more electronegative atom.
The vector sum of its bond dipoles (NCERT §4.3.6). If they cancel, it is nonpolar; if not, it is polar. So the shape decides.
Yes, if the shape is symmetric so the dipoles cancel: CO₂, BF₃, CCl₄.
It is linear, so the two equal C=O dipoles point in opposite directions: µ − µ = 0 (Table 4.5: 0 D).
It is bent, so the two O–H dipoles add instead of cancelling. The net is 2µ cos(104.5°/2) = 1.85 D.
NF₃. It is trigonal pyramidal, with a lone pair. BF₃ is trigonal planar and symmetric, so its dipoles cancel.
Swapping one Cl for H breaks the symmetry, so the dipoles no longer cancel: 0 D vs 1.04 D (NCERT Table 4.5).
1. Any polar bonds? If not, it is nonpolar. 2. Any lone pairs on the central atom? If so, it is polar. 3. Are all the outer atoms identical? If so, it is nonpolar; if not, it is polar.
Yes. It is linear, but the two ends are different (C–H is nonpolar and C≡N is polar), so nothing cancels the C≡N dipole.
No. Its C–H bonds are nonpolar (ΔEN 0.4), and it is symmetric anyway.
Yes. Use the same test: NO₂⁻ is bent, so its bond dipoles don't cancel, and it is polar. CO₃²⁻ and NH₄⁺ are symmetric, so they are nonpolar.
26 e⁻ · P with 3 bonds + 1 lone pair · 4 domains, tetrahedral · trigonal pyramidal, ≈107° · P–Br ΔEN 0.7, polar · polar molecule (lone pair).
18 e⁻ · [H–C(=O)–O]⁻ with 2 resonance forms · 3 domains, trigonal planar · trigonal planar, 120° · C–H nonpolar (0.4), C–O polar (1.0) · polar (H ≠ O).
18 e⁻ · Cl–N=O, with 1 lone pair on N · 3 domains, trigonal planar · bent, less than 120° · N–Cl nonpolar (0.0), N=O polar (0.5) · polar.
18 e⁻ · [O=N–O]⁻, 2 resonance forms, 1 lone pair on N · 3 domains, trigonal planar · bent, less than 120° · N–O polar (0.5) · polar.
32 e⁻ · C with 4 single bonds, no lone pair · 4 domains, tetrahedral · tetrahedral, 109.5° · C–F polar (1.5), C–Cl polar (0.5) · polar (F ≠ Cl).
NO₂⁻ has a lone pair on N (3 domains, 2 atoms), so it is bent. HCO₂⁻ has none on C (3 domains, 3 atoms), so it is trigonal planar. Count the domains every time.
With N=O every formal charge is 0. A Cl=N bond would put +1 on Cl and −1 on O, which is worse.
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