Second period non-metals (row 2)
\text{B} in Group 3A ⇒ 3 valence e⁻
\text{C} (4A) ⇒ 4 valence e⁻
\text{N} (5A) ⇒ 5 valence e⁻
\text{O} (6A) ⇒ 6 valence e⁻
\text{F} (7A) ⇒ 7 valence e⁻
Below-row analogues share the same counts (e.g.
\text{Cl}\,, \text{Br} also 7 e⁻; \text{Be}=2, \text{Li}=1).
Non-metals seek an octet (8 e⁻) by acquiring electrons.
\text{F} needs 1 ⇒ likes 1 bond
\text{O} needs 2 ⇒ likes 2 bonds
\text{N} needs 3 ⇒ likes 3 bonds
\text{C} needs 4 ⇒ likes 4 bonds
Metals (left side) prefer to lose e⁻.
\text{B} (3 e⁻) gives 3 ⇒ still 3 bonds but often incomplete octet.
\text{Be} gives 2 ⇒ 2 bonds
\text{Li} gives 1 ⇒ 1 ionic bond
Energy level 3 has subshells 3s, 3p, 3d (capacity 2+6+10=18 e⁻).
\text{P, S, Cl} may exceed 8 e⁻ (≥ 5 or 6 bonds).
e.g. \text{SF}6 (6 bonds), \text{PCl}5 (5 bonds), \text{ClO}_4^- (7 bonds overall on Cl).
1 bond ⇒ 3 lone-pairs (halogen pattern)
2 bonds ⇒ 2 lone-pairs (O in \text{H}_2\text{O})
3 bonds ⇒ 1 lone-pair (N in \text{NH}_3)
4 bonds ⇒ 0 lone-pairs (C in \text{CH}_4)
\boxed{\text{FC}=V-(B+D)}
V = valence e⁻
B = number of bonds (each counts as 1, not 2)
D = non-bonding electrons on atom.
Deviation from ideal bond count ⇒ non-zero FC.
If total valence e⁻ is a multiple of 8 and no hydrogens are attached:
The central atom will carry 0 lone pairs.
Quick locator for expanded-octet patterns.
For simple molecules the bond order equals the bond multiplicity.
\text{F}_2 single ⇒ \text{BO}=1
\text{O}_2 double ⇒ \text{BO}=2
\text{N}_2 triple ⇒ \text{BO}=3
Group count = # atoms attached + # lone pairs on center.
2 groups ⇒ Linear
3 groups ⇒ Trigonal planar
4 groups ⇒ Tetrahedral
5 groups ⇒ Trigonal bipyramidal
6 groups ⇒ Octahedral
If center owns lone pair(s) ⇒ molecular shape name changes (bent, trigonal-pyramidal, seesaw, square-pyramidal, square-planar, etc.).
Linear: 180^{\circ}
Trigonal planar: 120^{\circ}
Tetrahedral (no lone pair): 109.5^{\circ}
Trigonal pyramidal NH3: 107^{\circ}
Bent H2O: 104.5^{\circ}
Trigonal bipyramidal axes: axial–axial 180^{\circ}; axial–equatorial 90^{\circ}; eq–eq 120^{\circ}
Octahedral: 90^{\circ} (eq–eq & ax–eq); axial 180^{\circ}
Total groups → hybridisation set where exponents sum to groups.
2 groups ⇒ sp
3 groups ⇒ sp^2
4 groups ⇒ sp^3
5 groups ⇒ sp^3d
6 groups ⇒ sp^3d^2
• \text{F}2:7\times2=14 e⁻ → single bond, each F with 3 lone pairs ⇒ non-polar. • \text{O}2:6\times2=12 e⁻ → double bond, 2 lone pairs/atom ⇒ non-polar.
• \text{N}2:5\times2=10 e⁻ → triple bond, 1 lone pair/atom ⇒ non-polar. • \text{H}2 ⇒ single bond only; H satisfies duet.
6 total e⁻. Boron adopts 3 single bonds; octet incomplete (only 6 e⁻) but FC = 0.
Molecular geometry: trigonal planar (120°), non-polar.
\text{H}_2\text{O} → bent; FC(O)=0; polar; \angle H–O–H=104.5^{\circ}; sp^3.
\text{H}_3\text{O}^+ → trigonal pyramidal; FC(O)=+1; sp^3.
\text{OH}^- → linear fragment; FC(O)=−1; 3 lone pairs on O.
Polar bond: \Delta EN ≥ 0.5 (e.g.
\Delta EN_{\text{C−O}} = 3.5-2.5=1.0 ⇒ polar).
Whole-molecule polarity depends on vector sum of individual bond dipoles.
Symmetric shapes with identical outer atoms (e.g.
\text{CO}2 linear, \text{BF}3 trigonal planar, \text{SF}_6 octahedral) ⇒ net dipole =0.
Lone-pair asymmetry or mixed atoms yields net polarity (e.g.
\text{SO}2 bent, \text{NH}3 trigonal pyramidal, \text{SF}_4 seesaw).
\text{CO}_2 (linear) vs. \text{CO} (diatomic):
Both have polar C–O bonds.
Symmetry cancels dipoles in CO$_2$ ⇒ non-polar.
CO has unequal dipole; FC diagram yields C$^{−}$–O$^{+}$; polar.
\text{SO}2 is polar because of bent shape + lone pair; contrasts with non-polar CO$2$.
\text{SF}_6: 48 e⁻ (multiple of 8), octahedral, sp^3d^2, non-polar, S with 6 bonds.
\text{PCl}_5: trigonal bipyramidal, sp^3d, non-polar.
\text{IF}_5: square pyramidal, 1 lone pair → polar, sp^3d^2.
\text{XeF}_4: square planar, 2 lone pairs opposite → non-polar.
\text{SF}_4 (34 e⁻) → seesaw, 1 lone pair, polar, sp^3d.
\text{SF}_2 (20 e⁻) → bent, polar, sp^3.
\text{I}_3^-: linear (I–I–I), 3 lone pairs on center I; electron-pair geometry trigonal bipyramidal, sp^3d; net charge –1.
For \text{SO}2\text{Cl}2: 32 e⁻ → no lone pair on S; four single bonds (2 O, 2 Cl). Tetrahedral sp^3.
For \text{XeOF}_2: 28 e⁻ → 2 lone pairs, 3 bonds → T-shaped molecular geometry.
Occurs when electrons can shift without moving nuclei.
Curved arrows show 2-e⁻ (full) or 1-e⁻ (half) motion.
The resonance hybrid is the weighted average; bond order may be fractional.
Examples
Carbonate \text{CO}_3^{2-} → three equivalent structures with one C=O double bond rotating.
Nitrite \text{NO}_2^- → two equivalent structures (one N=O double bond each).
Nitrate \text{NO}_3^- → three equivalent.
\text{SO}_2 best structure has S=O double bond to each O (FC=0 on S).
\text{BF}_3 can draw B–F double-bond variant, but neutral all-single-bond form (incomplete octet) is more stable (all FC = 0).
To get FC = 0 on the center:
B = V - D
If dots found from 8-rule, number of bonds follows directly.
Choose the least electronegative or the atom capable of most bonds.
e.g.
\text{SF}_2 → S central (can handle 4 bonds) instead of O or F.
Hydrogen is never central; halogens rarely central unless expanded octet (e.g.
\text{ClO}_4^-).
Sulfate \text{SO}_4^{2-} → S with 6 bonds (2 double, 2 single), FC(S)=0, tetrahedral.
Phosphate \text{PO}_4^{3-} → P with 5 bonds; trigonal bipyramidal e-pair geometry → tetrahedral molecular shape (all singles, charge on O’s).
Chlorine oxy-anions
Perchlorate \text{ClO}_4^- → 7 bonds on Cl.
Chlorate \text{ClO}_3^- → 5 bonds; 1 lone pair.
Chlorite \text{ClO}_2^- → 3 bonds; 2 lone pairs.
Hypochlorite \text{ClO}^- → 1 bond; 3 lone pairs.
\text{NO}_2 (17 e⁻) → N has 3 bonds + 1 single e⁻; bent; reactive.
\text{NO} (11 e⁻) → N–O double; assign unpaired e⁻ to less-EN atom (N). Choose distribution giving smallest |FC|.
Two major forms:
\text{S}^{-}–\text{C}\equiv\text{N}
\text{S}=\text{C}=\text{N}^{-}
Greater stability putting negative charge on larger atom (S) despite lower EN.
Carbon family:
Alkane C–C single (ethane \text{C}2\text{H}6).
Alkene C=C (ethene \text{C}2\text{H}4).
Alkyne C≡C (acetylene \text{C}2\text{H}2).
Functional groups & drawing tips
Alcohol \text{–OH}: methanol \text{CH}_3\text{OH} (O 2 bonds + 2 LP).
Aldehyde: \text{CH}_3–\text{C}(=\text{O})H.
Ketone: \text{CH}3–\text{C}(=\text{O})–\text{CH}3 (acetone).
Ether: \text{CH}3–\text{O}–\text{CH}3 (dimethyl ether).
Carboxylic acid: \text{CH}_3–\text{C}(=\text{O})\text{OH} (acetic acid).
Ester: \text{CH}3–\text{C}(=\text{O})\text{O}–\text{CH}3.
Amine: \text{C}2\text{H}5\text{NH}_2; N 3 bonds + lone pair.
Amide: \text{CH}3–\text{C}(=\text{O})\text{NH}2.
Nitrile: \text{R–C}\equiv\text{N} (triple bond).
Molecular polarity informs solubility (e.g.
non-polar \text{SF}6 as electrical insulator; polar \text{NH}3 dissolves in water).
Formal charge & resonance guide reactivity sites in organic/biochemical reactions.
Expanded-octet hypervalent species (e.g.
\text{PCl}_5) are used in phosphorylation & chlorination industrial processes.
Always count valence e⁻ first; write total.
Identify central atom by minimal EN & maximal bonding capacity.
Apply multiple-of-eight & bond-dot=valence rule for center FC = 0.
Check octet/duet for each atom (remember B & Be exceptions).
Calculate FC quickly; adjust with multiple bonds for neutrality/minimum charge.
Determine group count → geometry → hybridisation.
Sketch dipoles to predict polarity.
For resonance, move only π or lone-pair electrons, never atoms.
Practise with oxy-anions & expanded-octet examples; they recur in exams.