Understanding Bonds
Bonds form when atoms achieve lower energy together than apart. This guide covers ionic and covalent bonding, Lewis structures, VSEPR geometry, and polarity — with diagrams at each step.
When a metal atom meets a nonmetal atom, there is a dramatic mismatch in how strongly each holds its electrons. The metal barely grips its outer electrons; the nonmetal desperately wants more. The result is a complete transfer — an electron leaves one atom and joins the other, creating two oppositely charged ions that attract each other strongly.
Metals have low ionization energies — it costs very little energy to remove their outer electrons. Nonmetals have high electron affinities — they release substantial energy when they gain electrons. The combined energy released by the electron transfer is greater than the energy required to remove the electron in the first place, so the ionic compound is lower in energy (more stable) than the separate atoms. Nature always gravitates toward the lower-energy state.
ΔEN > 1.7 → ionic bond. Cation = metal that lost e⁻ (positively charged). Anion = nonmetal that gained e⁻ (negatively charged). Ions arrange into a crystal lattice for maximum electrostatic stability.
ΔEN = 2.23 > 1.7 → ionic bond. The difference is large enough that Cl essentially takes Na's electron rather than sharing it.
Cl (Group 17) gains 1 e⁻ → Cl⁻ (anion)
Both ions now have 8 valence electrons (full octet). Na⁺ matches Ne; Cl⁻ matches Ar.
MgO: Mg²⁺ (charge 2+) and O²⁻ (charge 2−)
Electrostatic force is proportional to the product of the charges. For MgO: (2+)(2−) = 4 times the charge product of NaCl: (1+)(1−) = 1. The lattice forces in MgO are roughly 4× stronger.
Mg²⁺ is smaller than Na⁺ (higher charge pulls electrons closer), and O²⁻ is smaller than Cl⁻. Smaller ions get closer together, further increasing lattice energy.
b) Which compound has a higher melting point: NaCl or MgO? Why?
c) Why do ionic compounds conduct electricity when dissolved in water but not as solids?
a) ΔEN = 3.98 − 1.00 = 2.98 > 1.7 → ionic. Ca loses 2 electrons (one to each F), forming Ca²⁺ and 2 F⁻.
b) MgO has a higher melting point. Mg²⁺ and O²⁻ have double the charge of Na⁺ and Cl⁻, creating much stronger electrostatic attraction in the lattice. Higher lattice energy = more energy required to break → higher melting point.
c) In a solid, all ions are locked in fixed positions in the lattice and cannot move — so no charge flows. When dissolved in water (or melted), the ions are free to move independently and carry electrical current. Conductivity requires mobile charge carriers.
When two nonmetal atoms approach each other, neither is willing to give up electrons completely — both have relatively high electronegativities. Instead, they reach a compromise: they share electron pairs. Each shared pair is attracted by both nuclei simultaneously, holding the atoms together.
Consider two chlorine atoms (Cl–Cl). Both have EN = 3.16. Neither can pull electrons away from the other because the pull is identical. So instead of a winner and a loser, both atoms share equally — the electrons spend time between both nuclei, attracting both simultaneously. This shared pair is the covalent bond, and it is lower in energy than two separate Cl atoms.
More bonds = shorter bond length and higher bond energy. ΔEN < 1.7 → covalent.
H (Group 1): 1 valence e⁻ each
O forms 2 single bonds — one with each H. Each bond uses 1 e⁻ from O and 1 e⁻ from H = 2 shared e⁻.
Each H has 2 electrons (duet complete). O has 8 electrons (octet complete).
O × 2 (Group 16): 2 × 6 = 12 Ve⁻
Total: 4 + 12 = 16 Ve⁻
Fill outer O octets (6 e⁻ each): 12 e⁻ used. Remaining: 0
But C now has only 4 e⁻ (the 2 single bonds) — not 8! C needs an octet.
Result: O=C=O (two double bonds)
Now check: C has 4+4=8 e⁻ (two double bonds × 2 pairs each = 8). Each O has 4 bonding + 4 lone = 8. Total = 16 ✓
b) Which bond is shorter: C−O (single) or C=O (double)? Why?
c) Why does N₂ have a triple bond rather than a single or double bond?
a) N has 5 valence electrons. It needs 3 more to reach an octet, so it forms 3 covalent bonds (with 1 lone pair remaining). Example: NH₃ has 3 N−H bonds.
b) C=O (double bond) is shorter. More shared electron pairs pull the atoms closer together. Double bonds always have shorter bond lengths than the equivalent single bond.
c) N has 5 valence electrons — it needs 3 bonds to complete its octet. Two N atoms each contribute 3 electrons to bonding, forming 3 shared pairs = a triple bond. The triple bond also gives N₂ extremely high bond energy (945 kJ/mol), making it very stable and chemically unreactive.
A Lewis structure is a complete map of all valence electrons in a molecule — both the bonding pairs connecting atoms and the lone pairs sitting on individual atoms. Mastering Lewis structures is the key to understanding shape, polarity, and reactivity.
Lewis structures show every valence electron — bonding pairs and lone pairs. They predict shape (via VSEPR), polarity, reactivity, and even how molecules react. A lone pair on nitrogen is what makes NH₃ a base — it donates that pair to an H⁺. You cannot understand chemistry without understanding where the electrons are.
N is the central atom. Three H atoms surround it.
Check N: 3 bonding pairs (6 e⁻) + 1 lone pair (2 e⁻) = 8 e⁻ → full octet.
No — N already has a complete octet. Structure is final: N with 3 single bonds to H and 1 lone pair.
Fill each O with 3 lone pairs (6 e⁻ each): 12 e⁻ used. Remaining: 0
But C only has 2 bonding pairs = 4 e⁻. C needs 8 e⁻! Must form multiple bonds.
Result: O=C=O
C: 2 double bonds = 8 e⁻ (octet ✓). Each O: 1 double bond + 2 lone pairs = 4 + 4 = 8 e⁻ (octet ✓). Total e⁻ = 16 ✓
O: FC = 6 − 4 − ½(4) = 6 − 4 − 2 = 0
All formal charges = 0 → this is the best Lewis structure.
b) Draw the Lewis structure for O₂. (Hint: it needs a double bond.)
c) How do you know when to form a multiple bond in a Lewis structure?
a) HCl: H−Cl with 3 lone pairs on Cl. Total Ve⁻ = 1 + 7 = 8. Bond uses 2 e⁻; 6 remaining go as 3 lone pairs on Cl. H has duet; Cl has octet.
b) O₂: Total Ve⁻ = 6 + 6 = 12. If single bond: O−O, fill each O with 3 LP = 12 e⁻ used; each O has 2 + 6 = 8. But wait — with single bond + 3LP each: 2 + 12 = 14 e⁻ needed for 12 available. Move 1 LP from each O → shared pair → O=O double bond + 2 LP on each O. Total: 4 bond + 8 lone = 12 ✓. Each O: 4 bond + 4 lone = 8 ✓.
c) Form a multiple bond when: after completing outer atoms' octets and placing all remaining e⁻ on the central atom, the central atom still does not have 8 electrons. Convert a lone pair from an outer atom into a second bonding pair shared with the central atom. Repeat until the central atom has a full octet.
Once you have a Lewis structure, VSEPR lets you predict the three-dimensional shape. The key insight is simple: electron groups around the central atom repel each other and spread as far apart as possible, minimizing the repulsion energy.
Lone pairs are not constrained between two nuclei like bonding pairs are. A bonding pair is held in a relatively narrow region between two positive nuclei. A lone pair sits on just one nucleus and spreads out more, taking up more angular space. This extra volume means lone pairs push bonding pairs closer together. One lone pair compresses angles from 109.5° to ~107° (NH₃); two lone pairs compress further to ~104.5° (H₂O).
4 C−H single bonds, 0 lone pairs on C
4 electron groups, 0 lone pairs → tetrahedral geometry → bond angle 109.5°
4 electron groups → electron geometry is tetrahedral. This describes where all four electron groups point.
Molecular shape only considers atom positions (O and 2 H). The 2 lone pairs are invisible. With only 2 bonding pairs and 2 lone pairs, the molecular shape is bent.
2 lone pairs compress the H−O−H angle to ~104.5°
NH₃ has only 1 lone pair → 107°. H₂O has 2 → 104.5°. Each lone pair compresses further.
b) Predict the molecular geometry of CO₂. (Hint: count electron groups on C.)
c) Why is the bond angle in H₂O (104.5°) less than in NH₃ (107°)?
a) NH₃: N has 3 bonding pairs + 1 lone pair = 4 electron groups. Electron geometry = tetrahedral. Molecular shape = trigonal pyramidal (1 LP, 3 bonding pairs). Bond angle ≈ 107°.
b) CO₂: C has 2 double bonds (each double bond = 1 electron group) + 0 lone pairs = 2 electron groups. VSEPR: 2 groups → linear. Bond angle = 180°.
c) H₂O has 2 lone pairs around O; NH₃ has only 1 lone pair around N. Each lone pair takes up more angular space than a bonding pair and pushes the bonding pairs closer together. The second lone pair in H₂O adds extra compression beyond what 1 lone pair does in NH₃, resulting in 104.5° vs 107°.
Polarity operates at two levels: bond polarity (does a single bond have an uneven electron distribution?) and molecular polarity (does the whole molecule have a net dipole?). These are related but not the same — you must consider both electronegativity difference and molecular shape.
Think of a tug-of-war. In CO₂, two oxygen atoms each pull on carbon with equal force in exactly opposite directions. The pulls cancel — net force on carbon is zero. Similarly, the bond dipoles of the two C=O bonds point in opposite directions and cancel, giving CO₂ a net dipole of zero. In H₂O, the two O−H dipoles both point in roughly the same direction (toward the O, due to the bent shape) and add together, giving a net dipole that makes water polar.
ΔEN < 0.5 = nonpolar covalent; 0.5–1.7 = polar covalent; > 1.7 = ionic.
0.5 < 0.96 < 1.7 → polar covalent bond. Cl is δ− and H is δ+.
HCl is a diatomic molecule — only one bond. There is nothing to cancel with. The bond dipole IS the molecular dipole.
Tetrahedral shape with 4 identical substituents is perfectly symmetric. The 4 C−Cl dipoles point toward each Cl at tetrahedral angles — they cancel perfectly in 3D. Net dipole = 0 → nonpolar.
ΔEN(C−Cl) = 0.61 → polar
The geometry is still tetrahedral. But now one position has H (weak dipole) and three have Cl (stronger dipole). The molecule is no longer symmetric — the 3 C−Cl dipoles all pull in the same general direction (toward the Cl end). They do not cancel. Net dipole ≠ 0 → polar.
b) Is NH₃ polar? It has trigonal pyramidal geometry.
c) Explain why CO₂ is nonpolar even though each C=O bond is polar.
a) BF₃ is nonpolar. Although each B−F bond is polar (F is highly electronegative), the trigonal planar shape is perfectly symmetric: all three B−F dipoles are at 120° to each other and cancel exactly. Net dipole = 0 → nonpolar overall.
b) NH₃ is polar. The three N−H bonds are polar (N is more EN than H). The trigonal pyramidal shape is asymmetric — all three N−H dipoles point in roughly the same direction (toward N, which is on top of the pyramid). They add together rather than cancel. The lone pair on N also contributes to the net dipole. Net dipole ≠ 0 → polar molecule.
c) Each C=O bond is polar because O (EN = 3.44) is more electronegative than C (EN = 2.55), so electron density is shifted toward O. However, CO₂ is linear — the two C=O dipoles point in exactly opposite directions (180° apart) and are equal in magnitude. They cancel completely, giving a net molecular dipole of zero. A molecule can have very polar bonds and still be nonpolar if its shape places those dipoles to cancel.